calculus limits

CHAPTER

2 Limits

2.1 Limits, Rates of Change and Tangent Lines

Preliminary Questions

1. Average velocity is defined as a ratio of two quantities. Which quantities?

2. Average velocity is equal to the slope of a secant line through two points on a graph. Whichgraph?

3. Can instantaneous velocity be defined as a ratio? If not, how is instantaneous velocity com-puted?

4. What is the graphical interpretation of instantaneous velocity at a moment t = t0?

5. What is the graphical interpretation of the statement: average velocity approaches instanta-neous velocity as the time interval shrinks?

6. Does it make sense to speak of the (average or instantaneous) rate of change of y withrespect to x if the variable x is not a time variable? What is the graphical interpretation ofthe average or instantaneous rates of change?

7. The rate of change of atmospheric temperature with respect to altitude is equal to the slopeof the tangent line to a graph. Which graph? What are possible units for this rate?

Exercises

1. An object is released at time t = 0 and allowed to fall to earth. Use Galileo’s formula toanswer the following questions.(a) How far does the object travel during the time interval [2, 2.5]?(b) Compute the average velocity over [2, 2.5].(c) Compute the average velocity over time intervals [2, 2.01], [2, 2.005], [2, 2.001],

[2, 2.00001] and use this to estimate the object’s instantaneous velocity at t = 2.

1

2 Chapter 2 Limits

(a) Galileo’s formula is s(t) = 16t2. The object thus travels�s = s(2.5) − s(2) = 16(2.5)2 − 16(2)2 = 36 ft.

(b) The average velocity over [2, 2.5] is�s

�t= s(2.5) − s(2)

2.5 − 2= 36

.5= 72 ft/s.

(c)time interval [2, 2.01] [2, 2.05] [2, 2.001] [2, 2,00001]

average velocity 64.1600 64.0800 64.0160 64.0000The instantaneous velocity at t = 2 is 64 ft/s.

2.

If you toss a stone in the air from ground level with initial velocity 15 m/s, the height of thestone at time t is h = 15t − 4.9t2 m.

(a) Compute the average velocity of the stone over the time intervals [1, 1.001],[1, 1.0001], [.999, 1], [.9999, 1].

(b) Estimate the instantaneous velocity at t = 1.

3. After an initial deposit of $100, the balance in a bank account at time t (in years) isf (t) = 100(1.08)t dollars.(a) What are the units of the rate of change of f (t)?(b) Find the average rate of change over the time intervals [0, .5] and [0, 1].(c) Estimate the instantaneous rate of change at t = .5 to two decimal places by computing

the average rates of change over intervals of the form [.5, t] with t > .5 and [t, .5] witht < .5.

(a) The units of the rate of change of f (t) are dollars/year or $/yr.(b) The average rate of change of f (t) = 100(1.08)tover the time interval [t1, t2] is given

by� f

�t= f (t2) − f (t1)

t2 − t1.

time interval [0, .5] [0, 1]average rate of change 7.8461 8

(c)

time interval [.5, .51] [.5, .501] [.5, .5001] [.49, .5] [.499, .5] [.4999, .5]average rate of change 8.0011 7.9983 7.9981 7.9949 7.9977 7.998

The rate of change at t = .5 is approximately 8$ /yr.4.

Estimate the instantaneous velocity at t = 1 of a particle whose horizontal displacement(distance traveled) at time t is s(t) = t3 + t .

In Exercises 5–8, estimate the instantaneous rate of change (to two decimal places) at the pointindicated.

5. y = 1

x + 2at x = 2

x interval [2, 2.01] [2, 2.001] [2, 2.0001] [1.99, 2] [1.999, 2] [1.9999, 2]average rate of change -.0624 -.0625 -.0625 -.0627 -.0625 -.0625

The rate of change at x = 2 is approximately -0.06.6.

y = √3t + 1 at t = 1

7. P = 4x2 − 3 at x = −2

x interval [−2, −1.99] [−2, −1.999] [−2, −1.9999] [−2.01, −2] [−2.001, −2] [−2.0001, −2]average rate of change −15.96 −15.996 −15.9996 −16.04 −16.004 −16.0004

The rate of change at x = −2 is approximately 16.8.

Q = 3P − 5 at P = 9

2.1 Limits, Rates of Change and Tangent Lines 3

9. A small mass attached to a spring oscillates up and down. Suppose the height (in feet) of themass at time t (in seconds) is h(t) = .5 cos(8t).(a) Calculate the average velocity of the mass over the time intervals [0, 1] and [3, 5].(b) Estimate the instantaneous velocity at t = 3 by computing average rates of change over

small intervals.

(a) The average velocity over the time interval [t1, t2] is given by�h

�t= h (t2) − h (t1)

t2 − t1.

time interval [0, 1] [3, 5]average velocity -.5728 -.2728

(b)

time interval [3, 3.01] [3, 3.001] [3, 3.0001] [2.99, 3] [2.999,3] [2.9999,3]average velocity 3.55 3.6155 3.6216 3.6863 3.6291 3.6230

The velocity at t = 3 is approximately 3.62.10.

R & W The period T (in seconds) of a pendulum (the time required for a completeback-and-forth cycle) depends on the pendulum’s length L (in meters). Assume thatT = 3

2

√L .

(a) What are the units for the rate of change of T with respect to L? Explain in words whatthis rate measures.

(b) Which quantities are represented by the slopes of lines L and L ′ in Figure 1?(c) Calculate the average rate of change of the period T with respect to L over the interval

[2.9, 3].(d) Estimate the instantaneous rate at which T changes with respect to length, when L = 1.

Figure 1 The period T is the time required for a pendulum to swing back and forth.

11. Match the graphs in Figure 2 with the following statements:

(A) (B)x x

yy

Figure 2 Problem 11.

(a) For all x , the average rate of change over the interval [0, x] is greater than theinstantaneous rate of change at x .

(b) For all x , the average rate of change over the interval [0, x] is less than theinstantaneous rate of change at x .

(a) The average rate of change is greater than the instantaneous rate of change: (A).(b) The average rate of change is less than the instantaneous rate of change: (B)

12.

The percentage N (t) of susceptible children who were infected during the first three weeksof an outbreak of measles is recorded by an epidemiologist who finds that the followingformula fits the data reasonably well (t in days):

N (t) = 100t2

t3 + 5t2 − 100t + 380

(a) Draw the secant line whose slope is the average rate of increase in infected children (inunits of percent per day) over the interval between day 4 and day 6. Then compute thisaverage rate using the formula.

(b) Estimate the instantaneous rate of change of N (t) on day 12 (calculate average rates ofchange over intervals lying to the right and to the left of 12).

(c) On which time intervals are the instantaneous rates of change positive and on whichintervals are they negative?

(d) Over which interval does the largest average rate of change occur?

Figure 3 Graph of N (t).

13. A flu virus spreads through the population of a city. The percentage of the populationinfected is plotted as a function of time (in weeks) in Figure 4.(a) Describe the slopes of L and L ′ as rates of change (either average or instantaneous).(b) Estimate these slopes.(c) Is the flu spreading more rapidly at t = 4, 5 or 6?(d) Is the flu spreading more rapidly at t = 1, 2 or 3?

4 Chapter 2 Limits

2 3 4 5 6

0.1

0.2

0.3

(weeks)

(percentage infected)

L

L'

1

Figure 4

(a) Slope L is the average rate of change over the interval [4, 6], whereas slope L ′ is theinstantaneous rate of change at t = 6.

(b) Slope L ≈ (.28 − .19)/2 = .045, whereas slope L ′ ≈ (.28 − .15)/6 = .0217.(c) Among times t = 4, 5, 6, the flu is spreading most rapidly at t = 4 since the slope is

greatest at that instant.(d) Among times t = 1, 2, 3, the flu is spreading most rapidly at t = 3 since the slope is

greatest at that instant.14.

The graphs in Figures 5(A) and (B) represent the position of a moving particle as a functionof time.

(a) Are the instantaneous velocities at times t1, t2, t3 in (A) getting larger or smaller?(b) Is the particle speeding up or slowing down in (A)?(c) Is the particle speeding up or slowing down in (B)?

Figure 5

15. The graphs in Figures 6(A) through (D) represent the position of a moving particle as afunction of time. Match each graph with one of the following statements:(1) speeding up(2) speeding up and then slowing down(3) slowing down(4) slowing down and then speeding up

.5 1

2

4

.5 1

2

4

(A)

(D)

.5 1

2

4

.5 1

2

4

(B)

(C)

Figure 6

In graph (A), the particle is (3) slowing down.In graph (B), the particle is (2) speeding up and then slowing down.In graph (C), the particle is (4) slowing down and then speeding up.In graph (D), the particle is (1) speeding up.

2.1 Limits, Rates of Change and Tangent Lines 5

16.

The fungus fusarium exosporium infects a field of flax plants through the roots and causesthe plants to wilt. Eventually all plants in the field are infected. The percentage f (t) ofinfected plants in the field as a function of time t (in days) since planting is shown in Figure7.

(a) What are the units of the rate of change of f (t) with respect to t? What does this ratemeasure?

(b) Draw the tangent line at t = 40 and estimate its slope. (Hint: Use the points on the linewith t = 40 and t = 50 for the slope computation.)

(c) Use the graph to rank (from smallest to largest) the average infection rates over theintervals [0, 12], [20, 32], and [40, 52].

(d) Use the following table to compute the average rates of infection over the intervals[30, 40], [40, 50], [30, 50].

days 0 10 20 30 40 50 60percent infected 0 18 56 82 91 96 98

(e) Using Figure 7, determine which of the three average rates in (d) is closest to theinstantaneous infection rate at t = 40 days.

Figure 7

17. R & W Figure 8 shows the graph of atmospheric pressure (in pounds per square inch) asa function of altitude h (in feet).(a) Is the instantaneous rate of change (of pressure with respect to altitude) positive or

negative at h = 1000? What about other values of h? Explain using the graph.(b) Estimate the instantaneous rate of change at h = 1000 (draw the tangent line and

estimate its slope).(c) Is the following generalization correct: if the graph is decreasing, the instantaneous

rates of change are all negative? If so, explain.

Figure 8

(a) The instantaneous ROC is negative at h = 1000: the larger h grows, the lessatmospheric pressure becomes. In fact, the instantaneous ROC is negative for all valuesof h.

(b) Sketch the tangent line, estimate it passes through points (1000, 14) and (5000, 12).Estimated instantaneous rate of change is (14 − 12)/(1000 − 5000) = −.0005.

(c) Correct.

