Vantage Math 100/V1C,V1F
Exercises: L’Hôpital’s Rule
1. Use induction to show that for all n ≥ 1,
xn
= 0.
n→∞ ex
lim
2. Evaluate the following limits.
log x
= 0, p > 0
n→∞ xp
5x − tan(5x)
(b) lim
n→0
x3
1
1
(c) lim
−
n→1 log x
x−1
(a) lim
3. (Hard) Evaluate the following limits.
(a) lim+ xx
n→0
x
1
(b) lim 1 +
n→∞
x
(c) lim (1 + sin(2x))cot(3x)
x→0
Note: These will not be tested.
1
Solution.
1. Let’s check the base case, when n = 1. In this case, we have a limit of the form
by L’Hôpital’s rule,
x
1
= lim x = 0
x
x→∞ e
x→∞ e
lim
Thus the result is true for n = 1. Now let’s suppose there is a k ≥ 1 such that
xk
= 0.
x→∞ ex
lim
Now since
xk+1
ex
∞
∞
is of the form
as x goes to infinity, by L’Hôpital’s rule,
xk+1
(k + 1)xk
xk
=
lim
=
(k
+
1)
lim
= 0.
x→∞ ex
x→∞
x→∞ ex
ex
lim
Thus by induction, we have for all n ≥ 1,
xn
= 0.
x→∞ ex
lim
2. (a)
log x
xp
is of the form
∞
∞
as x goes to infinity, so by L’Hôpital’s rule,
1
1
log x
x
= lim
= lim p = 0
lim
x→∞ pxp−1
x→∞ x
x→∞ xp
The last line is true since p > 0.
(b) Our limit is of the form
0
0
as x goes to 0.
5x − tan(5x)
5 − 5sec2 (5x)
=
lim
x→0
x→0
x3
3x2
lim
Our new limit is again of the form 00 , so let’s use L’Hôpital’s Rule again,
−5 · 2sec(5x)sec(5x) tan(5x)5
5 − 5sec2 (5x)
= lim
2
x→0
x→0
3x
6x
2
−25sec (5x)
sin(5x)
= lim
lim
x→0 3 cos(5x) x→0
x
25
sin(5x)
=−
lim 5
3 x→0
5x
125
=−
3
lim
Where in the last line we used the fact that
sin x
=1
x→0 x
lim
2
∞
,
∞
so
(c) We begin by simplifying.
1
x − 1 − log x
1
−
=
log x x − 1
(x − 1) log x
Which is of the form
0
0
as x goes to 1, so let’s use L’Hôpital.
1 − x1
x − 1 − log x
= lim
x→1 (x − 1) log x
x→1 log x + x−1
x
x−1
= lim
x→1 x log x + x − 1
lim
Which is of the form 00 , so let’s use L’Hôpital’s rule again.
lim
x→1
1
x−1
= lim
x→1
x log x + x − 1
log x + x x1 + 1
1
= lim
x→1 log x + 2
1
=
2
3. (a) Again let L = limx→0+ xx , which is of the form 00 so we need to do some work.
Let’s begin by taking logarithms of both sides,
x
log L = log lim+ x
x→0
= lim+ log xx
x→0
= lim+ x log x
x→0
= lim+
log x
1
x
x→0
The second line is true by the continuity of log. So now we have a limit of the
form −∞
, so we can apply L’Hôpital’s rule.
∞
lim
log x
1
x
x→0+
= lim+
x→0
1
x
− x12
= lim+ −x
x→0
=0
So we have log L = 0 and thus L = 1, i.e.,
lim+ xx = 1.
x→0
3
(b) Let L = limx→∞ (1 + 1/x)x , which is of the form 1∞ . Again let’s take log of both
sides, and the continuity of logarithm gives us,
1
log L = lim x log 1 +
x→∞
x
1
log 1 + x
= lim
1
x→∞
x
Now we have a limit of the form 00 , so by L’Hôpitals rule, we have
lim
log 1 +
= lim
(− x12 )
1
1+ x1
− x12
1
= lim
x→∞ 1 + 1
x
1
x
x→∞
1
x
x→∞
=1
So log L = 1, and so L = e, i.e.,
lim
x→∞
1
1+
x
x
= e.
(c) Again let L = limx→0 (1 + sin(2x))cot(5x) , which is of the form 1∞ . Again as in (a),
(b), let’s take logarithms.
log L = lim cot(5x) log(1 + sin(2x))
x→0
log(1 + sin(2x))
x→0
tan(5x)
= lim
Which is now of the form 00 . So by L’Hôpital’s rule,
2 cos(2x)
log(1 + sin(2x))
1+sin(2x)
lim
= lim
x→0
x→0 5sec2 (5x)
tan(5x)
2 cos 0
= 1+sin20
5sec 0
2
=
5
Thus log L =
2
5
2
and therefore L = e 5 , i.e.,
2
lim (1 + sin(2x))cot(5x) = e 5
x→0
4
(wtf. . . )