Maclaurin’ s series 75
However, a knowledge of series does not help with
examples such as lim
x→1
x 2 + 3x − 4
x 2 − 7x + 6
L’Hopital’s rule will enable us to determine such
limits when the differential coefficients of the numerator
and denominator can be found.
L’Hopital’s rule states:
lim
x→a
f(x)
g(x)
= lim
x→a
f
(x)
g (x)
provided g (a) = 0
It can happen that lim
x→a
f (x)
g (x)
is still
0
0 ; if so, the
numerator 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 16. Determine lim
x→1
x 2 + 3x − 4
x 2 − 7x + 6
The first step is to substitute x = 1 into both numerator and denominator. In this case we obtain
0
0 . It is
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 + 3x − 4
x 2 − 7x + 6
= lim
x→1
2x + 3
2x − 7
i.e. both numerator and
denominator have
been differentiated
=
5
−5
= −1
Problem 17. Determine lim
x→0
sin x − x
x 2
Substituting x = 0 gives
lim
x→0
sin x − x
x 2
=
sin 0 − 0
0
=
0
0
Applying L’Hopital’s rule gives
lim
x→0
sin x − x
x 2
= lim
x→0
cos x − 1
2x
Substituting x = 0 gives
cos 0 − 1
0
=
1 − 1
0
=
0
0
again
Applying L’Hopital’s rule again gives
lim
x→0
cos x − 1
2x
= lim
x→0
−sin x
2
= 0
Problem 18. Determine lim
x→0
x − sin x
x − tan x
Substituting x = 0 gives
lim
x→0
x − sin x
x − tan x
=
0 − sin 0
0 − tan 0
=
0
0
Applying L’Hopital’s rule gives
lim
x→0
x − sin x
x − tan x
= lim
x→0
1 − cos x
1 − sec 2 x
Substituting x = 0 gives
lim
x→0
1 − cos x
1 − sec 2 x
=
1 − cos 0
1 − sec 2 0
=
1 − 1
1 − 1
=
0
0
again
Applying L’Hopital’s rule gives
lim
x→0
1 − cos x
1 − sec 2 x
= lim
x→0
sin x
(−2 sec x)(sec x tan x)
= lim
x→0
sin x
−2 sec 2 x tan x
Substituting x = 0 gives
sin 0
−2 sec 2 0 tan 0
=
0
0
again
Applying L’Hopital’s rule gives
lim
x→0
sin x
−2 sec 2 x tan x
= lim
x→0
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
cos x
(−2 sec
2 x)(sec
2 x)
+ (tan x)(−4 sec
2 x tan x)
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
using the product rule
Substituting x = 0 gives
cos 0
−2 sec 4 0 − 4 sec 2 0 tan 2 0
=
1
−2 − 0
= −
1
2
However, a knowledge of series does not help with
examples such as lim
x→1
x 2 + 3x − 4
x 2 − 7x + 6
L’Hopital’s rule will enable us to determine such
limits when the differential coefficients of the numerator
and denominator can be found.
L’Hopital’s rule states:
lim
x→a
f(x)
g(x)
= lim
x→a
f
(x)
g (x)
provided g (a) = 0
It can happen that lim
x→a
f (x)
g (x)
is still
0
0 ; if so, the
numerator 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 16. Determine lim
x→1
x 2 + 3x − 4
x 2 − 7x + 6
The first step is to substitute x = 1 into both numerator and denominator. In this case we obtain
0
0 . It is
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 + 3x − 4
x 2 − 7x + 6
= lim
x→1
2x + 3
2x − 7
i.e. both numerator and
denominator have
been differentiated
=
5
−5
= −1
Problem 17. Determine lim
x→0
sin x − x
x 2
Substituting x = 0 gives
lim
x→0
sin x − x
x 2
=
sin 0 − 0
0
=
0
0
Applying L’Hopital’s rule gives
lim
x→0
sin x − x
x 2
= lim
x→0
cos x − 1
2x
Substituting x = 0 gives
cos 0 − 1
0
=
1 − 1
0
=
0
0
again
Applying L’Hopital’s rule again gives
lim
x→0
cos x − 1
2x
= lim
x→0
−sin x
2
= 0
Problem 18. Determine lim
x→0
x − sin x
x − tan x
Substituting x = 0 gives
lim
x→0
x − sin x
x − tan x
=
0 − sin 0
0 − tan 0
=
0
0
Applying L’Hopital’s rule gives
lim
x→0
x − sin x
x − tan x
= lim
x→0
1 − cos x
1 − sec 2 x
Substituting x = 0 gives
lim
x→0
1 − cos x
1 − sec 2 x
=
1 − cos 0
1 − sec 2 0
=
1 − 1
1 − 1
=
0
0
again
Applying L’Hopital’s rule gives
lim
x→0
1 − cos x
1 − sec 2 x
= lim
x→0
sin x
(−2 sec x)(sec x tan x)
= lim
x→0
sin x
−2 sec 2 x tan x
Substituting x = 0 gives
sin 0
−2 sec 2 0 tan 0
=
0
0
again
Applying L’Hopital’s rule gives
lim
x→0
sin x
−2 sec 2 x tan x
= lim
x→0
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
cos x
(−2 sec
2 x)(sec
2 x)
+ (tan x)(−4 sec
2 x tan x)
⎫
⎪ ⎪ ⎬
⎪ ⎪ ⎭
using the product rule
Substituting x = 0 gives
cos 0
−2 sec 4 0 − 4 sec 2 0 tan 2 0
=
1
−2 − 0
= −
1
2