14 October 2015
L’Hospital’s Rule: solutions
ex − 1 − x
.
x→0
x2
(1) Use L’Hospital’s rule to evaluate lim
lim (ex − 1 − x) = e0 − 1 − 0 = 0 while lim x2 = 0. So, we have an indeterminate form of 0/0 and
x→0
can use l’Hospital’s rule. We have:
x→0
d x
(e − 1 − x)
ex − 1
e −1−x
dx
=
lim
lim
=
lim
d 2
x→0
x→0
x→0
x2
2x
(x )
dx
Since lim (ex − 1) = 0 and lim 2x = 0, we must apply l’Hospital’s rule again:
x
x→0
x→0
d x
(e − 1)
ex − 1
ex
1
lim
= lim dx
= lim
=
d
x→0
x→0
x→0 2
2x
2
(2x)
dx
x
x3 x2
and
lim
we
found
lim
in
lecture
.
x→∞ ex
x→∞ ex
x→∞ ex
(2) Find lim
In both cases, we have indeterminate forms ∞/∞. When we apply l’Hospital’s rule to x/ex , we
find we must evaluate lim 1/ex = 0. In the case of x3 /ex we find that, after applying l’Hospital’s rule
x→∞
3 times (as in the first question), we find that our limit of interest is just the limit lim 3 · 2/ex = 0.
x→∞
xn
? Explain.
x→∞ ex
Suppose that n is a positive integer. What can you conclude about lim
As we can see from the case of n = 1, 2 or 3 above, we can resolve this limit by applying l’Hospital’s
rule n times. The resulting limit we need to compute is lim n · (n − 1) · . . . 3 · 2/ex = 0.
x→∞
(3) Evaluate lim+ ln(x7 − 1) − ln(x5 − 1) by first putting it into the form
x→1
Recall a log law tells us that ln(x7 − 1) − ln(x5 − 1) = ln
x7 −1
x5 −1
0
0
or
∞
.
∞
.
d 7
(x − 1)
x −1
7x6
7
dx
lim+ 5
= lim+
= lim+ 4 =
d
x→1 x − 1
x→1
x→1 5x
5
(x5 − 1)
dx
Therefore, since ln is continuous, we may conclude that
7
x −1
x7 − 1
7
lim+ ln
= ln lim+ 5
= ln
5
x→1
x→1 x − 1
x −1
5
7
(4) Find the following limits by writing y = f (x) and finding the limit of log f (x).
(a) lim+ (4x + 1)cot x
x→0
Let f (x) = (4x + 1)cot(x) . Then
ln(f (x)) = cot(x) ln(4x + 1)
As x approaches 0 from the right, cot(x) goes to ∞ while ln(4x + 1) goes to 0. So, we rewrite
to obtain a ∞/∞ indefinite form. Now, we may apply l’Hospital’s
cot(x) ln(4x + 1) as ln(4x+1)
tan(x)
rule:
d
(ln(4x + 1))
ln(4x + 1)
4/(4x + 1)
lim+
= lim+ dx
= lim+
=4
d
x→0
x→0
x→0
tan(x)
sec2 (x)
(tan(x))
dx
Hence, lim+ ln(f (x)) = 4, so lim+ f (x) = e4 .
x→0
x→0
x
2
(b) lim 1 +
x→∞
x
x
2
. Then
Let f (x) = 1 +
x
2
ln(f (x)) = x ln 1 +
x
As x goes to ∞, ln 1 + x2 goes to 0. So we rewrite our product to the quotient
obtain a 0/0 indefinite form. Then, applying l’Hospital’s rule, we get
1
(−2/x2 )
ln 1 + x2
2
1+2/x
lim
= lim
= lim
=2
2
x→∞
x→∞
x→∞
1/x
−1/x
1 + 2/x
ln(1+ x2 )
1/x
to
So limx→∞ ln(f (x)) = 2, meaning limx→∞ f (x) = e2 .
ex + e−x
. What happens?
x→∞ ex − e−x
(5) Try to use L’Hospital’s rule to evaluate lim
d x
d x
(e + e−x ) = ex − e−x while
(e − e−x ) = ex + e−x . So attempting to solve the limit this way
dx
dx
leads to a “l’Hospital’s purgatory” of switching the numerator and denominator over and over again
(this is a very real possibility for quite a few quotients you should be on the look out for!)
Now use algebra to evaluate the limit.
We can instead split up our fraction into two pieces:
ex + e−x
ex
e−x
=
+
ex − e−x
ex − e−x ex − e−x
The advantage of this is that the second summand is no longer an indeterminate form: as x goes to
∞, this quotient goes to 0/∞, meaning the limit of this quotient is just 0. Of course, beware that
splitting this into two summands and evaluating the limit is only valid if we can show the limit of
the first summand exists! To see this, note that:
ex
ex
1
1
=
·
=
x
−x
x
−2x
e −e
e 1−e
1 − e−2x
This has limit 1. To conclude, we have
ex + e−x
ex
e−x
=
lim
+
lim
=1+0=1
x→∞ ex − e−x
x→∞ ex − e−x
x→∞ ex − e−x
lim
(6) Suppose that f is a positive function. If lim f (x) = 0 and lim g(x) = ∞, show that
x→a
x→a
g(x)
lim [f (x)]
x→a
= 0.
This shows that 0∞ is not an indeterminate form. (Hint: ab = eb ln a )
Following the hint, we have that f (x)g(x) = eg(x) ln(f (x)) . lim ln(f (x)) = ln(lim f (x)) = −∞. So
x→a
lim g(x) ln(f (x)) = −∞. Thus,
x→a
lim f (x)g(x) = lim eg(x) ln(f (x)) = 0
x→a
x→a
x→a