MACLAURIN’S SERIES 73A
L’Hopital’s rulewill enable us to determine such
limits when the differential coefficients of the numer-
ator and denominator can be found.
L’Hopital’s rule states:
lim
x→a{
f(x)
g(x)}
=lim
x→a{
f′(x)
g′(x)}providedg′(a)= 0It can happen that lim
x→a
{
f′(x)
g′(x)}
is still0
0; if so, thenumerator and denominator are differentiated again
(and again) until a non-zero value is obtained for the
denominator.
The following worked problems demonstrate how
L’Hopital’s rule is used. Refer to Chapter 27 for
methods of differentiation.
Problem 14. Determine lim
x→ 1{
x^2 + 3 x− 4
x^2 − 7 x+ 6}The first step is to substitutex=1 into both numer-
ator and denominator. In this case we obtain^00 .Itis
only when we obtain such a result that we then use
L’Hopital’s rule. Hence applying L’Hopital’s rule,
lim
x→ 1
{
x^2 + 3 x− 4
x^2 − 7 x+ 6}
=lim
x→ 1{
2 x+ 3
2 x− 7}i.e.both numerator and
denominator have
been differentiated=5
− 5=− 1Problem 15. Determine lim
x→ 0{
sinx−x
x^2}Substitutingx=0gives
lim
x→ 0{
sinx−x
x^2}
=sin 0− 0
0=0
0Applying L’Hopital’s rule gives
lim
x→ 0{
sinx−x
x^2}
=lim
x→ 0{
cosx− 1
2 x}Substitutingx=0gives
cos 0− 1
0=1 − 1
0=0
0againApplying L’Hopital’s rule again giveslim
x→ 0{
cosx− 1
2 x}
=lim
x→ 0{
−sinx
2}
= 0Problem 16. Determine lim
x→ 0{
x−sinx
x−tanx}Substitutingx=0giveslim
x→ 0{
x−sinx
x−tanx}
=0 −sin 0
0 −tan 0=0
0Applying L’Hopital’s rule giveslim
x→ 0{
x−sinx
x−tanx}
=lim
x→ 0{
1 −cosx
1 −sec^2 x}Substitutingx=0giveslim
x→ 0{
1 −cosx
1 −sec^2 x}
=1 −cos 0
1 −sec^20=1 − 1
1 − 1=0
0againApplying L’Hopital’s rule giveslim
x→ 0{
1 −cosx
1 −sec^2 x}
=lim
x→ 0{
sinx
(−2 secx)(secxtanx)}=lim
x→ 0{
sinx
−2 sec^2 xtanx}Substitutingx=0givessin 0
−2 sec^2 0 tan 0=0
0againApplying L’Hopital’s rule giveslim
x→ 0{
sinx
−2 sec^2 xtanx}=lim
x→ 0⎧
⎪⎪
⎨⎪⎪
⎩cosx
(−2 sec^2 x)(sec^2 x)
+(tanx)(−4 sec^2 xtanx)⎫
⎪⎪
⎬⎪⎪
⎭using the product ruleSubstitutingx=0givescos 0
−2 sec^40 −4 sec^2 0 tan^20=1
− 2 − 0=−1
2