Further Insights and Challenges18.

The height at time t (seconds) of a projectile fired in the air from ground level with initialvelocity 64 ft/s is h(t) = 64t − 16t2 ft.

(a) Compute h(2).(b) Show that the polynomial h(t) − h(2) can be factored with (t − 2) as a factor.(c) Use (b) to find a formula (in terms of t) for the average velocity of the ball over an

interval [2, t].(d) Use this formula to find the average velocity over several intervals [2, t] with t close to

2. Then estimate the instantaneous velocity at time t = 2.

19. Suppose that Q(t) = 2t2 − t . Find a formula (in terms of t as in the previous exercise)for the average rate of change of Q over the time interval [1, t] and use it to estimate theinstantaneous rate of change at t = 1.

(a) With Q(t) = 2t2 − t , we have Q(1) = 1.(b) The polynomial Q(t) − Q(1) = (

2t2 − t) − (1) = 2t2 − t − 1 may be factored as

(2t + 1)(t − 1).(c) The average velocity over the interval [1, t] is

Q(t) − Q(1)

t − 1= (2t + 1)(t − 1)

t − 1= 2t + 1.

(d) From the table of averages velocities below, the instantaneous velocity at t = 1 is 3.

time interval [.99, 1] [.999, 1] [.9999, 1] [1, 1.01] [1, 1.001] [1, 1.0001]average velocity 2.98 2.998 2.9998 3.02 3.002 3.0002

6 Chapter 2 Limits

20.

Sketch a graph of the function f (x) = x(1 − x) on the interval [0, 1].(a) For which values of x in this interval is the instantaneous rate of change positive?(b) Which is larger: the average rate of change over [0, .5] or [0, .75]?(c) What is the average rate of change over [0, 1]?(d) Consider the average rate of change over the intervals [a, 1

2 ] for 0 ≤ a < 12 . Does this

average rate get larger or smaller as a moves from 0 to 12 ?

(e) What is the instantaneous rate of change at x = 12 ? Justify your answer graphically,

without any computation.

21. R & W Assume as in Exercise 10 that the period T (in seconds) of a pendulum is givenby the formula T = 2

3

√L m where L is the length (in meters). Observe that the numbers in

the second column of Table 1 are increasing and those in the last column are decreasing.Explain why in terms of the graph of T .

Table 1 Average rates of change of T with respect to L .

lengthinterval

averagerate of change

lengthinterval

averagerate of change

[3, 3.2] .42603 [2.8, 3] .44048[3, 3.1] .42946 [2.9, 3] .43668[3, 3.001] .43298 [2.999, 3] .43305[3, 3.0005] .43299 [2.9995, 3] .43303[3, 3.0001] .43301 [2.9999, 3] .43302[3, 3.00001] .43301 [2.99999, 3] .43301

The first column underestimates the instantaneous ROC by secant slopes; this estimateimproves as L decreases toward L = 3. The second column overestimates theinstantaneous ROC by secant slopes; this estimate improves as L increases toward L = 3.

2.2 Numerical and Graphical Approach


1. What is the limit of f (x) = 1 as x approaches π?

2. What is the limit of g(t) = t as t → π?

3. Can f (x) approach a limit as x → a even if f (a) is undefined? If so, give an example.

4. What is limx→10

f (x) for the constant function f (x) = 8?

5. What does the following table suggest about limx→1−

f (x) and limx→1+

f (x)?

x f (x)

.9 7

.99 25

.999 43171.1 3.01261.01 3.00471.001 3.00011

6. Is it possible to tell if limx→a

f (x) exists by examining values f (x) for x close to but greater

than a? Explain.

7. If you know in advance that limx→5

f (x) exists, can you determine its value just knowing the

values of f (x) for all x > 5?

2.2 Numerical and Graphical Approach 7

8. For each of the following, state whether the information is sufficient to determine whetherlimx→5

f (x) exists. Explain your answers.

(a) the values of f (x) for all x .(b) the values of f (x) only for x in the interval [4.5, 5.5].(c) the values of f (x) for all x in the interval [4.5, 5.5] other than x = 5.(d) the values of f (x) for all x ≥ 5.(e) f (5).

Exercises

In Exercises 1–3, fill in the tables and make a guess for limx→a

f (x).

1. limx→1

f (x), where f (x) = x3 − 1

x2 − 1.

x f (x) x f (x)

1.002 .998

1.001 .999

1.0005 .9995

1.00001 .99999

x 0.998 0.999 0.9995 0.99999 1.00001 1.0005 1.001 1.002f (x) 1.498501 1.499250 1.499625 1.499993 1.500008 1.500375 1.500750 1.501500

The limit as x → 1 is 32.

2.limt→0

h(t) where h(t) = cos t − 1

t2. Note that h(t) is an even function, that is, h(t) = h(−t).

t ±.002 ±.0001 ±.00005 ±.00001

h(t)

3. limy→2

f (y), where f (y) = y2 − y − 2

y2 + y − 6.

y f (y) y f (y)

2.002 1.998

2.001 1.999

2.0001 1.9999

y 1.998 1.999 1.9999 2.0001 2.001 2.02f (y) .599840 .599920 .599992 .600008 .600080 .600160

The limit as y → 2 is 35 .

4.Determine lim

x→2+f (x) and lim

x→2−f (x) for the function shown in Figure 1.

Figure 1 Graph for Exercise 4.

8 Chapter 2 Limits

5. Determine one-sided limits at c = 1, 2, 4 of the function g(t) whose graph is shown inFigure 2 and state whether the limit exists at these points.

1

1

2

3

4

2 3 4 5


At c = 1, the left-hand limit is limt→1−

g(t) = 2.5, whereas the right-hand limit is

limt→1+

g(t) = 0.7. Accordingly, the two-sided limit does not exist at c = 1.


g(t) = 2, whereas the right-hand limit is

limt→2+

g(t) = 3. Accordingly, the two-sided limit does not exist at c = 2.


g(t) = 2, whereas the right-hand limit is

limt→4+

g(t) = 2. Accordingly, the two-sided limit exists at c = 2 and equals 2.

6.Determine the (one or two-sided) infinite limits in Figure 3.


7. Use the limit definition to show that limx→3

2x = 6. Hint: show that |2x − 6| can be made as

small as desired by taking x close to 3.

As x → 3, we have |2x − 6| = 2 |x − 3| → 0.

In Exercises 8–15, prove the limits have the indicated values using the definition of a limit.8.

limx→5

(x + 1) = 69. lim

x→3(5x − 7) = 8

As x → 3, note that |(5x − 7) − 8| = |5x − 15| = 5 |x − 3| → 0.10.

limx→9

(−2x) = −1811. lim

x→3(2x + 5) = 11

As x → 3, note that |(2x + 5) − 11| = |2x − 6| = 2 |x − 3| → 0.12.

limx→5

7 = 713. lim

x→−5(1 − 2x) = 11

As x → −5, we have

|(1 − 2x) − 11| = |−2x − 10| = |(−2)(x + 5)| = 2 |x − (−5)| → 0.

14.limx→0

x2 = 015. lim

x→0(x2 + 2x + 3) = 3

As x → 0, we have∣∣x2 + 2x + 3 − 3

∣∣ = ∣∣x2 + 2x∣∣ = |x ||x + 2| → 0.

16.Let f (x) = k (k constant). Using the limit definition show that lim

x→cf (x) = k for any c.


17. Prove that limx→0

|x | = 0. Hint: consider the left and right-hand limits separately.

As x → 0+, we have ||x | − 0| = |x | = x → 0.As x → 0−, we have ||x | − 0| = | − x | = −x → 0.

In Exercises 18–35, estimate the limit numerically or state that the limit does not exist.18.

limx→1

√x − 1

x − 119. limx→5

x2 − 7x + 10

x − 5

x 4.999 4.9999 5.0001 5.001f (x) 2.999 2.9999 3.0001 3.001

The limit as x → 5 is 3.20.

limx→−3

2x2 − 18

x + 321. limx→2

x2 + x − 6

x2 − x − 2

x 1.999 1.99999 2.00001 2.001f (x) 1.666889 1.666669 1.666664 1.666445

The limit as x → 2 is 53 .

22.

limh→0

cos

(1

h

)23. lim

h→0sin h cos

(1

h

)

h −.5 −.1 −.01 −.001 −.0001 .0001 .001 .01 .1 .5f (h) .199511 .083767 −.008623 −.000562 .000095 −.000095 .000562 .008623 −.083767 −.199511

The limit as x → 0 is 0.24.limθ→0

cos θ − 1

θ25. limθ→0

cos θ − 2

θ

θ −.01 −.001 −.0001 .0001 .001 .01f (θ) 100.0 1000.0 10000.0 −10000.0 −1000.0 −100.0

The limit does

not exist. As θ → 0−, f (θ) → ∞; similarly, as θ → 0+, f (θ) → −∞.26.limθ→0

cos θ − 1

θ 227. lim

x→0

sin 2x

x

x −.01 −.005 .005 .01f (x) 1.999867 1.999967 1.999967 1.99867

The limit as x → 0 is 2.28.

limx→0

sin x

x229. lim

x→0

sin x − x

x3

x −.05 −.001 .001 .05f (x) −.166646 −.166667 −.166667 −.166646

The limit as x → 0 is − 16.

30.

limh→0

2h − 1

h31. limx→0

2x − 3x

x

x −.05 −.001 .001 .05f (x) −.387710 −.405102 −.405829 −.424048

The limit as x → 0 is approximately −.405. (The exact answer is ln 2 − ln 3.)32.

limh→0

5h − 5−h

h33. limx→0

( sec x − cot x )

x −.05 −.001 .001 .05f (x) 20.98 1001.0 −999.0 −18.98

The limit does not exist. As x → 0−,

f (x) → ∞; similarly, as x → 0+, f (x) → −∞.

10 Chapter 2 Limits

34.

limx→0

cos x − 1 − 12 x2

x435. limx→0

|x |x

x −.05 −.001 −.00001 .00001 .001 .05f (x) 1.161586 1.006932 1.000115 .999885 .993116 .860892

The limit

as x → 0 is 1.36.

Draw a graph of

f (x) = x − 1

|x − 1|and use it to determine the one-sided limits lim

x→1+f (x) and lim

x→1−f (x).

37. Determine the one-sided limits numerically:

limx→0±

sin x

|x |

x −.2 −.02 .02 .2f (x) −.993347 −.999933 .999933 .993347

The left-hand limit is limx→0−

f (x) = −1, whereas the right-hand limit is limx→0+

f (x) = 1.38.

