Section 8.2, Other Indeterminate Forms Homework: 8.2 #1–39 odds

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Section 8.2, Other Indeterminate Forms
Homework: 8.2 #1–39 odds
In the last section, we will find limits with the indeterminate form 0/0. In this section, we will look
at limits of the form ∞/∞, 0 · ∞, ∞ − ∞, 00 , ∞0 , and 1∞ . It turns out that L’Hôpital’s Rule works
for all of these, too! For all of these forms except ∞/∞, we will need to rearrange the function in
the limit first.
f 0 (x)
exists in either the finite or
In the ∞/∞ case, if lim f (x) = lim g(x) = ∞ and if lim 0
x→c g (x)
x→c
x→c
infinite sense, then
f (x)
f 0 (x)
= lim 0
,
x→c g(x)
x→c g (x)
lim
where c may represent any real number, c− , c+ , or ±∞.
Examples
Calculate each of the following limits:
x9
x→∞ ex
This has the form ∞/∞. Applying L’Hôpital’s Rule, we get:
1. lim
9x8
9!
x9
= lim x = · · · = lim x = 0
x
x→∞ e
x→∞ e
x→∞ e
lim
(ln x)2
x→∞
2x
This also has the form ∞/∞, so
2. lim
2 · x1
(ln x)2
2(ln x)
=
lim
=
lim
=0
x→∞
x→∞ x2x ln 2
x→∞ 2x ln 2 + x2x (ln 2)2
2x
lim
3. lim 5x cot x
x→0
This has the form 0 · ∞. If we rearrange it, we can get a limit with either the form 0/0 or
∞/∞, then use L’Hôpital’s rule:
lim 5x cot x = lim
x→0
4.
x→0
tan x − sec x
lim
5x
5
= lim
=5
x→0
tan x
sec2 x
x→π/2
This has the form ∞ − ∞. We can rewrite this as one term using sines and cosines:
lim
x→π/2
5. lim cos x
sin x − 1
cos x
tan x − sec x = lim
= lim
=0
cos x
x→π/2
x→π/2 − sin x
1/x2
x→0
2
This has the form 1∞ . Let y = (cos x)1/x . Then, ln y =
∞ · 0). Taking the limit for ln y, we get:
sin x
− cos
ln cos x
− sin x
x
=
lim
= lim
x→0
x→0
x→0 2x cos x
x2
2x
− cos x
1
= lim
=−
x→0 2 cos x − 2x sin x
2
lim
1
x2
ln cos x (this will give us the form
Since this is the limit of ln y, not y, we need to exponentiate our answer to see that lim y =
x→0
e−1/2 .
6. lim+ xx
x→0
This has the form 00 , so we will use a similar approach to the last example. Let y = xx . Then,
ln y = x ln x, which will give us the form 0 · (−∞). The limit of ln y is:
ln x
x−1
= lim+
= lim+ −x = 0,
−1
x→0 −x−2
x→0
x→0
x→0 x
so lim xx = exp lim ln y = e0 = 1.
lim+ x ln x = lim+
x→0+
x→0+
Reminder: Limits with the forms 10 , 0∞ , ∞∞ , ∞ · ∞, ∞ + ∞, 0/∞ and ∞/0 are not indeterminate.
They can all be determined by methods from Calculus I.
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