Lecture 10

Limits (cont’d)

One-sided limits

(Relevant section from Stewart, Seventh Edition: Section 2.4, pp. 113.)

As you may recall from your earlier course in Calculus, we may define “one-sided limits,” where

the independent variable x is restricted to approach the point a from either the right (i.e., x > a) or

from the left (i.e., x < a).

• “Left-sided” or “Left-hand” limit:

The statement

limx→a−

f(x) = L,

means that for any ǫ > 0, there exists a δ > 0 such that

a − d < x < a implies that |f(x) − L| < ǫ.

• “Right-sided” or “Right-hand” limit:

The statement

limx→a+

f(x) = L,

means that for any ǫ > 0, there exists a δ > 0 such that

a < x < a + δ implies that |f(x) − L| < ǫ.

And for the limit to exist at x = a, both one-sided limits must be equal, i.e.,

limx→a

f(x) = limx→a−

f(x) = limx→a+

f(x) = L.

If the one-sided limits are not equal, then the limit limx→a

f(x) does not exist.

Example: Let’s return to the Heaviside function, defined in an earlier lecture:

H(x) =

1, x ≥ 0,

0, x < 0.(1)

We must consider three cases:

67

x

y

0

y = H(x)

1

Graph of Heaviside function H(x).

1. a > 0: In this case,

limx→a

H(x) = limx→a

1 = 1. (2)

The limit exists at all a > 0. No matter how close the point a is to 0, we can always approach

it from both sides.

2. a < 0: In this case,

limx→a

H(x) = limx→a

0 = 0. (3)

The limit exists at all a < 0. Once again, no matter how close the point a is to 0, we can always

approach it from both sides.

3. a = 0: In this case, we must consider right- and left-sided limits:

limx→0+

H(x) = limx→0+

1 = 1. (4)

limx→0−

H(x) = limx→0−

0 = 0. (5)

Since the two one-sided limits are not equal, limx→0

does not exist.

68

Infinite limits

(Relevant section from Stewart, Seventy Edition: Section 2.4, pp. 115-116.)

You may recall that we may treat infinity (or −∞) as a limit point: If a positive function f(x)

behaves as follows,

f(x) → ∞ as x → a+ andf(x) → ∞ as x → a−, (6)

then we may write

limx→a

f(x) = ∞. (7)

An example is the function f(x) =1

x2. As x → 0 from either side, f(x) → ∞, as the following sketch

of its graph shows.

y = 1

x2

y

0x

In contrast, the function f(x) =1

xdoes not have a limit as x → 0, since f(x) → −∞ as x → 0−

and f(x) → ∞ as x → 0+.

The question that remains is, “Can we translate the statement in (7) into an ǫ, δ-type statement

used for finite limits?” The answer is “Yes,” but with a little care. Recall that for finite limits, we

understand that the statement,

limx→a

f(x) = L, (8)

means the following: Given an ǫ > 0, there exists a δ > 0, such that

0 < |x − a| < δ implies that |f(x) − L| < ǫ. (9)

The problem we now face with infinite limits is that L is ∞. Does it make sense to write the

inequality,

“ |f(x) −∞| < ǫ ”? (10)

69

The answer is “No”: If we interpret the LHS as the distance between f(x) and ∞, then if f(x) is

finite, this distance is infinite. And no matter how much greater we made f(x), e.g., multiplying it by

10243, the distance is still infinite.

So how can we define ‘closeness’ to infinity in a reasonable way, if at all? The answer is to consider

the reciprocal of f(x): If f(x) is large, then1

f(x)is small. If we identify 0 as being the reciprocal of

∞, then for positive functions,∣

∣

∣

∣

1

f(x)− 0

∣

∣

∣

∣

=

∣

∣

∣

∣

1

f(x)

∣

∣

∣

∣

=1

f(x)(11)

may be viewed as the closeness of f(x) to ∞.

An ǫ, δ definition for (7) could then be phrased as follows: Given any ǫ > 0, there exists a δ > 0

such that

0 < |x − a| < δ implies that1

f(x)< ǫ. (12)

But this is perhaps a little cumbersome: We can rewrite the inequality on the right by taking recip-

rocals,1

f(x)< ǫ implies f(x) >

1

ǫ. (13)

And as ǫ → 0+,1

ǫ→ ∞. Instead of working with the reciprocal of ǫ, we define M =

1

ǫand re-express

the relation in (12) as follows:

“ limx→a

f(x) = ∞” means that given an M > 0, there exists a δ > 0, such that

0 < |x − a| < δ implies that f(x) > M. (14)

The idea is that we now let M become arbitrarily large. This means that f(x) can be made

arbitrarily large by taking x sufficiently close to a. (Note that we do not want to write that |f(x)| > M .

This could allow f(x) to approach −∞ on one side of a and ∞ on the other side, which is contrary to

the idea of f itself being arbitrarily large and positive.) A graphical sketch is given below.

70

x

M

a a + δa − δ

y = f(x)

A simple example of how δ may be determined in terms of M is given in Example 5 of Stewart’s

text, p. 116. Here, f(x) =1

x2and we find that (not surprisingly) δ = M−1/2.

A similar definition for limits to −∞ may be developed in the same way, replacing M > 0 with

N < 0, with the idea that |N | will become arbitrarily large.