Determine the one-sided limits numerically: limx→±0

|x |1/x

39. Determine the right- and left-hand limits of f (x) as x approaches 2 and 4, for the functionwhose graph is given in Figure 4.

2 4

–5

5

10

15


For c = 2, we have limx→2−

f (x) = ∞ and limx→2+

f (x) = ∞.

For c = 4, we have limx→4−

f (x) = −∞ and limx→4+

f (x) = 10.

40.Draw a graph of a function f such that lim

x→1f (x) = 2, lim

x→3−f (x) = 0, and lim

x→3+f (x) = 4.

41. Draw a graph of a function f such that limx→1

f (x) = ∞, limx→3−

f (x) = 0, and

limx→3+

f (x) = −∞.

–1 0 1 2 3 4 5–30

–20

–10

0

10

20

30

x

f(x)

42.Draw a graph of a function f such that lim

x→2+f (x) = f (2) = 3, lim

x→2−f (x) = −1, and

limx→4

f (x) = 2 �= f (4).


43. Draw a graph of a function f such that limx→1+

f (x) = ∞, limx→1−

f (x) = 3 and

limx→4

f (x) = −∞.

–10

–50

5

10

2 4

GU In Exercises 44–46, use a graphing calculator to estimate the value of the following limits.44.

limθ→0

sin 3θ

sin 2θ45. limx→0

2x − 1

4x − 1

x −.02 −.0002 .0002 .02f (x) .503466 .500035 .499966 .496534


46.limx→0

2x − cos x

x

Further Insights and Challenges

47. Investigate the limit

limθ→0

sin nθ

θ

numerically for several values of n and make a guess as to the value of this limit for gen-eral n.

For n = 3, we have

θ −.1 −.01 −.001 .001 .01 .1sin nθ

θ2.955202 2.999550 2.999996 2.999996 2.999550 2.955202

The limit as θ → 0 is 3.For n = −5, we have

θ −.1 −.01 −.001 .001 .01 .1sin nθ

θ−4.794255 −4.997917 −4.999979 −4.999979 −4.997917 −4.794255

The limit as θ → 0 is −5.

We surmise that, in general, limθ→0

sin nθ

θ= n.

48.

(a) Estimate limx→0

5x − 1

xnumerically.

(b) Compare your estimate to the value ln 5 obtained from a calculator.

(c) Estimate limx→0

3x − 1

xnumerically and compare your estimate to the value ln 3.

(d) Make a conjecture for the value of

limx→0

bx − 1

x

(b any positive number) and test your conjecture for two additional values of b. (Theseresults will be fully explained in Chapter 7.)

49. The greatest integer function is defined as follows: [x] = n, where n is the integer such thatn ≤ x < n + 1.

12 Chapter 2 Limits

(a) For which values of c do the following one-sided limits exist?

limx→c−

[x], limx→c+

[x]

(b) For which values of c are they different?(c) For which values of c does lim

x→c[x] exist?

–1 1 2 3

1

2

3

Figure 5 Graph of y = [x].

(a) The one-sided limits exist for all real values of c.(b) For each integer value of c, the one-sided limits differ. In particular, lim

x→c−[x] = c − 1,

whereas limx→c+

[x] = c. (For noninteger values of c, the one-sided limits both equal [c].)

(c) The limit limx→c

[x] exists when

limx→c−

[x] = limx→c+

[x] ,

namely for noninteger values of c: n < c < n + 1.50.

The function f (x) = (1 + x)1/x is not defined at x = 0.(a) Give numerical evidence that lim

x→0f (x) = e (the number e can be found to many

decimal places on any scientific calculator).(b) Investigate the limits numerically

limx→0

(1 + 2x)1/x, limx→0

(1 + 3x)1/x

and try to express them in terms of e.(c) Can you form a conjecture for the value of lim

x→0(1 + nx)1/x where n is a whole number?

51. Consider the limit

limx→1

xn − 1

xm − 1

(a) Investigate it numerically for the cases when (m, n) are (2, 1), (3, 1), (1, 2), and (1, 3).(b) Guess the value of the limit for general m and n.(c) Check your guess for at least one additional pair (m, n).

(a)x .99 .9999 1.0001 1.01

x − 1

x2 − 1.502513 .500025 .499975 .497512

The limit as x → 1 is 12 .

x .99 .9999 1.0001 1.01x − 1

x3 − 1.336689 .333367 .333300 .330022


x .99 .9999 1.0001 1.01x2 − 1

x − 11.99 1.9999 2.0001 2.01


x .99 .9999 1.0001 1.01x3 − 1

x − 12.9701 2.9997 3.0003 3.0301


(b) For general m and n, we have limx→1

xn − 1

xm − 1= n

m.


(c)x .99 .9999 1.0001 1.01

x3 − 1

x7 − 1.437200 .428657 .428486 .420058


52.(Adapted from Calculus Problems for a New Century.) Find by experimentation thepositive integers k such that

limx→0

sin(sin2 x)

xk

exists.

53. GU The function f (x) = 2x − 8

x − 3is not defined at x = 3.

(a) Use the zoom function on a graphing calculator to graph f (x) and estimateL = lim

x→3f (x).

(b) Observe that the graph of f (x) is increasing. Use this to conclude that

f (2.99999) ≤ L ≤ f (3.00001)

and determine L to three decimal places.(In Chapter 7, we will see that the exact value is 8 ln 2 where ln is the natural logarithm.)

(a)4

6

8

2 3 4

(b) Since f (x) = 2x − 8

x − 3is increasing and

f (2.99999) ≈ 5.545158 ≤ L ≤ 5.545197 ≈ f (3.00001), we conclude that L ≈ 5.545to three decimal places. (The exact limit is 8 ln 2.)

54.

GU The function

f (x) = 21/x − 2−1/x

21/x + 2−1/x

is defined for x �= 0.(a) Investigate lim

x→0+f (x) and lim

x→0−f (x) numerically.

(b) Produce a graph of f on a graphing utility and describe its behavior near x = 0.

55. R & W

(a) Show that the function f (x) = sin x/x has a geometric interpretation: f (x) is equal tothe slope of a secant line through the origin and the point (x, sin x) on the graph ofy = sin x (refer to Figure 6).

(b) What is the geometric interpretation of limx→0

f (x) discussed in the beginning of this

section?(c) Claim: as x moves off to the right (towards ∞), the secant lines discussed in (a)

oscillate up and down but the oscillations get progressively smaller. Explain this claimin words. How is this claim reflected in the graph of f (x) (Figure ??)?

p

1

x

sin x

(x, sin x)

Figure 6 Graph of y = sin x .

(a) The slope of a secant line through the points (0, 0) and (x, sin x) is given bysin x − 0

x − 0= sin x

x.

14 Chapter 2 Limits

(b) The slope of the secant lines approaches zero as the point (x, sin x) approaches theorigin.

(c) As x grows very large, the slope of the secant lines grows very small (since the heightvaries between -1 and 1 while the length (x) is increasing. This is reflected in the graphof f (x) by the indication that the slope of the secant lines from the right approacheszero as x approaches zero.

2.3 Basic Properties of Limits


1. Find limx→3

( f (x)2) assuming that limx→3

f (x) = 2.

2. Which of the following is a verbal version of the Rule for Products?

(a) the product of two functions has a limit(b) the limit of the product is the product of the limits(c) the product of a limit is a product of functions(d) a limit produces a product of functions

3. Which of the following statements are incorrect?(a) lim

x→ck = c for all constants c and k.

(b) limx→c

k = k for all constants c and k.

(c) limx→c2

x = c2 for all c.

(d) limx→c

x = x for all c.

4. Which of the following statements are incorrect?(a) The limit of a sum is the sum of the limits.(b) The Rule for Products may not hold if the limit of one of the functions is zero.(c) The Rule for Quotients only holds if the limit of the denominator is nonzero.

Exercises

In Exercises 1–32, evaluate the limits using the Limit Laws and the following two limit facts:

limx→c

x = c, limx→c

k = k

for all constants c and k.

1. limx→9

14

limx→9

14 = 14.2.

limx→−3

143. lim

x→9x

limx→9

x = 9.4.

limx→−3

x

2.3 Basic Properties of Limits 15

5. limy→9

y

limy→9

y = 9.6.

limt→3

t7. lim

y→−3(y + 14) (use Rule for Sums)

limy→−3

(y + 14) = limy→−3

y + limy→−3

14 = −3 + 14 = 11.8.

limy→−3

14y (use Rule for Constant Multiples)9. lim

y→−3y(y + 14) (use Rule for Products)

limy→−3

y(y + 14) =(

limy→−3

y

)(lim

y→−3(y + 14)

)= −3(11) = −33. (Here we used the result

from Exercise 7.)10.

limt→9

t

t + 1(use the Rule for Quotients)11. lim

x→24x

limx→2

4x = 4 limx→2

x = 4 · 2 = 8.12.

limt→4

(3t − 14)

13. limt→4

3t − 14

t + 1

limt→4

3t − 14

t + 1=

3 limt→4

t − limt→4

14

limt→4

t + limt→4

1= 3 · 4 − 14

4 + 1= −2

5.

14.limx→5

x2

15. limx→5

x3

limx→5

x3 =(

limx→5

x)3 = 53 = 125.

16.lim

x→−1x3

17. limx→−5

(x3 + 2x)

limx→−5

(x3 + 2x

) =(

limx→−5

x

)3

+ 2 limx→−5

x = (−5)3 + 2 · (−5) = −135.

18.limx→ 1

2

16x2

19. limx→ 1

2

(2x − 1)

limx→1/2

(2x − 1) = 2

(lim

x→1/2x

)− lim

x→1/21 = 2

(12

) − 1 = 0.

20.limx→ 1

2

(4x + 1)(2x − 1)

21. limx→3

4x3

limx→3

4x3 = 4(

limx→3

x)3 = 4(3)3 = 4 · 27 = 108.

22.limx→4

(x2 + 4x)23. lim

x→2(3x2 − 9)

limx→2

(3x2 − 9

) = 3(

limx→2

x)2 − lim

x→29 = 3(2)2 − 9 = 3.

24.limx→2

(x + 1)(3x2 − 9)

16 Chapter 2 Limits

25. limx→−1

(x + 2)(2x − 1)

limx→−1

(x + 2)(2x − 1) =(

limx→−1

x + limx→−1

2

)(2 lim

x→−1x − lim

x→−11

)= (−1 + 2)(2(−1) − 1) = 1 · (−3) = −3.

