Calculus- Early Transcendentals, 2021a

94 Limits
The easiest way to prove this is to use the L’Hôpital’s Rule, which we will introduce in a later chapter.
For now, one can plot and compare the graphs of an exponential function and a power function. Here is a
comparison between f (x)=x 2 and g(x)=2 x :
50
40
30
20
10
x
−6 −4 −2 2 4 6
Notice also that as x →−∞, x n grows in size but e x does not. More specifically, x n → ∞ or −∞
according as n is even or odd, while e x → 0. So, it is meaningless to compare their “growth” rates,
although we can still calculate the limit
e x
lim
x→−∞ x n = 0.
Let’s see an application of our theorem.
Example 3.32
Find the horizontal asymptote(s) of f (x)= x3 + 2e x
e x − 4x 2 .
Solution. To find horizontal asymptotes, we calculate the limits of f (x) as x → ∞ and x →−∞. Forx → ∞,
we divide the numerator and the denominator by e x , and then we take limit to get
x 3 + 2e x
lim
x→∞ e x − 4x 2 = lim x→∞
x 3
e x + 2
1 − 4(0) = 2.
1 − 4 x2
e x = 0 + 2
For x →−∞, we divide the numerator and the denominator by x 2 to get
x 3 + 2e x
lim
x→−∞ e x − 4x 2 = lim
x + 2 ex
x 2
x→−∞ e x
− 4 .
x 2
The denominator now approaches 0 − 4 = −4. The numerator has limit −∞. So, the quotient has limit ∞:
lim
x→−∞
x + 2 ex
x 2
e x
x 2 − 4 = ∞.