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Sullivan AP˙Sullivan˙Chapter01 October 8, 2016 17:4
Sullivan
130 Chapter 1 • Limits and Continuity
Teaching Tip
Ask your students to think through the
limits of rational functions near vertical
asymptotes mentally. For Example 3(a),
the thought process would be: We would
like to substitute a number in place of x
that is slightly larger than 3. When we do
so, the numerator will be essentially 7. The
denominator will be a very small positive
number because, for example, 3.00000001
− 3 = 0.00000001. Therefore,
x
lim 2 + 1
7
.
x − 3
≈ A very small positive number
=∞
→ +
x 3
Teaching Tip
Remind students of the graph of y = ln x
when working on Example 3b.
y
The graph of y = ln x crosses the x-axis at
x = 1. Therefore,
and
2
1
21
22
lim lnx
= a small negative number
x→1
−
lim lnx
= a small positive number.
x→ 1
+
2 4 6 8 x
EXAMPLE 3
Investigate: (a)
Investigating an Infinite Limit
2x + 1
lim
x→3 + x − 3
(b)
x 2
lim
x→1 − ln x
2x + 1
Solution (a) lim is the right-hand limit of the quotient of two functions,
x→3 + x − 3
f (x) = 2x + 1 and g(x) = x − 3. Since lim
x→3 +(x
− 3) = 0, we cannot use the Limit of a
Quotient. So we need a different strategy. As x approaches 3 + , we have
lim
x→3
+(2x
+ 1) = 7 and lim
+(x
− 3) = 0
From arithmetic, we know that a fraction p is arbitrarily large if p is a fixed nonzero
q
number and q is arbitrarily close to 0. Here, the limit of the numerator is 7. The
denominator is positive and approaching 0. So the quotient is positive and becoming
unbounded. We conclude that
2x + 1
lim
x→3 + x − 3 =∞
(b) As in (a), we cannot use the Limit of a Quotient because lim ln x = 0. Here we
x→1− have
lim x 2 = 1 lim ln x = 0
x→1 − x→1 −
We are seeking a left-hand limit, so for numbers close to 1 but less than 1, the denominator
ln x < 0. So the quotient is negative and becoming unbounded. We conclude that
x 2
lim
x→1 − ln x = −∞
■
x→3
The discussion below may prove useful when finding lim
x→c −
is a nonzero number and lim g(x) = 0.
x→c− • lim f (x) = L, L > 0, and lim g(x) = 0
x→c −
x→c −
If g(x) 0, and lim g(x) = 0
x→c −
x→c −
If g(x) >0 for numbers x close to c, but less than c, then
f (x)
lim
x→c − g(x) =∞
• lim f (x) = L, L < 0, and lim g(x) = 0
x→c −
x→c −
If g(x) 0 for numbers x close to c, but less than c, then
f (x)
lim
x→c − g(x) = −∞
f (x)
, where lim
g(x) f (x)
x→c −
Similar arguments are valid for right-hand limits.
NOW WORK Problem 27 and AP® Practice Problems 9 and 11.
AP® CaLC skill builder
for example 3
Investigating an Infinite Limit
⎛ −1
⎞
Find lim
⎝
⎜ − ⎠
⎟ .
x→5
− x 5
Solution
Think x ≈ 4.999999999
⎛ −1
⎞
⎝
⎜ − ⎠
⎟ ≈ −1
lim
→ − x 5 4.999999999 − 5
x 5
−1
=
a very small negative number
As the values of x get closer to 5 from the
left, the denominator decreases without
⎛ −1
⎞
bound and lim
⎝
⎜ − ⎠
⎟ =∞ .
x→5
− x 5
Graphical Approach
Solution
Also, consider how this limit may be
explained using the graph. Explain that since
⎛ −1
⎞
fx ( ) =
⎝
⎜ x − 5⎠
⎟ is a rational function, its graph
has asymptotes. Show the graph and observe that
there is a vertical asymptote at x = 5 because as
x approaches 5 from the right or from the left, the
function is infinite.
y
4
2
22
24
2 4 6 8
Also note that as x approaches infinity, the limit
approaches zero.
x
130
Chapter 1 • Limits and Continuity
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