Page 92 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 4] DERIVATIVES 83

from which we see that
0
f ðxÞ f ð Þ 1 gðx 1 Þ=gðxÞ
¼ ð1Þ
0
gðxÞ g ð Þ 1 f ðx 1 Þ=f ðxÞ
0
Let us now suppose that lim f ðxÞ ¼ L and write (1)as
0
x!x 0 þ g ðxÞ

0
f ðxÞ f ð Þ 1 gðx 1 Þ=gðxÞ 1 gðx 1 Þ=gðxÞ
L þ L
¼ ð2Þ
0
gðxÞ g ð Þ 1 f ðx 1 Þ=f ðxÞ 1 f ðx 1 Þ=f ðxÞ
We can choose x 1 so close to x 0 that j f ð Þ=g ð Þ Lj < . Keeping x 1 fixed, we see that
0
0

lim 1 gðx 1 Þ=gðxÞ ¼ 1 since lim f ðxÞ¼ 1 and lim gðxÞ¼ 1
x!x 0
x!x 0 þ
x!x 0 þ 1 f ðx 1 Þ=f ðxÞ
Then taking the limit as x ! x 0 þ on both sides of (2), we see that, as required,
0
lim f ðxÞ ¼ L ¼ lim f ðxÞ
0
x!x 0 þ gðxÞ x!x 0 þ g ðxÞ
Appropriate modifications of the above procedure establish the result if x ! x 0 , x ! x 0 ,
x !1, x ! 1.
e 2x 1 1 þ cos x
4.27. Evaluate (a) lim ðbÞ lim
2
x!0 x x!1 x 2x þ 1
All of these have the ‘‘indeterminate form’’ 0/0.
e 2x 1 2e 2x
lim ¼ lim ¼ 2
ðaÞ
x!0 x x!0 1
2
1 þ cos x sin x cos x 2
lim ¼ lim ¼ lim
ðbÞ 2 ¼
x!1 x 2x þ 1 x!1 2x 2 x!1 2 2
Note: Here L’Hospital’s rule is applied twice, since the first application again yields the ‘‘indeter-
minate form’’ 0/0 and the conditions for L’Hospital’s rule are satisfied once more.
2
3x x þ 5
2 x
4.28. Evaluate (a) lim ðbÞ lim x e
2
x!1 5x þ 6x 3 x!1
All of these have or can be arranged to have the ‘‘indeterminate form’’ 1=1.
2
3x x þ 5 6x 1 6 3
lim ¼ lim ¼ lim
x!1 5x þ 6x 3 x!1 10x þ 6 x!1 10 5
ðaÞ 2 ¼
x 2 2x 2
2 x
lim x e ¼ lim ¼ lim ¼ lim ¼ 0
x!1 e x!1 e x!1 e
ðbÞ x x x
x!1
2
4.29. Evaluate lim x ln x.
x!0þ
ln x 1=x x 2
2
lim x ln x ¼ lim ¼ lim ¼ lim ¼ 0
x!0þ 1=x 2 x!0þ 2=x 3 x!0þ 2
x!0þ
The given limit has the ‘‘indeterminate form’’ 0 1.Inthe second step the form is altered so as to give
the indeterminate form 1=1 and L’Hospital’s rule is then applied.
1=x 2
4.30. Find limðcos xÞ .
x!0
2
Since lim cos x ¼ 1 and lim 1=x ¼1,the limit takes the ‘‘indeterminate form’’ 1 .
1
x!0 x!0