26.limx→3

1 − x

1 + x27. limx→−1

x

x3 + 4x

limx→−1

x

x3 + 4x=

limx→−1

x(lim

x→−1x

)3

+ 4 limx→−1

x

= −1

(−1)3 + 4(−1)= −1

−1 − 4= 1

5 .

28.

limx→2

(x + 1)3x2

x3 + 4x29. limx→−2

x

x + 1

x2

x2 + 4

limx→−2

x

x + 1

x2

x2 + 4=

limx→−2

x

limx→−2

x + limx→−2

1

(lim

x→−2x

)2

(lim

x→−2x

)2

+ limx→−2

4

= −2

−2 + 1

(−2)2

(−2)2 + 4=

−8

−8= 1.

30.

limx→−3

x2 + 4x

3x2 − 931. limx→−1

x(x + 2)(x − 3)

limx→−1

x(x + 2)(x − 3) =(

limx→−1

x

)(lim

x→−1x + 2

)(lim

x→−1x − 3

)= (−1)(1)(−4) = 4.

32.limx→2

x(x3 + 2)(x2 + 3)33. Use the Rule for Quotients to prove that the limit of a reciprocal is the reciprocal of a limit.

More precisely: if limx→c

f (x) exists and is nonzero, then

limx→c

1

f (x)= 1

limx→c

f (x)

limx→c

(1

f (x)

)=

(limx→c

1)

(limx→c

f (x)) = 1

limx→c

f (x).

34.Evaluate lim

x→0x/ sin x = 1 assuming the fact

limx→0

sin x/x = 1.In Exercises 35–45, evaluate the limit assuming that

limx→−4

f (x) = 3, limx→5

f (x)= −6

limx→−4

g(x) = 1, limx→5

g(x) = 2

35. limx→−4

f (x)g(x)

limx→−4

f (x)g(x) = limx→−4

f (x) limx→−4

g(x) = 3 · 1 = 3.36.

limx→−4

f (x)2

2.3 Basic Properties of Limits 17

37. limx→−4

(2 f (x) + 3g(x))

limx→−4

(2 f (x) + 3g(x)) = 2 limx→−4

f (x) + 3 limx→−4

g(x) = 2 · 3 + 3 · 1 = 6 + 3 = 9.38.

limx→−4

f (x)

g(x)39. limx→−4

f (x)2g(x)3

limx→−4

f (x)2g(x)3 =(

limx→−4

f (x)

)2 (lim

x→−4g(x)

)3

= 32 · 13 = 9.40.

limx→−4

x f (x)

41. limx→5

1

f (x)

limx→5

1

f (x)= 1

limx→5

f (x)= 1

−6= −1

6.

42.limx→5

(2 f (x) + 5g(x))2

43. limx→5

f (x) + 1

3g(x) − 9

limx→5

f (x) + 1

3g(x) − 9=

limx→5

f (x) + limx→5

1

3 limx→5

g(x) − limx→5

9= −6 + 1

3 · 2 − 9= −5

−3= 5

3.

44.limx→5

(3 f (x) − 11g(x))

45. limx→5

(x · f (x) + g(x)

x

)

limx→5

(x · f (x) + g(x)

x

)=

(limx→5

x)(

limx→5

f (x))

+limx→5

g(x)

limx→5

x= 5 · (−6) + 2

5= −29 3

5 = −148

5.

46.Can the Rule for Quotients be applied to evaluate lim

x→0

sin x

x? Explain.47. Show that the Rule for Products cannot be used to evaluate the lim

x→π/2(x − π/2) · tan x

The limit Rule for Products cannot be applied to evaluate limx→π/2

(x − π/2) tan x since

limx→π/2

tan x does not exist (for example, as x → π/2−, tan x → ∞). This violates a

hypothesis of the Rule for Products. Accordingly, the rule cannot be employed.


Show that limt→3

g(t) = 4 if limt→3

tg(t) = 12.

Hint: Use the Product Rule to write

limt→3

tg(t) =(

limt→3

t)(

limt→3

g(t))49. Prove that if lim

t→3

h(t)

t= 5, then lim

t→3h(t) = 15.

Given that limt→3

h(t)

t= 5, observe that lim

t→3t = 3. Now use the Rule for Products:

limt→3

h(t) = limt→3

th(t)

t=

(limt→3

t) (

limt→3

h(t)

t

)= 3 · 5 = 15.

50.

(Adapted from Calculus Problems for a New Century.) Suppose that limx→0

f (x)

x= 1. Which

of the following statements is necessarily true? Why?(a) f (0) = 0.(b) lim

x→0f (x) = 0.

51. Give an example of functions f (x) and g(x) such that limx→0

( f (x) + g(x)) exists but the

limits limx→0

f (x) and limx→0

g(x) do not exist.

Let f (x) = 1/x and g(x) = −1/x . Then limx→0

f (x) + g(x) = limx→0

0 = 0 However,

limx→0

f (x) = limx→0

1/x andlimx→0

g(x) = limx→0

−1/x do not exist.

18 Chapter 2 Limits

52.Suppose that lim

x→cf (x)g(x) exists and that lim

x→cg(x) exists and is nonzero. Show that

limx→c

f (x) also exists. Hint: apply the Rule for Quotients.53. Give an example of a function such that limx→0

f (x)2 exists but limx→0

f (x) does not exist.

Let f (x) ={

1, x > 0−1, x < 0

. Then f (x)2 = 1, x �= 0. Hence limx→0

f (x)2 = 1, but limx→0

f (x)

does not exist.

2.4 Limits and Continuity


1. Which property of x3 allows us to conclude that limx→2

x3 = 8?

2. What can be said about f (3) if f is a continuous function such that limx→3

f (x) = 12?

3. Suppose that f (x) < 0 if x is positive and f (x) > 1 if x is negative. Explain why f cannotbe continuous.

4. What is f (5) if f is continuous and f (x) = x2 for all x > 5? Is it possible to determinef (4)?

5. Is it possible to determine f (7) if f (x) = 3 for all x < 7 and f is right-continuous atx = 7?

6. Can the Continuity Laws be used to evaluate limt→3

t + 1

t − 3? If not, which hypothesis is violated?

7. True or false: f is continuous at x = a if the left- and right-hand limits of f (x) as x → aexist and are equal.

8. True or false: f is continuous at x = a if the left- and right-hand limits of f (x) as x → aexist and equal f (a).

9. True or false: if the left- and right-hand limits of f (x) as x → a exist, then f has a remov-able discontinuity at x = a.

10. Let f (x) and g(x) be functions. Which of the following statements are true?(a) If f (x) and g(x) are continuous at x = a, then f (x) + g(x) is continuous at x = a.(b) If f (x) and g(x) have a discontinuity at x = a, then f (x) + g(x) has a discontinuity at

x = a.

2.4 Limits and Continuity 19

Exercises

1. Find the points of discontinuity of the function f shown in Figure 1 and state whether f isleft- or right-continuous (or neither) at these points.

1

2

3

4

5

1 2 3 4 5 6


The function f is discontinuous at x = 1; it is left-continuous there.The function f is discontinuous at x = 3; it is neither left-continuous norright-continuous there.The function f is discontinuous at x = 5; it is left-continuous there.

In Exercises 2–4, refer to the function in Figure 2.

1

2

3

4 53

4

5

1 2

Figure 2 Graph for Exercises 2–4.

2.Find the points of discontinuity of the function f and state whether f is left- or right-

continuous (or neither) at these points.3. (a) Find the point c at which f has a removable discontinuity.(b) What value should be assigned to f (c) to make f continuous at x = c?

(a) The function f has a removable discontinuity at x = 3.(b) Assigning f (3) = 4.5 makes f continous at x = 3.

4.

(a) Find the point c′ at which f has a jump discontinuity but is left-continuous.(b) What value should be assigned to f (c′) to make f right-continuous at x = c′?

5. This exercise refers to Figure 3.(a) Determine the one-sided limits at the points where f (x) is discontinuous.(b) Which of these discontinuities is removable and how should f be redefined to make it

continuous at this point?

20 Chapter 2 Limits

–2 2 4

2

6


(a) The function f is discontinuous at x = 0, at which limx→0−

f (x) = ∞ and

limx→0+

f (x) = 2.

The function f is discontinuous at x = 2, at which limx→2−

f (x) = 6 and

limx→2+

f (x) = 6.

(b) The discontinuity at x = 2 is removable. Assigning f (2) = 6 makes f continuous atx = 2.

6.Determine the infinite one- and two-sided limits in Figure 4.


In Exercises 7–12, use the Continuity Laws to show that the function is continuous assumingonly that the functions x, sin x, and cos x are continuous.

7. x + sin x

Since x and sin x are continuous, so is x + sin x by Continuity Law (i).8.

x sin x9. 3x + 4 sin x

Since x and sin x are continuous, so are 3x and 4 sin x by Continuity Law (iii). Thus3x + 4 sin x is continuous by Continuity Law (i).

10.3x3 + 8x2 − 20x

11.1

x2 + 1

Since x is continuous, so is x2 by Continuity Law (ii).Recall that constant functions, such as 1, are continuous. Thus x2 + 1 is continuous.

Finally,1

x2 + 1is continuous by Continuity Law (iv).

12.x2 − cos x

3 + cos xIn Exercises 13–16, determine the points at which the function is discontinuous and state thetype of the discontinuity: removable, jump, infinite, or none of these.

13.1

x

The function 1/x is discontinuous at x = 0, at which there is an infinite discontinuity.14.

cos

(1

x

)15.

x − 2

|x − 1|The function

x − 2

|x − 1| is discontinuous at x = 1, at which there is an infinite discontinuity.


16. [1

2x

]In Exercises 17–26, determine the domain of the function and prove that it is continuous on itsdomain using the Laws of Continuity and the facts quoted in this section.

17. x + sin x

Since x and sin x are continuous, so is x2 + sin x by Continuity Laws (i) and (ii).18. x cos x

19.√

x2 + 9

Since√

x and the polynomial x2 + 9 are both continuous, so is the composite function√x2 + 9.

20. √9 − x2

21.x2

x2 − 9

Since x is continuous, so are x2 and x2 − 9 by Continuity Laws (ii) and (i); hence, byContinuity Law (iv), so is x2

x2−9whenever x �= ±3.

22.x · sin x · cos x

23. x − tan x

By Continuity Law (iv), tan x = sin xcos x

is continuous if cos x �= 0; hence by Continuity Law(i), x + tan x is continuous on its domain, x �= (2k + 1)π/2 for integers k.

