Solutions to Worksheet for Section 2.5

Limits Involving Infinity

Math 1a

February 4, 2008

1. Sketch the graph of a function f that satisfies all of these:

• limx→2

f(x) = −∞

• limx→∞

f(x) =∞

• limx→−∞

f(x) = 0

• limx→0+

f(x) =∞

• limx→0−

f(x) = −∞

Solution. Here is one:

Find the limits.

2. limx→1

2− x(x− 1)2

Solution. As x → 1, the numerator tends to 1, while the denominator tends to zero whileremaining positive. So the quotient consists of increasingly large positive numbers, hencetends to ∞.

3. limx→π−

cotx

Solution. Remember that cotx iscosxsinx

. As x → π but x < π, then cosx → −1 whilesinx → 0, but remains positive. So the quotients are large and negative, hence tend to−∞.

4. limx→∞

x3 + 5x2x3 − x2 + 4

Solution.

limx→∞

x3 + 5x2x3 − x2 + 4

= limx→∞

��x3 (1 + 5/x)��x3(2− 1/x + 4/x3)

=12

5. limt→−∞

t2 + 2t3 + t2 − 1

Solution.

limt→−∞

t2 + 2t3 + t2 − 1

= limt→∞

t2(1 + 2/t2)t3(1 + 1/t− 1/t3

= limt→∞

1t· limt→∞

1 + 2/t2

1 + 1/t− 1/t3

= 0 · 1 = 0.

6. limx→∞

(√9x2 + x− 3x

)Solution. Multiply by the conjugate radical:

limx→∞

(√9x2 + x− 3x

)·√

9x2 + x+ 3x√9x2 + x+ 3x

= limx→∞

x√9x2 + x+ 3x

= limx→∞

1√9 + 1/x + 3

=1√

9 + 3=

16

7. limx→∞

(x−√x)

Solution. Same manipulation:

limx→∞

(x−√x)· x+

√x

x+√x

= limx→∞

x2 − xx+√x

= limx→∞

x2(1− x−1)x(1 + x−1/2)

= limx→∞

x · limx→∞

1− x−1

1 + x−1/2=∞ · 1 =∞

8. limx→∞

sin2 x

x2

Solution. We can use a version of the squeeze theorem. Notice that

0 ≤ sin2 x

x2≤ 1x2

for all x. Since1x2→ 0 as x→∞, lim

x→∞

sin2 x

x2= 0.

9. Consider the function

f(x) =√

2x2 + 13x− 5

Sketch the graph (without using your calculator) by finding all its asymptotes and filling it in.

Solution. We can show:

limx→5/3+

f(x) = +∞ limx→∞

f(x) =√

2/3

limx→5/3−

f(x) = −∞ limx→∞

f(x) = −√

2/3

So we can sketch in at lest this much of the graph:

It turns out the graph is a little bit more complicated; it makes a “hump” on the negative branchand as x→∞ it approaches −

√2/3 from above. But we can see at least this much with the limits

alone.