Continuity of functions

(Relevant section from Stewart, Seventh Edition: Section 2.5)

We have now discussed, in some detail, the concept of the limit of a function, i.e.,

limx→a

f(x) = L. (15)

This statement implies, very roughly, that as x approaches the value a the value f(x) approaches the

value L. (As we’ve seen, the ǫδ definition is a precise formulation of this idea.)

L

ax

y

y = f(x)

The function f possesses a limit at a: limx→a

f(x) = L

Of course, this says absolutely nothing about what f does at x = a. The function f may or may

71

not even be defined at x = a, i.e., f(a) may or may not even exist. As far as we’re concerned, the

above limit statement leaves a “hole” in the graph of f .

As you already know, if f is nice and continuous at x = a, then the limit L is precisely the value

f(a). Formally, then

Definition: A function f is continuous at a if

limx→a

f(x) = f(a). (16)

Note that there are three pieces of information in the above equation, namely,

1. f(a) is defined (i.e., the function f is defined at a),

2. limx→a

f(x) exists (i.e., the limit L of f exists),

3. The limit L is equal to f(a).

As such, the “hole” in the above graph is now filled in, as sketched below. You’ve probably heard

or read this before: If f is continuous at all x in some interval [c, d], then the graph of f over this

interval can be drawn “without taking your pen/pencil” off the paper.”

ax

y

y = f(x)

f(a)

The function f is continuous at a: limx→a

f(x) = f(a)

A more formal ǫδ definition of continuity would be as follows:

Definition: A function f is continuous at a if, given any ǫ > 0, there exists a δ > 0, such

that

|f(x) − f(a)| < ǫ for all x such that 0 < |x − a| < δ. (17)

In other words, we have replaced L in the ǫδ limit definition with f(a).

72

Another way to write Eq. (16) is as follows,

limh→0

f(a + h) = f(a). (18)

We have simply let x = a+h here. This formulation will be convenient in our discussion of derivatives.

An important consequence from the definition of continuity is the following:

If f is continuous at a, then for x close to a, f(x) is close to f(a).

This sounds rather simple, but it is a profound concept that is the basis of many methods in both

theory and applications.

73

Lecture 11

Continuous functions (cont’d)

One-sided continuity

Because there are one-sided limits, we can define “one-sided continuity”:

1. A function f is left-continuous or continuous from the left at a if

limx→a−

f(x) = f(a). (19)

2. A function f is right-continuous or continuous from the right at a if

limx→a+

f(x) = f(a). (20)

Of course, for the limit in Eq. (16) to exist, i.e., for f to be continuous at a, the two one-sided limits

must be equal:

f is continuous at a if and only if limx→a+

f(x) = limx→a−

f(x) = f(a).

Example: The Heaviside function discussed earlier,

H(x) =

1, x ≥ 0,

0, x < 0.(21)

Once again, we must consider three cases:

x

y

0

y = H(x)

1

Graph of Heaviside function H(x).

74

1. a > 0: In this case,

limx→a

H(x) = limx→a

1 = 1 = H(a). (22)

Therefore H(x) is continuous at all a > 0. We can also write that H(x) is continuous at all

x > 0.

2. a < 0: In this case,

limx→a

H(x) = limx→a

0 = 0 = H(a). (23)

Therefore H(x) is continuous at all x < 0.

3. a = 0: In this case, we must consider right- and left-sided limits:

limx→0+

H(x) = limx→0+

1 = 1 = H(0). (24)

limx→0−

H(x) = limx→0−

0 = 0 6= H(0). (25)

Therefore H(x) is not continuous at x = 0. However, it is right continuous at x = 0.

In summary: The Heaviside function H(x) is continuous at all x 6= 0 and right-continuous at x = 0.

Each of the results of the following Theorem (Theorem 4 from Stewart, p. 122) follows from the

appropriate Limit Law from a previous lecture.

Theorem: If f and g are continuous at a, and c ∈ R is a constant, then the following functions are

continuous at a:

1. f ± g

2. cf

3. fg

4.f

g, provided that g(a) 6= 0.

For example, to prove the first result,

limx→a

[f(x) ± g(x)] = limx→a

f(x) ± limx→a

g(x) = f(a) ± g(a). (26)

75

We are now in a position to determine the continuity properties of some important functions. For

example, from the limit relation,

limx→a

x = a, (27)

it follows from the Limit Law for Products that

limx→a

x2 =(

limx→a

x)(

limx→a

x)

= a2. (28)

We can continue this procedure to obtain the result

limx→a

xn = an. (29)

In other words, the functions f(x) = xn are continuous for n = 0, 1, 2, . . . . (The case n = 0 is trivial.)

From the above Theorem, it follows that any linear combination of these functions is also continuous.

In other words,

Any polynomial,

P (x) = c0 + c1x + · · · + cnxn, (30)

where the ci are constants and n ≥ 0, is continuous at any x ∈ R.

From the above Theorem, it also follows that any ratio of polynomials, i.e.,

R(x) =P (x)

Q(x)=

c0 + c1x + · · · + cnxn

d0 + d1x + · · · + dmxm, (31)

for m,n ≥ 0, is continuous at any x for which Q(x) 6= 0. This is the same as stating that the rational

function R(x) is continuous at all x in its domain of definition.

