Section 8.1, Indeterminate Forms of the Type 0/0
Homework: 8.1 #1–23 odds
f (x)
where lim f (x) = lim g(x) = 0. In some
x→c
x→c
g(x)
cases, you have been able to find these. For example,
In this section, we will find limits of the form lim
x→c
(x − 4)(x − 3)
x−3
1
x2 − 7x + 12
= lim
= lim
=
2
x→4 (x − 4)(x + 4)
x→4 x + 4
x→4
x − 16
8
lim
What happens when the numerator and denominator are not polynomials?
To handle this case, we can use L’Hôpital’s Rule, which says that if lim f (x) = lim g(x) = 0 and
f 0 (x)
if lim 0
exists (in either the finite or infinite sense), then
x→c g (x)
x→c
x→c
f 0 (x)
f (x)
= lim 0
x→c g (x)
x→c g(x)
lim
There is a proof of this in the book.
Examples
Find each of the following limits. Make sure that you have an indeterminate form of 0/0 before
applying L’Hôpital’s Rule.
x2 − 7x + 12
x→4
x2 − 16
1. lim
x2 − 7x + 12
2x − 7
1
= lim
=
2
x→4
x→4
x − 16
2x
8
lim
sin x
x→0 x
2. lim
lim
x→0
3. lim
x→0
2x + 1 − cos x
tan x
lim
x→0
4. lim
x→0
sin x
cos x
= lim
=1
x→0
x
1
2 + sin x
2x + 1 − cos x
= lim
=2
x→0
tan x
sec2 x
sin x
− 4x
x2
lim
x→0
sin x
cos x
1
= lim
=−
x2 − 4x x→0 2x − 4
4
x2 − x − 6
x→−2 x2 + 4x + 4
5. lim
lim +
x→−2
x2 − x − 6
2x − 1
= lim
= −∞
x2 + 4x + 4 x→−2+ 2x + 4
sinh x + tanh x
x→0
ex − 1
6. lim
sinh x + tanh x
cosh x + sech2 x
=2
=
lim
x→0
x→0
ex − 1
ex
lim
ex − ln(1 + x) − 1
x→0
x2
7. lim
1
ex − 1+x
ex − ln(1 + x) − 1
ex + (1 + x)ex
(1 + x)ex − 1
2
=
lim
= lim
= lim
= =1
2
2
x→0
x→0
x→0
x→0
x
2x
2x + 2x
2 + 4x
2
lim