24.3x2 − 4x sin x

25.√

x sin x

Since√

x and sin x are continuous, so is√

x sin x by Continuity Law (ii).26.

sin(tan x)27. Draw a graph of a function f such that lim

x→1+f (x) and lim

x→1−f (x) both exist and are equal,

but f is not continuous at x = 1.

–1

0

1

2

3

–1 1 2 3

28.Draw the graph of a function on [0, 5], which is left-continuous but not continuous at x = 2and right-continuous but not continuous at x = 3.29. Draw a graph of a function f such that f (2) = lim

x→2+f (x) and f (2) �= lim

x→2−f (x).

–1

0

1

2

3

–1 1 2 3

30.

Draw the graph of a function on the interval [0, 5] with(a) a removable discontinuity at x = 1(b) a jump discontinuity at x = 2(c) lim

x→3−f (x) = −∞ and lim

x→3+f (x) = 2

22 Chapter 2 Limits

31. Draw the graph of a function with the following properties:(a) f is right-continuous but not left-continuous at x = 1(b) f is left-continuous but not right-continuous at x = 2(c) f is neither left-continuous nor right-continuous at x = 3

–1

0

1

2

3

4

–1 1 2 3 4 5

In Exercises 32–47, find the points where the function is discontinuous.32.

1

x2 − 133.x + 1

4x − 2

The function f (x) = x + 1

4x − 2is discontinuous at x = 1

2.

34.4x9 − 7x5 + x + 4

35. 3x3/2 − 9x3

The function f (x) = 3x3/2 − 9x3 is continuous for x > 0. At x = 0 it is right-continuous.(It is not defined for x < 0.)

36.

g(t) = cos

(1

t

)37. 3x−3/2 − 9x3

The function f (x) = 3x−3/2 − 9x3 is continuous for x > 0. (It is not defined for x ≤ 0.)38.

cos

(1

x

)39.

1 − 2z

z2 − z − 6

The function f (z) = 1 − 2z

z2 − z − 6= 1 − 2z

(z + 2)(z − 3)is discontinuous at z = −2 and z = 3.

40.1 − 2z

z2 + 941. tan 2t

The function f (t) = tan 2t = sin 2t

cos 2tis discontinuous whenever cos 2t = 0; i.e., whenever

2t = (2n + 1)π

2or

t = (2n + 1)π

4, where n is an integer.

42.csc x2

43. tan(sin x)

The function f (x) = tan(sin x) is continuous everywhere. Reason: sin x is continuouseverywhere and tan u is continuous on

(− π

2, π

2

)—and in particular on −1 ≤ u = sin x ≤ 1.

Continuity of tan(sin x) follows by the continuity of composite functions.44.

sec(t + 1)

45. [√x]The function f (x) = [√

x]

is discontinuous wherever√

x = n or x = n2, where n is anonnegative integer.

46. [z + 1

2

]


47. x − |x |The function f (x) = x − |x | is continuous everywhere.

In Exercises 48–53, evaluate the limit using the continuity of f at x = c.48.

limx→π

sin( x

2− π

)49. lim

x→5x2

limx→5

x2 = 52 = 25.50.

limx→8

x3

51. limx→0

x + 9

x − 9

limx→0

x + 9

x − 9= 0 + 9

0 − 9= −1.

52.limx→π

sin(x2)

53. limx→π

1

cos x

limx→π

1

cos x= 1

cos π= 1

−1= −1.

In Exercises 54–57, sketch a graph of the given function. Find all points of discontinuity anddetermine whether f is left-continuous, right-continuous, or neither.54.

f (x) ={

x2 for x ≤ 1

2 − x for x > 155. f (x) =

x + 1 for x < 1

1

xfor x ≥ 1

–1

0

1

2

–2 –1 1 2 3The function f is right-continuous at x = 1.

56.

f (x) ={ |x − 3| for x ≤ 3

x − 3 for x > 357. f (x) =

x3 + 1 for − ∞ < x ≤ 0

− x + 1 for 0 < x < 2

− x2 + 10x − 15 for x ≥ 2

–5

0

5

10

5

The function f is left-continuous at x = 0 and right-continuous at x = 2.58.

Find a value of c for which the following function is continuous:

f (x) ={

x2 − c for x < 5

4x + 2c for x ≥ 5

59. Find the constant c for which the following function is continuous.

f (x) ={

2x + 9 for x ≤ 3

− 4x + c for x > 3

As x → 3−, we have 2x + 9 → 15 = L . As x → 3+, we have −4x + c → c − 12 = R.Match the limits: L = R or 15 = c − 12 implies c = 27.

24 Chapter 2 Limits

60.

Which of the following quantities would be represented by continuous functions of timeand which would have one or more discontinuities?

(a) velocity of an airplane during a flight from Boston to Chicago(b) temperature in a room under ordinary conditions(c) value of a bank account with interest paid quarterly(d) the salary of a teacher(e) the population of the world

61. Suppose that f (x) = 2 for all x > 0 and f (x) = −4 for x < 0. What is f (0) if f isleft-continuous at x = 0? What is f (0) if f is right-continuous at x = 0?

Let f (x) = 2 for positive x and f (x) = −4 for negative x .

If f is left-continuous at x = 0, then f (0) = −4.If f is right-continuous at x = 0, then f (0) = 2.

62.Sawtooth Function Draw the graph of f (x) = x − [x]. At which points is fdiscontinuous? Is it left- or right-continuous at those points?63. Does f (x) = |x | have any discontinuities?

The function f (x) = |x | is continuous everywhere.64.

Let T (x) be the amount of federal income tax owed on an income of x dollars in 1993.(a) Would you expect this function to have any discontinuities?(b) One year the actual function was determined by the formula

T (x) =

.15x for 0 ≤ x < 21450

3217.50 + .28(x − 21450)

for 21450 ≤ x < 51900

11743.50 + .31(x − 51900)

for x ≥ 51900

Sketch the graph of T (x) and determine if it has any discontinuities.(c) Explain why, if T (x) had a jump discontinuity, it might be advantageous in some

situations to earn less money.

Further Insights and Challenges

65. By definition, f (x) has a removable discontinuity at x = c if it is possible to define f (c) insuch a way that f (x) is continuous at x = c. Can there exist more than one way of doingthis?

In order for f (x) to have a removable discontinuity at x = c, limx→c

f (x) = L must exist. To

remove the discontinuity, we define f (c) = L . Then f is continuous at x = csincelim

x→cf (x) = L = f (c). Now assume that we may define f (c) = M �= L and still have

f continuous at x = c. Then limx→c

f (x) = f (c) = M . Therefore M = L , a contradiction.

Roughly speaking, there’s only one way to fill in the hole in the graph of f !66.Function Continuous at Only One Point Let f (x) = x for x rational, f (x) = −x for xirrational. Show that f is continuous at x = 0 and discontinuous at all points x �= 0.67. Find all constants a, b such that the following function has no discontinuities.

f (x) =

ax + cos x for x ≤ π

4

bx + 2 for x >π

4

As x → π

4−, we have ax + cos x → aπ

4+ 1√

2= L . As x → π

4+, we have

bx + 2 → bπ

4+ 2 = R. Match the limits: L = R or aπ

4+ 1√

2= bπ

4+ 2. This implies

π

4(a − b) = 2 − 1√

2or a − b = 8 − 2

√2

π.

68.Prove that the following function is continuous:

f (x) =

1

qif x is a fraction

p

q(in lowest terms)

0 if x is irrational2.5 Evaluating Limits Algebraically


1. Which of the following functions is indeterminate at x = 1?

x2 − 1

x − 1,

x + 1

x + 2,

x − 1√x + 3 − 1

,

x − 1√x + 3 − 2

, tan(π

2x)

2.5 Evaluating Limits Algebraically 25

2. True or False:(a) If f (c) is indeterminate, then the right- and left-hand limits of f (x) as x → c are not

equal.(b) If f (c) is indeterminate but lim

x→cf (x) = L , then we can make f (x) continuous at x = c

by setting f (c) = L .(c) If f (c) is indeterminate, then lim

x→cf (x) must exist.

(d) If f (c) is undefined but f (x) = g(x) for all x �= c, where g(x) is a continuous function,then lim

x→cf (x) = g(c).

Exercises

In Exercises 1–4, show that the limit leads to an indeterminate form. Then carry out the two-stepprocedure: transform the function algebraically and evaluate using continuity.

1. limx→5

x2 − 25

x − 5

limx→5

x2 − 25

x − 5= lim

x→5

(x − 5)(x + 5)

x − 5= lim

x→5(x + 5) = 10.

2.lim

x→−1

x + 1

x2 + 2x + 13. limt→7

2t − 14

5t − 35

limt→7

2t − 14

5t − 35= lim

t→7

2(t − 7)

5(t − 7)= lim

t→7

2

5= 2

5.

4.lim

h→−3

h + 3

h2 − 9In Exercises 5–43, evaluate the limit or state that it does not exist.

5. limx→8

x2 − 64

x − 8

limx→2

x2 − 64

x − 8= −60

−10= 6.

6.

limx→8

x2 − 64

x − 97. limx→8

x2 − 65

x − 8

limx→8

x2 − 65

x − 8does not exist.

8.

limx→2

x2 − 3x + 2

x − 29. limx→2

x3 − 4x

x − 2

limx→2

x3 − 4x

x − 2= lim

x→2

x(x − 2)(x + 2)

x − 2= lim

x→2x(x + 2) = 8.

10.

limx→2

x3 − 5x

x − 211. limz→1

1

z − 1

limz→1

1

z − 1does not exist.

26 Chapter 2 Limits

12.limz→3

1

z − 113. limx→2

x3 − 4

x − 2

limx→2

x3 − 4


14.limx→2

x − 2

x3 − 4x15. limx→2

x

x3 − 4x

limx→2

x

x3 − 4x= lim

x→2

1

x2 − 4does not exist.

16.

limx→4

x2 − 16

x − 417. limx→−4

x2 − 16

x + 4

limx→−4

x2 − 16

x + 4= lim

x→−4

(x − 4)(x + 4)

x + 4= lim

x→−4(x − 4) = −8.

18.

limx→2

3x2 − 4x − 4

2x2 − 819. limy→5

y2 − 4y

y − 5

limy→5

y2 − 4

y − 5does not exist.

As y → 5+, we havey2 − 4

y − 5→ ∞.

As y → 5−, we havey2 − 4

y − 5→ −∞.

20.lim

t→−2

2t + 4

12 − 3t221. lim

y→2

(y − 2)3

y3 − 5y + 2

limy→2

(y − 2)3

y3 − 5y + 2= lim

y→2

(y − 2)3

(y − 2) (y2 + 2y − 1)= lim

y→2

(y − 2)2

y2 + 2y − 1= 0.