We can continue to build functions in this way to produce a list of functions that are continuous

on their domain of definition (see Theorem 7 of Stewart, p. 124). In addition, we would expect that

if a function were continuous and one-to-one, then its inverse function would also be a continuous

function. As a result, the following functions are continuous on their domains of definition:

1. polynomials

2. rational functions

3. root functions (e.g.,√

x)

76

4. trigonometric functions

5. inverse trigonometric functions

6. exponential functions

7. logarithmic functions

Composition of functions

We now consider the continuity properties of composite functions, f ◦ g(x) = f(g(x)). The following

important result may be found in Stewart, Theorem 8, p. 125:

Theorem: If limx→a

g(x) = b and f is continuous at b, then limx→a

f(g(x)) = f(b). In other words,

limx→a

f(g(x)) = f(

limx→a

g(x))

. (32)

An ǫδ proof of this theorem is given in Appendix F of Stewart’s text. We can view this result

graphically as follows: By letting x approach b, we can make the value of g(x) approach b. (But we

y

x t

y

.b

b g(x)

f(t)

g(x)

a x

f(b) .

.

.

don’t even consider x = b.) As g(x) approaches b, f(g(x)) approaches f(b). And because f is assumed

to be continuous, the limit of the latter process is f(b).

Example: Let f(x) =√

x and g(x) = 1 +sin x

x. We have that

limx→0

g(x) = limx→0

(

1 +sin x

x

)

= 1 + 1 = 2. (33)

77

(In a few lectures, we’ll prove that limx→0

sinx

x= 1.) Note that g(0) is not defined. The function

f(x) =√

x is continuous at all x > 0, in particular, x = 2. From the above theorem,

limx→0

√

1 +sinx

x=

√

limx→0

(

1 +sinx

x

)

=√

2. (34)

In the following Theorem, we make the additional assumption that limx→a

g(x) = g(a), i.e., g is

continuous at a.

Theorem: If g is continuous at a and f is continuous at g(a), then the composite function f ◦ g is

continuous at a, i.e.,

limx→a

(f ◦ g)(x) = (f ◦ g)(a). (35)

In other words, a continuous function of a continuous function is a continuous function.

The proof of this result is quite straightforward, following from the previous Theorem.

limx→a

f(g(x)) = f(

limx→a

g(x))

= f(g(a)). (36)

78

Continuity of functions (cont’d)

Intermediate Value Theorem

We start with a definition:

Definition: A function f is said to be continuous over an interval I if it is continuous at

all x ∈ I. In the case that I = [a, b], it is understood that f is continuous at all x ∈ (a, b),

right-continuous at a and left-continuous at b.

Continuous functions on closed intervals [a, b] exhibit some special properties, one of which we

now discuss. A possible case is sketched below. Note that we consider the case that f(a) 6= f(b). Of

course, the graph of f is a continuous curve that starts at the point (a, f(a)) and ends at the point

(b, f(b)). As such, if we consider the line y = N , where N is any value that lies between f(a) and

f(b), the graph of f will have to cross this line at least once.

N

f(b)

f(a)

y = f(x)

a c b

x

y

We now state this property more formally as a theorem.

Intermediate Value Theorem: Suppose that f is continuous on [a, b], with f(a) 6= f(b).

Let N be any real number between f(a) and f(b). (In other words, N 6= f(a) and

N 6= f(b).) Then there exists a number c ∈ (a, b) such that f(c) = N .

Notes:

1. The number N is an “intermediate value”, i.e., it lies between f(a) and f(b).

2. The phrase “there exists a number c” could be read as “there exists at least one number c”.

There could be several distinct numbers c1, c2, · · · , cn at which f assumes the value N .

79

3. The theorem does not tell you where c is located. It is only an existence theorem.

4. This theorem relies on the assumption that f is continuous on [a, b]. If it were not continuous,

it could have “jumps” that would allow the graph to avoid values between f(a) and f(b).

5. In the case that f(a) = f(b), no such c value need exist. Can you sketch an example?

You may well encounter a proof of the IVT in a course on real analysis, e.g., AMATH 331.

Application of IVT: Estimating zeros of functions

Once again, assume that f is a continuous function over the closed interval [a, b]. Now assume that

f(a) and f(b) are nonzero and have opposite signs, i.e., f(a)f(b) < 0. This means that N = 0 is

an intermediate value. The IVT guarantees the existence of at least one number c ∈ (a, b) at which

f(c) = 0, i.e., c is a zero of the function f . This is sketched below.

0 a b

x

f(a)

f(b)

y

y = f(x)

c

Many applications rely on the numerical estimation of zeros of functions. There are various pow-

erful methods to accomplish this estimation, including the Newton-Raphson method, to be discussed

later. For the moment, however, we can see the use of the IVT in narrowing down the location of the

zeros of a function. If we sample a continuous function at various x-values, call them xn, and if it

changes sign at two consecutive x-values, say xK and xK+1, then a zero of f is located in the interval

(xK , xK+1).

Example: Consider the function f(x) = 4x3 − 6x2 + 3x − 2. Note that

f(0) = −2, f(1) = −1, f(2) = 12. (37)

80

We may conclude that a zero of f – we’ll call it x̄ – lies in the open interval (1, 2), i.e., 1 < x̄ < 2. Of

course, we may not be content with this observation, being interested in a more accurate estimate of

x̄. In other words, we require a more refined search.

One way would be to simply divide the interval [1, 2] into n subintervals of length 1/n and then

evaluating f(x) at the partition points. If we do this for n = 10, we find that a sign change occurs

here:

f(1.2) = −0.128, f(1.3) = 0.548. (38)

As a result, we now know that 1.2 < x̄ < 1.3. And if we now divide [1.2, 1.3] into n = 10 subintervals,

we find that a sign change occurs here:

f(1.22) = −0.007008, f(1.23) = 0.056068. (39)

We now know that 1.22x̄ < 1.23. And the procedure can be continued until we have estimated x̄ to a

desired degree of accuracy.

