sin x − x
Find lim
.
x→0
x3
1
Worksheet 7,
Math 10560
sin x − x
is indeterminate of type 00 . By L’Hopital’s rule it equals
x→0
x3
Solution: lim
lim
x→0
(cos x) − 1
.
3x2
This is still an indeterminate form of type 00 , so we apply L’Hopital’s rule again. Our limit is now
equal to
− sin x
lim
.
x→0
6x
Applying L’Hopitals rule again or using the fact that limx→0 sinx x = 1, we get that the above limit
.
equals −1
6
2
x2
.
x→∞ ex
Find lim
(a) 0
(b) e2
(c) ∞
(d)
2
e
(e) Does no exist
3
Find lim
t→0
arcsin(t)
.
t
(a) 1
(b) 0
(c) +∞
(d)
π
6
(e) Does not exist and is not +∞
4
Example Evaluate the limit
lim x2 ln x.
x→0+
Solution: We have that
lim+ x2 ln x = lim+
x→0
x→0
x−1
1
ln x
=
lim
= lim+ − x2 = 0
−2
−3
+
x→0 −2x
x→0
x
2
where we used l’Hopital’s rule for the second equality. We can do that since both ln x and x−2 go
to ±∞ when x → 0+ .
Page 2
5
Evaluate the limit
lim (e2x + 1)1/x .
x→∞
Solution: We have that
(e2x + 1)1/x = exp
and
ln(e2x + 1)
x
2e2x
ln(e2x + 1)
2
2x
lim
= lim e +1 = lim
=2
x→∞
x→∞
x→∞ 1 + e−2x
x
1
where we used l’Hopital’s rule for the second equality and we divided both numerator and denominator by e2x at the third equality. Since exp is continuous we conclude that
lim (e2x + 1)1/x = exp(2) = e2 .
x→∞
6
Evaluate
“
lim (2x)
1
x−1
”
x→1
Solution: This is not an indeterminate form.
“
lim (2x)
”
1
x−1
= lim e
x→1
ln(2x)
x−1
x→1
limx→1
=e
ln(2x)
x−1
Since limx→1 ln(2x) = ln(2) which is finite and limx→1 ln(x − 1) = 0, we have
lim+
ln(2x) x−1
x→1
and
lim−
ln(2x) x−1
x→1
Hence
“
lim (2x)
x→1
1
x−1
= +∞
= −∞
”
does not exist
Page 3
7
Evaluate the following limit.
(ln x)2
.
x→∞
x
lim
Please show your work.
Solution:
Use l’Hospital’s Rule twice:
(ln x)2
=
x→∞
x
lim
=
=
=
=
8
2(ln x)(1/x)
[l’Hospital for “∞/∞”]
x→∞
1
lim
2 ln x
x→∞
x
lim
2(1/x)
x→∞
1
2
lim
x→∞ x
lim
[l’Hospital for “∞/∞”]
0.
Compute the limit
lim
x→2
1
x x−2
2
.
1
Solution: Solution: We have an indeterminate form 1∞ . Let L = limx→2 x2 x−2 . Then
1
x x−2
ln x2
1
1
ln L = lim ln
= lim
(l’Hospital’s rule) = lim = .
x→2
x→2 x − 2
x→2 x
2
2
√
1
Therefore L = e 2 = e.
Page 4
9
Evaluate the limit
lim cos x
1/x2
x→0
.
1/x2
Solution: Solution: Solution: The limit has indeterminate form 1∞ . Let L = limx→0 cos(x)
.
1/x2 ln cos(x)
ln L = lim ln cos(x)
= lim
(Indet. form 00 . Use L’Hospital’s Rule)
2
x→0
x→0
x
1
(−
sin(x))
− tan(x)
cos(x)
= lim
= lim
(Indet. form 00 . Use L’Hospital’s Rule)
x→0
x→0
2x
2x
1
-sec2 (x)
=− .
= lim
x→0
2
2
1
Therefore L = e− 2 .
Page 5