22.

limx→16

√x − 4

x − 1623. limh→0

√2 + h − 2

h

limh→0

√h + 2 − 2

hdoes not exist.

As h → 0+, we have√h + 2 − 2

h=

(√h + 2 − 2

)(√

h + 2 + 2)

h(√

h + 2 + 2)= h − 2

x(√

h + 2 + 2)→ −∞.

As h → 0−, we have√h + 2 − 2

h=

(√h + 2 − 2

)(√

h + 2 + 2)

h(√

h + 2 + 2)= h − 2

h(√

h + 2 + 2)→ ∞.

24.

limx→10

√x − 6 − 2

x − 1025. limx→25

√x − 5√

x − 16 − 3

limx→25

√x − 5√

x − 16 − 3·√

x − 16 + 3√x − 16 + 3

= limx→25

(√

x − 5)(√

x − 16 + 3)

x − 25=

limx→25

(√

x − 5)(√

x − 16 + 3)

(√x − 5

) (√x + 5

) =√

9 + 3√25 + 5

= 6

10.

2.5 Evaluating Limits Algebraically 27

26.

limx→5

√x + 4 − 3

x − 527. limx→0

√1 + x − √

1 − x

x

limx→0

√1 + x − √

1 − x

x= lim

x→0

(√1 + x − √

1 − x) (√

1 + x + √1 − x

)x

(√1 + x + √

1 − x)

= limx→0

2x

x(√

1 + x + √1 + x

) = limx→0

2(√1 + x + √

1 − x) = 1.

28.

limx→9

√x − 5 − 2

x − 729. limx→2

√x2 − 1 − √

x + 1

x − 3

limx→2

√x2 − 1 − √

x + 1

x − 3=

√3 − √

3

−1= 0.

30.

limx→7+

√x + 2 − 3√

x − 731. limx→4

(1√

x − 2− 4

x − 4

)

limx→4

(1√

x − 2− 4

x − 4

)= lim

x→4

√x + 2 − 4(√

x − 2) (√

x + 2) = lim

x→4

√x − 2(√

x − 2) (√

x + 2) = 1

4.

32.

limx→0+

(1√x

− 1√x2 + x

)33. lim

x→0

cot x

csc x

limx→0

cot x

csc x= lim

x→0

cos x

sin x· sin x = cos 0 = 1.

34.limx→ π

2

cot x

csc x35. limx→ π

4

sin x − cos x

tan x − 1

limx→ π

4

sin x − cos x

tan x − 1· cos x

cos x= lim

x→ π4

(sin x − cos x) cos x

sin x − cos x= cos

π

4=

√2

2.

36.

limx→ π

3

2 cos2 x + 3 cos x − 2

2 cos x − 137. limθ→ π

2

(sec θ − tan θ

)(Hint: write as a fraction and multiply numerator and denominator by (1 + sin θ).)

limθ→ π

2

(sec θ − tan θ

) = limθ→ π

2

1 − sin θ

cos θ· 1 + sin θ

1 + sin θ= lim

θ→ π2

1 − sin2 θ

cos θ (1 + sin θ)=

limθ→ π

2

cos θ

1 + sin θ= 0

2= 0.

The following identity a3 − b3 = (a − b)(a2 + ab + b2) is useful in Exercises 38–43.

38.

limx→1

x3 − 1

x − 139. limx→2

x3 − 8

x2 − 4

limx→2

x3 − 8

x2 − 4= lim

x→2

(x − 2)(x2 + 2x + 4

)(x − 2)(x + 2)

= limx→2

(x2 + 2x + 4

)x + 2

= 12

4= 3.

40.

limx→1

x2 − 3x + 2

x3 − 1

28 Chapter 2 Limits

41. limx→2

x3 − 8

x2 − 6x + 8

limx→2

x3 − 8

x2 − 6x + 8= lim

x→2

(x − 2)(x2 + 2x + 4

)(x − 2)(x − 4)

= limx→2

(x2 + 2x + 4

)x − 4

= 12

−2= −6.

42.

limx→3

x3 − 27

x − 343. limx→27

x − 27

x1/3 − 3

limx→27

x − 27

x1/3 − 3= lim

x→27

(x1/3 − 3

) (x2/3 + 3x1/3 + 9

)x1/3 − 3

= limx→27

(x2/3 + 3x1/3 + 9

) = 27.

In Exercises 44–57, evaluate the limits in terms of the constants involved.44.

limx→0

(2a + x)

45. limh→0

(4ah + 7a)

limh→0

(4ah + 7a) = 7a.46.

limh→−2

(4ah + 7a)

47. limt→2

(4t − 2at + 3a)

limt→2

(4t − 2at + 3a) = 8 − a.48.

limt→−1

(4t − 2at + 3a)

49. limh→0

(3a + h)2 − 9a2

h

limh→0

(3a + h)2 − 9a2

h= lim

h→0

6ah + h2

h= lim

h→0(6a + h) = 6a.

50.

limx→0

2(x + h)2 − 2x2

h51. limh→4

(h + 2)2 − 9h

h − 4

limh→4

(h + 2)2 − 9h

h − 4= lim

h→4

h2 − 5h + 4

h − 4= lim

h→4

(h − 1)(h − 4)

h − 4= lim

h→4(h − 1) = 3.

52.

limx→a

(x + a)2 − 4x2

x − a53. limx→a

√x − √

a

x − a

limx→a

√x − √

a

x − a= lim

x→a

√x − √

a(√x − √

a) (√

x + √a) = lim

x→a

1√x + √

a= 1

2√

a.

54.

limh→0

√a + 2h − √

a

h55. limx→0

(x + a)3 − a3

x

limx→0

(x + a)3 − a3

x= lim

x→0

x3 + 3x2a + 3xa2 + a3 − a3

x= lim

x→0

(x2 + 3xa + 3a2

) = 3a2.56.

limx→a

(x + a)3 − 8a3

x2 + ax − 2a2

(Hint: Show that (x − a) is a factor of the numerator and the denominator.)57. lim

h→0

13+h

− 13

h

limh→0

13+h

− 13

h= lim

h→0

1

h

3 − (3 + h)

3(3 + h)= lim

h→0

−1

3(3 + h)= −1

9.

2.6 Trigonometric Limits 29


(a) Show that limx→1

x2 − 5x − 6


(b) For which values of c does the following limit exist:

limx→c

x2 − 5x − 6

x − c

59. Find a value of c for which the following limit exists:

limx→3

x2 + 3x + c

x − 1.

limx→3

x2 + 3x + c

x − 1= c + 18

2exists for all values of c.

60.For which value of the sign ± does the following limit exist?

limx→0

(1

x± 1

x(x − 1)

)61. For which value of the sign ± does the following limit exist?

limx→0+

(1

x± 1

|x |)

The limit limx→0+

(1

x− 1

|x |)

= limx→0+

(1

x− 1

x

)= lim

x→0+0 = 0.

The limit limx→0+

(1

x+ 1

|x |)

= limx→0+

(1

x+ 1

x

)= lim

x→0+

(2

x

)= ∞.

62.Which of the following limits exist?

(a) limx→2

x2 + 5x − 6

x2 − 6x + 5

(b) limx→5

x2 + 5x − 6

x2 − 6x + 5

(c) limx→1

x2 + 5x − 6

x2 − 6x + 5

2.6 Trigonometric Limits


1. What two hypotheses on the functions g(x) and h(x) must be satisfied in order to apply theSqueeze Theorem to evaluate a limit lim

x→af (x)?

2. What is the value limx→0

f (x), assuming that −x4 ≤ f (x) ≤ x2?

3. Suppose that −x4 ≤ f (x) ≤ x2. Can this information be used to evaluate limx→1

f (x)?

4. What is limx→0

sin x + 8x

x?

Exercises

In Exercises 1–35, evaluate the limits using the fact that limx→0

(sin x/x) = 1.

1. limx→0

sin x cos x

x

limx→0

sin x cos x

x= lim

x→0

sin x

x· lim

x→0cos x = 1.

2.limx→0

4 sin x

x

30 Chapter 2 Limits

3. limt→0

√t + 1 sin t

t

limt→0

√t + 1 sin t

t= lim

t→0

√t + 1 · lim

t→0

sin t

t= 1.

4.

limt→0

sin2 t

t5. limt→0

sin2 t

t2

limt→0

sin2 t

t2= lim

t→0

sin t

t· lim

t→0

sin t

t= 1.

6.limx→0

tan x

x7. limx→0

x

sin x

limx→0

x

sin x= lim

x→0

1

(sin x) /x= 1.

8.

limx→0

x2

sin2 x9. limh→0

6 sin h

6h

limh→0

6 sin h

6h= lim

h→0

sin h

h= 1.

10.limh→0

sin 6h

6h11. limh→0

sin h

6h

limh→0

sin h

6h= 1

6.

12.limh→0

sin 6h

h13. limx→0

sin 10x

x

limx→0

sin 10x

x= lim

x→010 · sin 10x

10x= 10.

14.limx→0

sin 7x

3x15. limx→0

x

sin 11x

limx→0

x

sin 11x= lim

x→0

1

11· 11x

sin 11x= 1

11.

16.limx→0

tan 4x

9x17. limt→0

sec(9t) sin t

t

limt→0

sec(9t) sin t

t= lim

t→0

1

cos(9t)· lim

t→0

sin t

t= 1.

18.limt→0

sec t sin(9t)

t19. limx→0

x

csc x

limx→0

x

csc x= lim

x→0

x

1/ sin x= lim

x→0x sin x = 0.

20.limx→0

x

csc 25x21. lim

t→0

tan 4t

t

limt→0

tan(4t)

t= lim

t→0

4 sin(4t)

cos(4t) · 4t= lim

t→0

4

cos(4t)· sin(4t)

4t= 4.

22.limt→0

sec 9t tan 4t

t

2.6 Trigonometric Limits 31

23. limt→0

tan 4t

t sec t

limt→0

tan 4t

t sec t= lim

t→0

4 sin 4t

4t cos(4t) sec(4t)= lim

t→0

4 cos t

cos 4t· sin 4t

4t= 4.

24.limh→0

sin 9h sin 8h

h225. lim

t→0

7t2

sin2 8t

limt→0

7t2

sin2 8t= lim

t→0

7

82

(8t

sin 8t

)2

= 7

64.

26.

limt→0

7t2

sin 7t sin 8t27. limx→0

x sin 4x

sin 2x sin 2x

limx→0

x sin 4x

sin 2x sin 2x= lim

x→0

x · 2 sin 2x cos 2x

sin 2x sin 2x= lim

x→0

2x

sin 2x· cos 2x = 1.