The above method is rather inefficient: Each subdivision will increase the accuracy by possibly

one digit but we may have to search over many subintervals in order to locate the one over which there

is a sign change. A more efficient method is obtained if we perform a binary search over subintervals.

This is the basis of the bisection method: If we start with an interval [a, b] that is known to contain

a zero x̄ of a function f , we then split the interval in two and then check each subinterval. We then

take the one at which there is a sign change and repeat the procedure, splitting it in two, etc.. This

iterative algorithm is outlined below:

Bisection method for estimating zeros of a continuous function:

We start with an interval [a1, b1] for which f(a1)f(b1) < 0, implying that x̄ ∈ (a1, c1). This

algorithm will produce a set of subintervals [an, bn] in the following way: At each step n,

1. Compute the midpoint of the interval [an, bn], i.e., c = 12(an+bn). Check if f(c) = 0 to

the desired accuracy, i.e., to m digits of accuracy. If so, then you can stop. You may

also wish to stop if bn −an < ǫ, where ǫ represents some desired accuracy. Otherwise:

2. (a) If f(an)f(c) < 0, set an+1 = an and bn+1 = c .

(b) If f(c)f(bn) < 0, set an+1 = c and bn+1 = bn .

Then replace n with n + 1 and go to Step No. 1.

81

We illustrate the bisection method with the function f(x) = x2 − 2 as a way of estimating√

2,

one of its zeros. We’ll start with the interval [a1, b1] = [1, 2], noting that f(1) = −1 and f(2) = 2.

The results of the first 20 iterations are listed in the table below.

n an bn

1 1.0000000 2.0000000

2 1.0000000 1.5000000

3 1.2500000 1.5000000

4 1.3750000 1.5000000

5 1.3750000 1.4375000

6 1.4062500 1.4375000

7 1.4062500 1.4218750

8 1.4140625 1.4218750

9 1.4140625 1.4179688

10 1.4140625 1.4160156

11 1.4140625 1.4150391

12 1.4140625 1.4145508

13 1.4140625 1.4143066

14 1.4141846 1.4143066

15 1.4141846 1.4142456

16 1.4141846 1.4142151

17 1.4141998 1.4142151

18 1.4142075 1.4142151

19 1.4142113 1.4142151

20 1.4142132 1.4142151

21 1.4142132 1.4142141

22 1.4142132 1.4142137

23 1.4142134 1.4142137

Results of bisection method applied to f(x) = x2 − 2 to estimate the zero√

2. The actual value of√

2, to 7 digits, is 1.4142135.

82

Lecture 12

Limits at infinity

Note: This section was accidentally omitted from the Friday, October 5 lecture,

but will be included here. It will be covered at the beginning of Lecture 13 on

Wednesday, October 10

(Relevant section from Stewart, Seventh Edition: Section 2.6)

We are concerned with the behaviour of a function f(x) in two particular cases: (i) x → ∞ and

(ii) x → −∞. For the moment, we shall consider the particular case x → ∞.

It is certainly possible that a function f behaves in the following way:

“ limx→∞

f(x) = L”. (40)

In other words, as x > 0 gets larger and larger, the values f(x) gets closer and closer to L. This

implies that the line y = L is a horizontal asymptote of the graph of f . A couple of possibilities

are sketched below.

L

x

y

y = f(x)

L

x

y

y = f(x)

limx→∞

f(x) = L

A slightly more mathematical description of this limiting behaviour is that “f(x) approaches the

limit L arbitrarily closely for x sufficiently large.” An ǫδ-type definition is as follows:

Given any ǫ > 0, there exists an M > 0 such that

|f(x) − L| < ǫ for all x such that x > M. (41)

83

Note that our δ is replaced by M , with the idea that M → ∞. Recall that the “closeness” of a

number x to infinity is measured by how large it is. This idea is illustrated in the figure below.

M

y

x

LL − ǫ

L + ǫ

Example: The function f(x) =1

x. As x → ∞, it seems clear that

1

xis getting smaller and smaller.

In other words, we conjecture that

limx→∞

1

x= 0. (42)

Let us prove this with the above ǫ definition. With an eye to (41), given an ǫ > 0, can we find an

M > 0 such that

|f(x) − L| =

∣

∣

∣

∣

1

x− 0

∣

∣

∣

∣

< ǫ for all x such that x > M ? (43)

Keeping in mind that x > 0, we have that

1

x< ǫ implies that x >

1

ǫ. (44)

Therefore, we can choose M =1

ǫ. (As ǫ → 0, M → ∞.) We have therefore proved Eq. (42).

In the same way, we can show that

limx→∞

1

xr= 0 for any r > 0. (45)

Question: For a given r, what is the largest M -value that corresponds to an ǫ > 0?

Rational functions

Consider the rational function,

f(x) =5x2 + 2

2x2 + 4x + 3. (46)

84

We ask the question, “How does f(x) behave as x → ∞?”

Stepping back for a moment, we simply examine the numerator and denominator:

1. Numerator: For x very large, e.g., x = 1020, the term 5x2 will dominate, i.e., the term “2”

will be insignificant.

2. Denominator: For x very large, e.g., x = 1020, the term 2x2 will dominate, i.e., 1040 vs. 1020

and 3.

