MATH 100 V1A
October 31st – Practice problems
Hints and Solutions
1. Find the horizontal asymptotes of the following functions, if they have any:
(a) f (x) =
sin( x1 )
cos( x1 )−1
.
Solution: To calculate the horizontal asymptotes, we want to calculate limx→∞ f (x)
and limx→−∞ . Since x1 goes to 0 as x goes to ±∞ and trigonometric functions are
continuous, the numerator will go to sin(0) = 0 and the denominator will go to
d
1 − cos(0)
= 0 as x goes to infinity. Since − sin(x) = dx
(cos(x) − 1) is nonzero on
π π
− 2 , 2 except at 0, we can use L’Hospital’s Rule to evaluate the limits. So,
sin x1
cos x1
cos x1 · −1
2
L’H
x
= −∞.
lim
= lim
= lim
x→∞ cos 1 − 1
x→∞ − sin 1
x→∞ − sin 1 · −1
2
x
x
x
x
Similarly, limx→−∞ f (x) = ∞, and so we see that the function does not have any
horizontal asymptotes.
(b) f (x) = x2 ex
Hint: First show that limx→∞ f (x) = ∞ (you do not need to use L’Hospital’s
2
Rule for this). Then, for limx→−∞ f (x), rewrite f as f (x) = ex−x so that both
the numerator and denominator go to infinity as x goes to negative infinity. Use
L’Hospital’s Rule to show that the function has a horizontal asymptote at y = 0.
(c) f (x) =
x log(x2 )
ex
Hint: Use L’Hospital’s Rule to show that both limx→∞ f (x) = 0 and limx→−∞ f (x) =
0.
(d)
(
f (x) =
1+
0
1 x
x
if x > 0
.
otherwise
Hint: First show that limx→−∞ f (x) = 0. Next, use L’Hospital’s Rule to show that
lim x → ∞x log 1 + x1 = 1. Use this calculation to show that the function also has a
horizontal asymptote at y = e.
2. Find the following limits:
(a) limx→0
cos(x)−cos(2x)
.
ex −x−1
Solution: Using L’Hospital’s Rule (twice!) we get
cos(x) − cos(2x) L’H
− sin(x) + 2 sin(2x)
= lim
x
x→0
x→0
e −x−1
ex − 1
− cos(x) + 4 cos(2x)
L’H
= lim
= 3.
x→0
ex
lim
(b) limx→0
1
x
− csc(x) .
1
Solution: First we rewrite x1 − csc x = x1 − sin(x)
= sin(x)−x
. Since both sin(x) − x
x sin(x)
and x sin(x) go to zero as x goes to zero, we can use L’Hospital’s Rule to compute
the limit.
1
sin(x) − x L’H
cos(x) − 1
lim
− csc(x) = lim
= lim
x→0
x→0 x sin(x)
x→0 sin(x) + x cos(x)
x
− sin(x)
L’H
=0
= lim
x→0 2 cos(x) − x sin(x)
(c) limx→∞
2x−3 2x+1
.
2x+5
2x−3
2x+5
Solution: First we evaluate limx→∞ (2x + 1) log
= limx→∞
L’Hospital’s Rule.
lim
x→∞
log
2x−3
2x+5
1
2x+1
1
L’H
= lim
x→∞
( 2x−3
2x+5 )
·
2(2x+5)−2(2x−3)
(2x−3)2
−2
(2x+1)2
2
−8(2x + 1)
x→∞ (2x − 3)(2x + 5)
−8x2 (2 + x1 )2
= lim 2
= −8.
x→∞ x (2 − 3 )(2 + 5 )
x
x
= lim
Hence, limx→∞
2x−3 2x+1
2x+5
= e−8 .
2
log( 2x−3
2x+5 )
1
2x+1
using
3. Let f be twice differentiable. What is limh→0
f (x+h)−2f (x)+f (x−h)
?
h2
Hint: Use L’Hospital’s Rule to show that this limit is equal to
f 0 (x + h) − f 0 (x − h)
h→0
2h
lim
(remember what you’re differentiating with respect to!). Then, rewrite this limit as
1
f 0 (x + h) − f 0 (x) f 0 (x − h) − f 0 (x)
lim
+
.
2 h→0
h
−h
Use this to show that the original limit is equal to f 00 (x). Note that you cannot use
L’Hospital’s Rule twice because we are not told that the second derivative is continuous.
4. Explain why L’Hospital’s Rule cannot be used to evaluate limx→0
sin(x)
.
x
d
sin(x) = cos(x) was showing that limx→0 sin(x)
=
Solution: Recall that part of showing that dx
x
1 using a geometric argument. We cannot use L’Hospital’s Rule to evaluate the limit
because we need to know the derivative of sin(x) in order to use L’Hospital’s Rule, but
we found the derivative by evaluating the limit! That is, using L’Hospital’s Rule to
evaluate the limit would be a circular argument; we would use the calculation of the
limit to calculate the limit!
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