28.Evaluate lim

h→0

sin 2h

sin 5h.

Hint: writesin 2h

sin 5h= 2

5· sin 2h

2h· 5h

sin 5h.

29. limh→0

sin 3h

sin 5h

limh→0

sin 3h

sin 5h= lim

h→0

3

5

(sin 3h) / (3h)

(sin 5h) / (5h)= 3

5.

30.limz→0

sin(z/3)

sin z31. limx→0

sin(−3x)

sin(4x)

limx→0

sin(−3x)

sin(4x)= lim

x→0

− sin(3x)

3x· 3

4· 4x

sin(4x)= −3

4.

32.limx→0

tan 4x

tan 9x33. limt→0

csc 8t

csc 4t

limt→0

csc 8t

csc 4t= lim

t→0

sin 4t

sin 8t· 8t

4t· 1

2= 1

2.

34.limx→0

sin 5x sin 2x

sin 3x sin 5x35. limx→0

sin 3x sin 2x

x sin 5x

limx→0

sin 3x sin 2x

x sin 5x= lim

x→0

(3

sin 3x

3x· 2

5

(sin 2x) / (2x)

(sin 5x) / (5x)

)= 6

5.

36.limh→0

sin 6h sin 8h

h37. limh→0

sin 6h sin 8h

h2

limh→0

sin 6h sin 8h

h2= lim

h→0

(6

sin 6h

6h· 8

sin 8h

8h

)= 48.

38.

limx→0

cos2 x − 1

x39. Calculate the right- and left-hand limits

(a) limx→0+

sin x

|x |(b) lim

x→0−sin x

|x |

limx→0+

sin x

|x | = limx→0+

sin x

x= 1.

32 Chapter 2 Limits

limx→0−

sin x

|x | = limx→0−

sin x

−x= −1.

In Exercises 40–43, use the Squeeze Theorem to evaluate the limit.40.

limx→0

x2 sin1

x41. limx→0

tan x cos

(sin

1

x

)

limx→0

tan x cos

(sin

1

x

)= 0. For as x → 0, we have∣∣tan x cos

(sin 1

x

) − 0∣∣ = |tan x | · ∣∣cos

(sin 1

x

)∣∣ ≤ |tan x | → 0.42.

limx→1

(x − 1) sin

(π

x − 1

)43. lim

x→0cos

(π

x

)sin x

Since −1 ≤ cos(

π

x

) ≤ 1, observe that − sin x ≤ cos(

π

x

)sin x ≤ sin x . Since

limx→0

sin x = 0 = limx→0

− sin x , the Squeeze Theorem shows that limx→0

cos(π

x

)sin x = 0.

44.(a) Draw the graphs of h(x) = sin x and g(x) = 1 + |x − π

2 | on the same set of axes.(b) Determine lim

x→π/2f (x), where f (x) is a function having h(x) as a lower bound and g(x)

as an upper bound.

45. GU

(a) Investigate limh→0

cos h − 1

h2numerically.

(b) Confirm your conclusion graphically.(c) Evaluate the limit exactly.

Hint: multiply numerator and denominator by cos h + 1.

(a)h −.1 −.01 .01 .1

cos h − 1

h2−.499583 −.499996 −.499996 −.499583

The limit is − 12.

(b) –0.6

–0.5

–0.4

–1 0 1

(c) limh→0

cos h − 1

h2= lim

h→0

cos2 h − 1

h2(cos h + 1)= lim

h→0

(sin h

h

)2 −1

cos h + 1= −1

2.


This exercise gives an alternate proof of the second basic trigonometric limit.(a) Use a diagram of the unit circle to show that the following inequality holds

(1 − cos θ)2 + sin2 θ ≤ θ 2

(b) Conclude that 2(1 − cos θ) ≤ θ 2.

(c) Prove that limθ→0

1 − cos θ

θ= 0.

(d) Use (b) to prove that cos θ is continuous at θ = 0.

47. Continuity of Sine Although the continuity of sin x seems intuitively clear, it requiresproof. A proof is given in the following steps.(a) Show that lim

θ→0sin θ = 0 (this shows that sin θ is continuous at θ = 0).

Write sin θ = θ(sin θ/θ) and apply the Rule for Products.

2.7 The Formal Definition of a Limit 33

(b) Show that limθ→0

cos θ = 1 (this shows that cos θ is continuous at θ = 0).

Write (cos θ − 1) = θ((cos θ − 1)/θ) and apply the Rule for Products.(c) Use the results of (a) and (b) together with the addition formula

sin(a + h) = sin a cos h + cos a sin h

to prove that limh→0

sin(a + h) = sin a for all a.

(d) Set x = a + h and conclude that limx→a

sin x = sin a.

(a) Write sin θ = θ sin θ

θ. By the Rule for Products, lim

θ→0sin θ = lim

θ→0θ

sin θ

θ= 0 · 1 = 0.

(b) Write (cos θ − 1) = θ((cos θ − 1)/θ). By the Rule for Products,limθ→0

(cos θ − 1) = limθ→0

θ((cos θ − 1)/θ) = 0 · 0 = 0.

(c) For all a, we have limh→0

sin(a + h) = limh→0

(sin a cos h + cos a sin h) =sin a lim

h→0cos h + cos a lim

h→0sin h = sin a · 1 + cos a · 0 = sin a

(d) limx→a

sin x = lima+h→a

sin(a + h) = limh→0

sin(a + h) = sin a.48.

As shown in the text, the following inequality holds for − π

2< θ < π

2:

cos θ ≤ sin θ

θ≤ 1

(a) Show, by squaring each term, that

cos2 θ ≤ 1 − cos2 θ

θ 2≤ 1.

(b) Use the Squeeze Theorem to show that

limθ→0

1 − cos2 θ

θ 2= 1

(c) Use (b) to show that limθ→0

1 − cos θ

θ 2= 1

2.

49. Find limx→0

√cos x − 1

x

limx→0

√cos x − 1

x= lim

x→0

√cos x − 1

x

√cos x + 1√cos x + 1

= limx→0

cos x − 1

x

1√cos x + 1

= 0 · 12 = 0

via ??.50.

(a) Investigate the limit

limx→c

sin x − sin c

x − c

numerically for the four values c = 1, π

3, π

4, π

2.

(b) Can you guess the answer for general c? (We’ll see why in Chapter 3.)(c) Check if your answer to (b) works for two other values of c.

2.7 The Formal Definition of a Limit


Which of the following statements follows from the fact that limx→0

cos x = 1? Answer True or

False and explain.

1. If | cos x − 1| is very small, then x must be close to 0.

2. There is a number ε > 0 such that x is at most 10−5 if| cos x − 1| < ε.

3. There is a number δ > 0 such that

| cos x − 1| < 10−5 if |x | < δ

4. There is a number δ > 0 such that

| cos x | < 10−5 if |x − 1| < δ

34 Chapter 2 Limits

Exercises

1. Consider limx→2

(4x − 1).

(a) Use the fact that |(4x − 1) − 7| = 4|x − 2| to conclude that

|(4x − 1) − 7| < 4δ if |x − 2| < δ.

(b) Find a δ such that the statement

|(4x − 1) − 7| < .01 if |x − 2| < δ

is true.(c) Give a rigorous proof that lim

x→24x − 1 = 7 by showing that for any ε > 0, the statement

|(4x − 1) − 7| < ε if |x − 2| < δ

where δ = ε

4 .

(a) If 0 < |x − 2| < δ, then |(4x − 1) − 7| = 4|x − 2| < 4δ.(b) If 0 < |x − 2| < δ = .025, then |(4x − 1) − 7| = 4|x − 2| < 4δ = .01.(c) Let ε > 0 be given. Then whenever 0 < |x − 2| < δ = ε/4, we have

|(4x − 1) − 7| = 4|x − 2| < 4δ = ε. Since εwas arbitrary, we conclude thatlimx→2

(4x − 1) = 7.2.

With regard to the limit limx→2

x2 = 4,

(a) Show that the gap |x2 − 4| is less than .05 if 0 < |x − 2| < .01.(b) Show that |x2 − 4| is less than .0002 if 0 < |x − 2| < .0009.

3. With regard to the limit limx→5

x2 = 25,

(a) How large can the gap |x2 − 25| be if 0 < |x − 5| < .2 (the case δ = .2)?(b) Show that |x2 − 25| < 11|x − 5| if 4 < x < 6.

Hint: write |x2 − 25| = |x + 5| · |x − 5|.(c) Find a δ such that |x2 − 25| < 10−3 is |x − 5| < δ.(d) Show that |x2 − 25| < ε if |x − 5| < δ, where δ is any number smaller than ε

11and 1.

(a) If 0 < |x − 5| < δ = .2, then x < 5.2 and∣∣x2 − 25∣∣ = |x − 5||x + 5| ≤ |x − 5| (|x | + 5) < 10.2|x − 5| < 2.04.

(b) If 4 < x < 6, then |x − 5| < δ = 1 and∣∣x2 − 25∣∣ = |x − 5||x + 5| ≤ |x − 5| (|x | + 5) < 11|x − 5|.

(c) If 0 < |x − 5| < δ = min{1, .001

11

}, then x < 6 and∣∣x2 − 25

∣∣ = |x − 5||x + 5| ≤ |x − 5| (|x | + 5) < 11|x − 5| < .001.

(d) If 0 < |x − 5| < δ = min{1, ε

11

}, then x < 6 and∣∣x2 − 25

∣∣ = |x − 5||x + 5| ≤ |x − 5| (|x | + 5) < 11|x − 5| < ε.4.

Consider limx→2

1x.

(a) Show that ∣∣∣∣ 1

x− 1

2

∣∣∣∣ = |x − 2|2x

< |x − 2|

if x > 12.

(b) Show that the statement ∣∣∣∣ 1

x− 1

2

∣∣∣∣ < δ if |x − 2| < δ

holds for all δ < 32

(this condition insures that x > 12).

(c) Find a δ such that ∣∣∣∣ 1

x− 1

2

∣∣∣∣ < .01 if |x − 2| < δ.

(d) Prove that limx→2

1x

= 12 by showing that for any ε > 0, the statement

∣ ∣

5. Consider limx→1

√x + 3.

(a) Show that if x > −3, then

|√x + 3 − 2| <1

2|x − 1|

Hint: multiply the inequality by |√x + 3 + 2| and observe that |√x + 3 + 2| > 2.(b) Find δ such that |√x + 3 − 2| < 10−4 for |x − 1| < δ.