From these observations, we suspect that

f(x) ≈ 5x2

2x2=

5

2for x very large. (47)

In other words, we conjecture that

limx→∞

f(x) =5

2. (48)

This is fine, but can we make this argument more mathematically rigorous, in terms of limit laws?

The answer is “Yes.” Let’s divide the numerator and denominator of f(x) by x2,

f(x) =5x2 + 2

2x2 + 4x + 3=

5 + 3x2

2 + 4x + 3

x2

. (49)

Then we can write that

limx→∞

f(x) = limx→∞

5x2 + 2

2x2 + 4x + 3

= limx→∞

5 + 3x2

2 + 4x + 3

x2

=limx→∞(5 + 3

x2 )

limx→∞(2 + 4x + 3

x2 )(by limit law for quotients – both limits exist)

=5

2. (50)

Infinite limits at infinity

It is certainly possible that functions become arbitrarily large in magnitude as |x| gets very large. A

simple example is f(x) = x. Recalling that we can consider the value ∞ or −∞ as a limit, we can

write that

limx→∞

x = ∞. (51)

As well, we can write that

limx→∞

x2 = ∞. (52)

85

Can we define the statement,

limx→∞

f(x) = ∞, (53)

more precisely, in an “ǫδ way? In this statement, f(x) is getting closer to ∞ as x gets closer to ∞.

Recalling that “closeness” of “something” to ∞ is defined by how large “something” is. We apply this

idea to both f(x) and x. As a result, we state that Eq. (53) means the following:

Given any N > 0, there exists an M > 0 such that

f(x) > N for all x > M. (54)

In the particular case, f(x) = x, we have that M = N .

So what about the function f(x) = x2 − x? What is

limx→∞

(x2 − x) ? (55)

Can we do the following operation, invoking “limit laws”?

limx→∞

(x2 − x) = limx→∞

x2 − limx→∞

x = ∞−∞. (56)

The question is: What is “∞−∞”?

The answer is that “∞−∞” is meaningless. The above procedure is incorrect. The use of limit laws

is incorrect when we obtain meaningless expressions such as the above, or “0 · ∞”. What we can do

is the following:

limx→∞

(x2 − x) = limx→∞

x(x − 1) = ∞. (57)

If x is large, then x − 1 is also large, implying that the product x(x − 1) is large.

Here is another example, taken from Stewart’s textbook. Does the following limit exist:

limx→∞

(

√

x2 + 1 − x)

? (58)

First of all, we cannot write that

limx→∞

(

√

x2 + 1 − x)

= limxto∞

√

x2 + 1 − limx→∞

x = ∞−∞, (59)

for reasons mentioned earlier: “∞−∞” is meaningless.

86

If we do the following,

√

x2 + 1 − x = (√

x2 + 1 − x)

√x2 + 1 + x√x2 + 1 + x

=(x2 + 1) − x2

√x2 + 1 + x

=1√

x2 + 1 + x. (60)

As x → ∞, it looks like the final expression goes to zero. We’ll divide numerator and denominator by

x:1√

x2 + 1 + x=

1x

√

1 + 1x2 + 1

. (61)

We can now write that

limx→∞

(√

x2 + 1 − x) = limx→∞

1√x2 + 1 + x

= limx→∞

1x

√

1 + 1x2 + 1

=limx→∞

1x

limx→∞(√

1 + 1x2 + 1)

(limit law for quotients)

=0

2

= 0. (62)

Finally, the case x → −∞ can be treated in a similar way – but we must be careful about the

signs of terms. For example x is negative, x2 is positive, x3 is negative, etc.. Please read Section 2.6

of Stewart’s textbook for more details.

87

Derivatives and rates of change

Note: Because much of this material is review, it was delivered by means of the

data projector, which explains the larger-than-normal number of pages covered.

(Relevant section from Stewart, Seventh Edition: Section 2.7)

We have arrived at another important concept in this course – the derivative, with which you are

already familiar. Only a few minutes were available at the end of this lecture period, so the discussion

was quite brief. We shall return to some of these ideas in more detail in the next lecture.

The following discussion will refer to the figure below.

∆x

f(a)

∆y

f(x)

a xx

tangent line at x = a

y = f(x)

P

Q

In this discussion, we consider the point a as a reference point. For a given x 6= a, consider the

points P at (a, f(a)) and Q at (x, f(x)). The slope of the line segment PQ is given by

mPQ =f(x) − f(a)

x − a=

∆y

∆x. (63)

The slope of the tangent line to the curve y = f(x) at x = a is given by the following limit (provided

that it exists),

m = limQ→P

mPQ = lim∆x→0

∆y

∆x= lim

x→a

f(x) − f(a)

x − a. (64)

Quite often in practice, we let x − a represent the displacement from a and define

h = x − a, hence x = a + h, (65)

so that Eq. (64) becomes

m = limh→0

f(a + h) − f(a)

h, (66)

where it is understood that h can be positive or negative (but not zero).

88

As you know, m represents the instantaneous rate of change of f with respect to x and is

called the derivative of f at a and denoted as

f ′(a) = limh→0

f(a + h) − f(a)

h. (67)

The important quantity,f(a + h) − f(a)

h, (68)

is known as the Newton quotient of f at a. (It is also called the difference quotient.)

An important application to Physics: In fact, it was this problem that spawned the development

of the Calculus. The following is simply a translation of the previous discussion into one involving

physical quantities.

