Mathematical Methods for Physics and Engineering : A Comprehensive Guide

4.7 EVALUATION OF LIMITS

Therefore we find

lim

x→a

f(x)

g(x)

=

f′(a)

g′(a)

,

providedf′(a)andg′(a) are not themselves both equal to zero. If, however,

f′(a)andg′(a)areboth zero then the same process can be applied to the ratio

f′(x)/g′(x) to yield

lim

x→a

f(x)

g(x)

=

f′′(a)

g′′(a)

,

provided that at least one off′′(a)andg′′(a) is non-zero. If the original limit does

exist then it can be found by repeating the process as many times as is necessary

for the ratio of correspondingnth derivatives not to be of the indeterminate form

0 /0, i.e.

lim

x→a

f(x)

g(x)

=

f(n)(a)

g(n)(a)

.

Evaluate the limit

lim

x→ 0

sinx

x

.

We first note that ifx= 0, both numerator and denominator are zero. Thus we apply
l’Hopital’s rule: differentiating, we obtainˆ

lim

x→ 0

(sinx/x) = lim

x→ 0

(cosx/1) = 1.

So far we have only considered the case wheref(a)=g(a)=0.Forthecase

wheref(a)=g(a)=∞we may still apply l’Hopital’s rule by writingˆ

lim

x→a

f(x)

g(x)

= lim

x→a

1 /g(x)

1 /f(x)

,

whichisnowoftheform0/0atx=a. Note also that l’Hopital’s rule is stillˆ

valid for finding limits asx→∞,i.e.whena=∞. This is easily shown by letting

y=1/xas follows:

lim

x→∞

f(x)

g(x)

= lim

y→ 0

f(1/y)

g(1/y)

= lim

y→ 0

−f′(1/y)/y^2

−g′(1/y)/y^2

= lim

y→ 0

f′(1/y)

g′(1/y)

= lim

x→∞

f′(x)

g′(x)

.