2.7 The Formal Definition of a Limit 35

(c) Prove that the limit is equal to 2.

(a) If x > −3, then√

x + 3 is defined (and positive),

whence∣∣∣√x + 3 − 2

∣∣∣ =∣∣∣∣∣(√

x + 3 − 2)

1

(√x + 3 + 2

)(√

x + 3 + 2)∣∣∣∣∣ = |x − 1|√

x + 3 + 2<

|x − 1|2

.

(b) Choose δ = .0002. Then provided 0 < |x − 1| < δ, we have x > −3 and therefore∣∣∣√x + 3 − 2∣∣∣ =<

|x − 1|2

<δ

2= .0001 by part (a).

(c) Let ε > 0 be given. Then whenever 0 < |x − 1| < δ = min {2ε,4}, we have x > −3and thus∣∣∣√x + 3 − 2

∣∣∣ =∣∣∣∣∣(√

x + 3 − 2)

1

(√x + 3 + 2

)(√

x + 3 + 2)∣∣∣∣∣ = |x − 1|√

x + 3 + 2<

2ε

2= ε.

Since ε was arbitrary, we conclude that limx→1

√x + 3 = 2.


Uniqueness of the Limit Show that a function f (x) converges to at most one limitingvalue L . In other words, show that if the limit definition can be used to show that lim

x→cf (x) =

L1 and limx→c

f (x) = L2, then L1 = L2.

7. Here is a function with strange continuity properties. Define

f (x) =

1

qif x is the rational number

p

qin

lowest terms

0 if x is an irrational number

It turns out that f (x) is continuous at every irrational number, but discontinuous at everyrational number. Can you think of a reason why this is so? Needless to say, it is impossibleto graph this function.

Let c be irrational and let ε > 0 be given. Choose a positive integer N > 1/ε and setδ = 1/N . Whenever 0 < |x − c| < δ we have| f (x) − f (c)| = | f (x) − 0| = | f (x)| < 1/N < ε. (REMARK: Why is | f (x)| < 1/N?Well, if x is irrational, then | f (x)| = |0| < 1/N . On the other hand, if x = p/q (inlowest terms) is rational, then | f (x)| = 1/q < 1/N because x is within 1/N units of theirrational number c.) Since ε was arbritrary, we conclude that lim

x→cf (x) = 0 = f (c).

Therefore, f is continuous at c.Let r = p/q (in lowest terms) be any rational number. Let {x1, x2, . . . } be a sequence ofirrational points that approach r ; i.e., as n → ∞, the xn get arbitrarily close to r . Notice

that as n → ∞, we have | f (xn) − f (r)| =∣∣∣0 − 1

q

∣∣∣ = 1/q → 1/q �= 0. Therefore, it is

not true that limx→r

f (x) = f (r). Accordingly, f is not continuous at r .

CONCLUSION: Since the choices of c and r were arbitrary, we conclude that f iscontinuous at all irrational numbers and discontinuous at all rational numbers.

8.Prove the Squeeze Theorem using the ε–δ definition.

36 Chapter 2 Limits

2.8 Intermediate Value Theorem


1. Explain why the equation x3 − 8x + 4 = 0 has a root in the interval [0, 1].2. The temperature in Los Angeles was 46◦ at 6 AM and rose to 74◦ at noon. What must we

assume about temperature (as a function of time) to conclude that the temperature was 60◦

at some moment of time between 6 AM and noon?

In Questions 3-5, answer True or False and explain.

3. If f (a) > 0 and f (b) > 0, then f (x) does not have a root in the interval [a, b].4. If f (x) has a root in [a, b], then f (a) and f (b) have opposite signs.

5. If f (x) has a root in [a, b], then f (x) takes on both positive and negative values on theinterval [a, b].

Exercises

1. Use the Intermediate Value Theorem to show that x3 + x − 9 has a root in the interval [1, 2].Let f (x) = x3 + x − 9. Observe that f (1) = −7 and f (2) = 1. Since f is a polynomial, itis continuous everywhere; in particular on [1, 2]. Therefore, by the IVT there is a c ∈ [1, 2]such that f (c) = 0.

2.Show that cos x = x has a solution in the interval [0, 1]. Hint: consider the functionf (x) = x − cos x .3. Show that sin(nx) = cos x has a solution in [0, π ] for every integer n.

For each integer n, let f (x) = sin(nx) − cos x . Observe that f is continuous withf (0) = −1 and f (π) = 1. Therefore, by the IVT there is a c ∈ [0, π ] such thatf (c) = sin(nc) − cos(nc) = 0. Thus sin(nc) = cos(nc) and hence the equationsin nx = cos x has a solution c in the interval [0, π ].

4.Show that there exists a number x such that

√x + √

x + 1 = 2.5. Use the IVT to show that

√2 exists.

Let f (x) = x2 − 2. Observe that f is continuous with f (1) = −1 and f (2) = 2.Therefore, by the IVT there is a c ∈ [1, 2] such that f (c) = c2 − 2 = 0. Thus c2 = 2 andhence the equation x2 = 2 has a solution c in [1, 2]. This proves the existence of

√2, a

number whose square is 2.6.

Let f (x) = x3 − 8x − 1. Use the IVT and inspection to find a root of f in the interval[2, 3] to two decimal places.7. Use the IVT and inspection to find a root of f (x) = x5 − 5x + 1 in the interval [0, 1] totwo decimal places.

Let f (x) = x5 − 5x + 1. Observe that f is continuous with f (0) = 1 and f (1) = −3.Therefore, by the IVT there is a c ∈ [0, 1] such that f (c) = 0. By inspection we havef (.19) ≈ .0502, f (.20) ≈ .0003, and f (.21) ≈ −.0496. Hence c ≈ .20.

8.

Let f (x) = x3 + 2x + 1 = 0.(a) Can the IVT be applied to show that f (x) has a root in the interval [1, 2]?(b) Does it have a root in [−1, 0]?(c) Find an interval of width 1

4containing a root of f (x).

9. Show that for all positive integers k, the equation cos x = xk has a solution for somex ∈ [0, π

2].

For each positive integer k, let f (x) = xk − cos x . Observe that f is continuous on[0, π

2

]with f (0) = −1 and f ( π

2) = (

π

2

)k> 0. Therefore, by the IVT there is a c ∈ [

0, π

2

]such

that f (c) = ck − cos(c) = 0. Thus cos c = ck and hence the equation cos x = xk has asolution c in the interval

[0, π

2

].

2.8 Intermediate Value Theorem 37

10.Draw the graph of a function on the interval [0, 4] with a jump discontinuity at x = 2 thatdoes not satisfy the conclusion of the Intermediate Value Theorem.11. Draw the graph of a function on the interval [0, 4] that satisfies the conclusion of theIntermediate Value Theorem, even though it has a jump discontinuity at x = 2.

The function graphed below has a jump discontinuity at x = 2. Note that for every value Mbetween f (0) = 2 and f (4) = 4, there is a point c in the interval [0, 4] such thatf (c) = M . Accordingly, the conclusion of the IVT is satisfied.

0

2

4

2 412.

Draw the graph of a function on the interval [0, 4] that has infinite one-sided limits at x = 2and does not satisfy the conclusion of the Intermediate Value Theorem.13. Draw the graph of a function on the interval [0, 4] that satisfies the conclusion of theIntermediate Value Theorem, even though it has infinite one-sided limits at x = 2.

The function graphed below has infinite one-sided limts at x = 2. Note that for every valueM between f (0) = 0 and f (4) = 4, there is a point c in the interval [0, 4] such thatf (c) = M . Accordingly, the conclusion of the IVT is satisfied.

0

2

4

6

2 4


Define

f (x) = sin

(1

x

)for x �= 0

0 for x = 0

Show that f satisfies the conclusion of the Intermediate Value Theorem on every interval[−a, a], even though f is discontinuous at x = 0.

15. Take any map (for example, of the United States) and draw any circle on it. Prove that at anymoment in time there exists a pair of diametrically opposite points on that circle correspond-ing to locations where the current temperatures are equal. Hint: fix a coordinate system andconsider the function f (θ) defined as the difference in temperatures at the locations corre-sponding to θ and θ + π .

Say the circle has (fixed but arbitrary) radius r and use polar coordinates with the pole atthe center of the circle. For 0 ≤ θ ≤ 2π , let T (θ) be the temperature at the point(r cos θ, r sin θ). We assume this temperture varies continuously. For 0 ≤ θ ≤ π , define fas the difference f (θ) = T (θ) − T (θ + π). Then f is continous on [0, π ]. There are threecases.

If f (0) = T (0) − T (π) = 0, then T (0) = T (π) and we have found a pair ofdiametrically opposite points on the circle at which the temperatures are equal.If f (0) = T (0) − T (π) > 0, then f (π) = T (π) − T (2π) = T (π) − T (0) < 0. [Notethat the angles 0 and 2π correspond to the same point, (x, y) = (r, 0).] Since f iscontinuous on [0, π ], we have by the IVT that f (c) = T (c) − T (c + π) = 0 for somec ∈ [0, π ]. Accordingly, T (c) = T (c + π) and we have again found a pair ofdiametrically opposite points on the circle at which the temperatures are equal.

38 Chapter 2 Limits

If f (0) = T (0) − T (π) < 0, then f (π) = T (π) − T (2π) = T (π) − T (0) > 0. Since fis continuous on [0, π ], we have by the IVT that f (d) = T (d) − T (d + π) = 0 for somed ∈ [0, π ]. Accordingly, T (d) = T (d + π) and once more we have found a pair ofdiametrically opposite points on the circle at which the temperatures are equal.

CONCLUSION: There is always a pair of diametrically opposite points on the circle atwhich the temperatures are equal.

16.Use the IVT to show that tan x = x has infinitely many solutions.

17. Show that nth roots exist. More precisely, use the IVT to show that if n is a positive integerand c is a positive real number, then xn = c has a solution.

Let n be a positive integer. Let c > 0 be a real number. We know xn is defined for all realnumbers x . 0n = 0 < c, and, as xn → ∞ as x → ∞, there exists some real b such thatbn > c. (To be more specific, set b equal to the greater of c and 2; if c > 1, then cn > c. Ifc < 1, then cn < c < 2.) We therefore have a closed interval [0, b] inside the open interval(−∞, ∞) on which the function f (x) = xn − c is defined where f (0) < 0 and f (b) > 0.By the intermediate value theorem, there is some d ∈ [0, b] such that f (d) = 0. We canrefer to d as n

√x .

calculus limits

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