Suppose that a particle is moving in one dimension, e.g., along the x-axis, and that its position

as a function of time is denoted by the function x(t), as plotted below.

slope = x′(t1)

t1 t2t

tangent line at t = t1

y = x(t)

x(t1)

x(t2)

y

∆x

∆t

Now consider a time interval [t1, t2] for t1 < t2. The quotient,

x(t2) − x(t1)

t2 − t1=

∆x

∆t, (69)

defines the average velocity of the particle over the time interval [t1, t2]. We now let t2 approach

t1, in an effort to get a better estimate of the rate of change of position with respect to time of the

particle at t = t1. In the limit, provided that it exists, we have

limt2→t1

x(t2) − x(t1)

t2 − t1= lim

∆t→0

∆x

∆t. (70)

Once again, we may let t2 = t1 +h. The instantaneous rate of change of the position with respect

to time of the particle at t1 is given by

x′(t1) = limh→0

x(t1 + h) − x(t1)

h, (71)

89

the derivative of x(t) with respect to t at t1. As you well know, this is the velocity of the particle at

time t1, denoted as “v(t1).”

Historically, the idea of letting the time t2 approach t1 in an effort to describe the motion of

the particle over smaller and smaller time represented a monumental jump in thought which led to

the development of Calculus. Some more comments on this procedure will be given in this week’s

tutorials. We now return to “do some mathematics.”

The derivative as a function

Given a function f(x), we now define its derivative at a general point x (as opposed to an “anchor

point” a):

f ′(x) = limh→0

f(x + h) − f(x)

h, (72)

provided that the limit exists. The derivative is now a function of x, i.e., we could write,

g(x) = f ′(x), (73)

to emphasize that it is a new function.

Important note: In response to an excellent question in class, it is worthwhile to mention that for

f ′(x) to exist, the function f must be defined at x, i.e., f(x) must exist! Otherwise, the Newton

quotient of f at x, i.e.,f(x + h) − f(x)

h

would not be defined, since f(x) is not defined.

Let us now apply the Newton quotient method to some simple functions.

Example 1: The “world’s simplest function” f(x) = C, where C is a constant. (You know the answer

to this, but let’s do it anyway.) The Newton quotient of f at an arbitrary x is

f(x + h) − f(x)

h=

1 − 1

h= 0. (74)

90

As a result,

f ′(x) = limh→0

f(x + h) − f(x)

h= lim

h→00 = 0. (75)

Example 2: The “world’s next-to-simplest function” f(x) = x. (Yes, once again, you know the

answer.) The Newton quotient of f at an arbitrary x is

f(x + h) − f(x)

h=

(x + h) − x

h=

h

h= 1. (76)

As a result,

f ′(x) = limh→0

f(x + h) − f(x)

h= lim

h→01 = 1. (77)

Example 3: The “world’s next-to-next-to-simplest function” f(x) = x2. The Newton quotient of f

at an arbitrary x is

f(x + h) − f(x)

h=

(x + h)2 − x2

h=

2xh + h2

h= 2x + h. (78)

As a result,

f ′(x) = limh→0

f(x + h) − f(x)

h

= limh→0

(2x + h)

= limh→0

2x + limh→0

h

= 2x. (79)

Example 4: The general function f(x) = xn for n = 1, 2, · · · . The Newton quotient of f at an

arbitrary x is

f(x + h) − f(x)

h=

(x + h)n − xn

h

=xn + nxn−1h + · · · + nxhn−1 + hn − xn

h

= nxn−1 + A1h + A2h2 + · · ·An−1h

n−1 (where the Ak are constants). (80)

91

As a result,

f ′(x) = limh→0

f(x + h) − f(x)

h

= limh→0

(nxn−1 + A1h + A2h2 + · · ·An−1h

n−1)

= nxn−1. (81)

Example 5: The absolute value function f(x) = |x|. We must go back to its fundamental definition,

|x| =

x, x ≥ 0

−x, x < 0.(82)

Because of the special nature of the point x = 0, we’ll have to consider three cases:

1. Case 1: x > 0. In this case, the Newton quotient of f is (for sufficiently small h, i.e., |h| < x),

f(x + h) − f(x)

h=

(x + h) − x

h=

h

h= 1. (83)

As a result,

f ′(x) = limh→0

f(x + h) − f(x)

h= lim

h→01 = 1, (84)

as expected. (We could have simply used our result for f(x) = x from before.)

2. Case 2: x < 0. In this case, the Newton quotient of f is (for sufficiently small h, i.e., |h| < |x|),

f(x + h) − f(x)

h=

−(x + h) + x

h=

−h

h= −1. (85)

As a result,

f ′(x) = limh→0

f(x + h) − f(x)

h= lim

h→0−1 = −1. (86)

3. Case 3: x = 0. In this case, the Newton quotient of f at 0 is

f(h) − f(0)

h=

|h|h

. (87)

We have examined this function before, but let’s go through the details again. We have to

consider two cases, i.e., the right- and left-sided limits as h → 0:

(a) h > 0: limh→0+

|h|h

= limh→0+

h

h= lim

h→0+1 = 1.

(b) h < 0: limh→0−

|h|h

= limh→0−

−h

h= lim

h→0−−1 = −1.

92

Since the two limits are not equal, it follows that f ′(0) does not exist.

In summary, the derivative of the absolute value function is given by

f ′(x) =

1, x > 0

−1, x < 0

undefined, x = 0.

