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Lecture

# Growth rate

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University of British Columbia

Mathematics

MATH 104

Jacob Levy

Fall

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Math 104 Section 101 Notes for 2013/10/17
November 6, 2013
Di▯erent functions have di▯erent growth rates, which are important for eval-
uating limits at in▯nity. The function x grows faster than x: we have
2
x
lim = lim x = 1
xto1 x x!1
and x
lim = 0
x!1 x2
n m
and similarly, if n > m, then x grows faster than x . The growth rate of a
polynomial comes from the growth rate of the top-degree term, and in general
any sum grows as fast as its fastest-growing term.
Example 1. Let f(x) and g(x) be polynomials, with f(x) = d x +:::+a a0d
e
g(x) = bex + ::: + b0; by convention, we assume the top-degree termsda and
beare nonzero. We write their fraction as
d d▯e d▯e▯1 ▯e
f(x) = adx + ::: + a0 = adx + ad▯1x + ::: + 0 x
g(x) b x + ::: + b b + ::: + b xe
e 0 e 0
e
where we obtain the second equality by dividing each term by x .
If d = e, then we get
▯e
f(x) ad+ ::: + a0x ad
x!▯1m = x!▯1 ▯e =
g(x) be+ ::: + b0x be
since all other terms in the fraction have a negative power of x, which then
becomes 0 as x ! ▯1. Moreover, a =d ie a ▯nite, nonzero number; we say f
and g grow at approximately the same rate. If in contrast e > d, then
d▯e ▯e
f(x) adx + ::: + 0 x 0
x!▯1m g(x) = x!▯1 b + ::: + b xe = b = 0
e 0 e
since all terms in the fraction exceet b have negative powers of x; in this case,
g(x)
lim = ▯1
x!▯1 f(x)
and g(x) grows faster than f(x).
1 Exponential growth is faster than any polynomial growth. This can be shown
using L’Hopital’s rule, but it can also be proven directly, only using elementary
di▯erentiation. While the statements of the following results are important
for the purposes of this class and may be tested, the proofs are included for
completeness only.
Proposition 2. Exponential growth is faster than linear growth. In other words,
x
e x
x!1m x = 1 and x!1m ex = 0
Proof. We will show the second equality using the squeeze theorem. Since x
and e are both positive for x > 0, the function x=e is squeezed from below
by 0; we just need to ▯nd a function that is bigger than x=ewhose limit as
x ! 1 is zero. We will show that e is bigger than a quadratic function, and
then x=(ax ) = 1=ax ! 0 as x ! 1 as long as a 6= 0.
I claim that e > x =2 for all x > 0. To show this, ▯rst note that e = 1 >
2 x 2
0 = 0 =2. We will now show that e ▯ x =2 is an increasing function for all x.
We have (e ▯ x

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