Calculus- Early Transcendentals, 2021a

3.5. Infinite Limits and Limits at Infinity 93
Solution. Since m > n, m − n is a positive integer. Therefore,
f (x)
lim
x→∞ g(x) = lim x m
x→∞ x n = lim
x→∞ xm−n = ∞.
♣
Example 3.29
Recall that a polynomial is monic if its leading coefficient is 1. Show that if m > n are two positive
integers, then any monic polynomial P m (x) of degree m grows faster than any monic polynomial
P n (x) of degree n as x → ∞.
Solution. By assumption, P m (x)=x m + terms of degrees less than m = x m + a m−1 x m−1 + ...,andP n (x)=
x n + terms of degrees less than n = x n + b n−1 x n−1 + .... Dividing the numerator and denominator by x n ,
we get
f (x)
lim
x→∞ g(x) = lim x m−n + a m−1 x m−n−1 + ...
x→∞ 1 + b = lim x m−n ( 1 + a m−1
x
+ ...
n−1
x
+ ... x→∞ 1 + b n−1
x
+ ... )=∞,
since the limit of the bracketed fraction is 1 and the limit of x m−n is ∞, as we showed in Example 3.28.
♣
Example 3.30
Show that a polynomial grows exactly as fast as its highest degree term as x → ∞ or −∞. Thatis,if
P(x) is any polynomial and Q(x) is its highest degree term, then both limits
are finite and nonzero.
P(x)
lim
x→∞ Q(x) and
lim
x→−∞
P(x)
Q(x)
Solution. Suppose that P(x) =a n x n + a n−1 x n−1 + ...+ a 1 x + a 0 ,wherea n ≠ 0. Then the highest degree
term is Q(x)=a n x n .So,
P(x)
(
lim
x→∞ Q(x) = lim a n + a n−1
+ ... + a 1
x→∞ x x n−1 + a )
0
x n = a n ≠ 0.
♣
Let’s state a theorem we mentioned when we discussed the last example in the last subsection:
Theorem 3.31
Let n be any positive integer and let a > 1. Thenf (x)=a x grows faster than g(x)=x n as x → ∞:
In particular,
lim
x→∞
a x
= ∞, lim
xn x→∞
x n
a x = 0.
e x x n
lim = ∞, lim
x→∞ xn x→∞ e x = 0.