(88)

We may represent this result graphically as follows,

y = f ′(x)

x

y

Example 6: The Heaviside function H(x) examined earlier:

H(x) =

1, x ≥ 0

0, x < 0.(89)

Once again, we’ll have to consider three cases:

1. x > 0: Here H(x) = 1. We can simply state the result that H ′(x) = 0 from Example 1,

where C = 1. (Or we could have used the Newton quotient – it is always zero, implying that

H ′(x) = 0.)

2. x < 0: Here H ′(x) = 0. Again, from Example 1, it follows that H ′(x) = 0.

3. x = 0: Before going on to examine the Newton quotient, let’s step back and consider if the

following procedure is valid: We have found that H ′(x) = 0 for x > 0 and for x < 0. Can we

conclude that

H ′(0) = limx→0−

H ′(x) = limx→0+

H ′(x) = 0 ? (90)

93

The answer is NO!!!!! Just because the two one-sided limits exist and are equal to each other,

implying the existence of the limit, we may not conclude that H ′(0) = 0. That would imply

that the function H ′(x) is continuous at x = 0, and we have no guarantee that this is true.

To determine whether or not H ′(0) exists, we must examine the Newton quotient at x = 0,

i.e.,H(0 + h) − H(0)

h=

H(h) − 1

h. (91)

Case 1: (right-sided limit) For h > 0,

limh→0+

H(0 + h) − H(0)

h= lim

h→0+

H(h) − 1

h= lim

h→0+

0

h= 0. (92)

Case 2: (left-sided limit) For h < 0,

limh→0−

H(0 + h) − H(0)

h= lim

h→0−

H(h) − 1

h= lim

h→0−

−1

h= ∞ (since h < 0). (93)

Before we go on, let’s try to understand these results graphically: For h > 0, the slope of the

line segment PQ is always zero, implying that the right-sided limit of the Newton quotient is

zero. But for h < 0, the slope of the line segment PQ is1

|h| which goes to ∞ as h → 0.

Q

x

y = H(x)y

P

Q

hh

Because the two one-sided limits are not equal, we conclude that H ′(0) does not exist. In

summary, we have

H ′(x) =

0, x 6= 0

undefined, x = 0.(94)

A plot of H ′(x) is given below.

Example 7: At this point it is instructive to examine the function f(x) = x|x| for x ∈ R. Because

of the presence of the absolute value function, we may suspect that the point x = 0 is a “bad point.”

94

y = H ′(x)

y

x

From the definition of the absolute value function, we have

f(x) =

x2, x > 0

−x2, x < 0

0, x = 0.

(95)

The graph of f(x) is sketched below.

x

y y = f ′(x)

For x 6= 0, we may use the piecewise-defined formulas of f(x) to compute f ′(x): f ′(x) = 2x when

x > 0 and f ′(x) = −2x when x < 0. It remains to see whether f ′(0) exists. Once again, we must

examine the Newton quotient at x = 0,

f(0 + h) − f(0)

h=

f(h)

h. (96)

Case 1: (right-sided limit) For h > 0,

limh→0+

f(0 + h) − f(0)

h= lim

h→0+

f(h)

h= lim

h→0+

h2

h= lim

h→0+h = 0. (97)

Case 2: (left-sided limit) For h < 0,

limh→0−

f(0 + h) − f(0)

h= lim

h→0−

f(h)

h= lim

h→0−

−h2

h= lim

h→0−(−h) = 0. (98)

95

Since both limits are equal to zero, it follows that f ′(0) = 0. We may therefore summarize our results

as follows,

f ′(x) =

2x, x > 0

−2x, x < 0

0, x = 0.

(99)

The graph of f ′(x) is sketched below.

y y = f ′(x)

x1-1 0

4

2

You’ll notice that the graph of f ′(x) looks something like that of the absolute value function. In

fact, from the piecewise definition of f ′(x), it is not difficult to see that it may be rewritten in terms

of the absolute value function as follows,

f ′(x) = 2|x|. (100)

The moral of this story is that just because an absolute value term appears in a function, it doesn’t

necessarily mean that the function is nondifferentiable. The reader may wish to investigate the more

general family of functions f(x) = xn|x| for n = 1, 2, · · · . As n increases, the function f(x) will have

even higher order derivatives, e.g., f ′′(x).

Differentiability implies continuity

We have now arrived at an important concept involving differentiability of a function.

Theorem: If f is differentiable at a, then it is continuous at a.

Let’s rewrite this theorem slightly, emphasizing the difference between “hypothesis” and “conclusion”:

96

Theorem: If

Hypothesis: f is differentiable, then

Conclusion: f is continuous at a.

Proof: From the hypothesis, i.e., that f is differentiable at a, we have that

f ′(a) = limx→a

f(x) − f(a)

x − a. (101)

We want to show that

limx→a

f(x) = f(a). (102)

This implies that f(x) approaches f(a), so we’ll examine the term f(x) − f(a). We’ll rewrite it as

follows,

f(x) − f(a) =f(x) − f(a)

x − a· (x − a) for x 6= a. (103)

We now take limits, using the Product Law:

limx→a

[f(x) − f(a)] = limx→a

[

f(x) − f(a)

x − a· (x − a)

]

= limx→a

[

f(x) − f(a)

x − a

]

limx→a

(x − a) (since both limits exist)

= f ′(a) · 0

= 0. (104)

We’re essentially done. It follows that

limx→a

f(x) = limx→a

[f(a) + f(x) − f(a)]

= limx→a

f(a) + limx→a

[f(x) − f(a)]

= f(a) + 0

= f(a). (105)

This implies that f is continuous at a and the theorem is proved.

