MATH 100 V1A
October 29th – Practice problems
Hints and Solutions
1. Do the following functions have any vertical asymptotes? If so, where?
(a) f (x) =
√1 .
x−2
Solution: The function has a vertical asymptote at x = 2. This is because
1
= ∞. We can show this rigorously: let N > 0 be any positive
limx→2+ √x−2
integer and let δ = N12 . Then for any x ∈ (2, 2 + δ),
1
1
√
>q
= N.
x−2
2 + N12 − 2
This means that f (x) is unbounded on the interval (2, 2 + δ) for every δ > 0, and
so limx→2+ f (x) = ∞.
(b) f (θ) = sec2 θ.
Hint: Show that limx→c+ sec2 θ = ∞ for c =
(2n+1)π
2
for every integer n.
2. Can a function have an infinite number of vertical asymptotes? Can a function have
an infinite number of vertical asymptotes in a finite interval [a,b]?
Hint: The answer to both of these questions is yes. We’ve already seen an example of
a function that has an infinite number of vertical asymptotes: tan θ.
For the second question, use a piecewise function to construct a function on [0, 1] that
has a vertical
at x = n1 for every integer n ≥ 1. As a first step, construct a
asymptote
1
1
function on n+1 , n which has a vertical asymptote at x = n1 . Then “glue” all of these
functions together into one piecewise function.
3. Do the following functions have any horizontal asymptotes? If so, where?
(a) f (x) =
3−2x
.
x−5
Solution: The function has one horizontal asymptote at y = −2. To see this, we
calculate the limit as x tends to infinity:
3
x x3 − 2
−2
3 − 2x
x
lim
= lim
=
lim
= −2.
5
x→∞ x − 5
x→∞ x 1 −
x→∞ 1 − 5
x
x
Similarly, limx→−∞ f (x) = −2.
(b) f (x) =
1
.
log(x+2)
Hint: Show that f (x) has a horizontal asymptote at y = 0 by showing that limx→∞ = 0.
Note that we cannot limx→−∞ f (x) does not exist since the domain of the function is
{x ∈ R | x > −2}.
4. Draw the graph of a function which has no vertical asymptotes, has a horizontal asymptote at y = −π, and satisfies f (x) = −π exactly twice. Now come up with an algebraic
expression for such a function.
Solution: The following is the graph of one example of such a function.
y
−3
−2
−1
x
1
2
3
−1
−2
−3
−4
Of course, it is fairly easy to come up with examples that are also continuous.
The function given by the equation
( 2
e−x − π
f (x) =
−π
if x 6= −1, 1
if x = −1, 1
is an example of such a function (similar to the function graphed above).
5. Use L’Hospital’s Rule to calculate the following limits.
(a) limx→∞
x
.
ex
Solution: Since both x and ex go to infinity as x goes to infinity, we can use
L’Hospital’s Rule to evaluate the limit:
x L’H
1
=
lim
= 0.
x→∞ ex
x→∞ ex
lim
2
(b) limx→∞
xn
,
ex
where n is a positive integer.
Solution: Again, since xn and ex both go to infinity (for every positive integer n),
we can use L’Hospital’s Rule to evaluate the limit. In fact, we can and will use it
multiple times:
xn L’H
nx(n−1)
=
lim
x→∞ ex
x→∞
ex
n(n − 1)x(n−2)
L’H
= lim
x→∞
ex
..
.
lim
L’H
n!
=0
x→∞ ex
= lim
3