Recall that the contrapositive to the statement A ⇒ B is

not B ⇒ not A. (106)

The contrapositive to the above theorem is not C ⇒ not H, i.e.,

97

If f is not continuous at a, then f is not differentiable at a.

This explains the fact that the Heaviside function is not differentiable at x = 0 – it’s not even

continuous there.

What about the “reverse” of the theorem: If f is continuous at a, is it differentiable at a? The

answer is NO – the absolute value function f(x) = |x| illustrates that this statement is not true.

Differentiability of a function is a stronger requirement than continuity

The fact that differentiability implies continuity, but not the other way around, indicates that differen-

tiability is a stronger requirement. A differentiable function is generally “smoother” than a continuous

function. Just compare the graphs of f(x) = x2 and g(x) = |x| at the point x = 0: f is differentiable

there but g is not.

Some differentiation rules

(Relevant section from Stewart, Seventh Edition: Sections 3.1-3.3)

In the previous lecture, we derived the well known “Power rule” of differentiation:

d

dxxn = nxn−1, n = 0, 1, 2, · · · . (107)

In a little while, we’ll show that this rule extends to all real values of n.

From the limit definition of the derivative (i.e., via Newton quotient) and the “Limit Laws”, the

following familiar rules for differentiation may be derived (for more details, see Stewart, Section 3.1):

Constant multiple rule:d

dx[cf(x)] = cf ′(x) for any constant c ∈ R

Sum/difference rule:d

dx[f(x) ± g(x)] = f ′(x) ± g′(x)

There are two other well-known differentiation rules which require a little more effort to prove:

98

Product rule:d

dx[f(x)g(x)] = f ′(x)g(x) + f(x)g′(x)

Quotient rule:d

dx

[

f(x)

g(x)

]

=f ′(x)g(x) − f(x)g′(x)

g(x)2

Here we shall prove the Product Rule - you can find a proof of the Sum Rule in Stewart’s textbook.

Proof of Product Rule: In what follows, we assume that f and g are differentiable at x. First

define

p(x) = f(x)g(x). (108)

We’ll have to consider the numerator of the Newton quotient of p(x), namely,

p(x + h) − p(x) = f(x + h)g(x + h) − f(x)g(x). (109)

It’s convenient to define the following,

∆f = f(x + h) − f(x), ∆g = g(x + h) − g(x), (110)

so that

f(x + h) = f(x) + ∆f, g(x + h) = g(x) + ∆g. (111)

From Eq. (109),

p(x + h) − p(x) = f(x + h)g(x + h) − f(x)g(x)

= [f(x) + ∆f ][g(x) + ∆g] − f(x)g(x)

= f(x)g(x) + f(x)∆g + g(x)∆f + ∆f∆g − f(x)g(x)

= f(x)∆g + g(x)∆f + ∆f∆g. (112)

The Newton quotient for p then becomes

p(x + h) − p(x)

h= f(x)

[

g(x + h) − g(x)

h

]

+ g(x)

[

f(x + h) − f(x)

h

]

+

[

f(x + h) − f(x)

h

]

[g(x + h) − g(x)]. (113)

We now take the limit of both sides as h → 0. Since f(x) and g(x) are constants, we may write

limh→0

p(x + h) − p(x)

h= f(x) lim

h→0

[

g(x + h) − g(x)

h

]

+ g(x) limh→0

[

f(x + h) − f(x)

h

]

+

limh→0

[

f(x + h) − f(x)

h

]

limh→0

[g(x + h) − g(x)]. (114)

99

In the last term, the limit of the product is the product of the limits, since the limits exist. These

limits become

p′(x) = f(x)g′(x) + g(x)f ′(x) + f ′(x) · 0. (115)

The zero in the last term follows from the fact that g is continuous at x, which follows from the

assumption that g is differentiable at x. Eq. (115) is identical to the product rule, i.e.,

[f(x)g(x)]′ = f ′(x)g(x) + f(x)g′(x). (116)

Exponential functions

We now wish to find the derivatives of the exponential functions, defined as follows,

fa(x) = ax, a > 0. (117)

We must consider the Newton quotient definition for the derivative of fa(x),

f ′

a(x) = limh→0

fa(x + h) − fa(x)

h

= limh→0

ax+h − ax

h

= ax limh→0

ah − 1

h. (118)

If we consider the special value x = 0 in the above equation, then a0 = 1 and we have

f ′

a(0) = limh→0

ah − 1

h. (119)

As a result, we can rewrite Eq. (118) as follows,

f ′

a(x) = f ′

a(0)ax, (120)

ord

dxax = f ′

a(0)ax. (121)

In other words, the derivative of the function ax is proportional to ax. The coefficient of proportionality,

f ′

a(0) is given by Eq. (119). A couple of values of this proportionality constant are given in the table

below.

100

a limh→0

ah − 1

h

2 0.693147

3 1.098612

It is a fact that this factor f ′

a(0) increases with a. From a look at the table, there is a number a

between 2 and 3 such that the proportionality constant is 1. This special value of a is called “e”.

Definition: e is the number such that

f ′

e(0) = limh→0

eh − 1

h= 1. (122)

Consequently, from Eq. (121),d

dxex = ex. (123)

In other words, ex is its own derivative. The slope of the graph of ex is 1 at x = 0. As x increases,

the slope of the graph increases.

101