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x0 <  < x
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f x f 0  f 0 x lim 0 lim 0 L g x x!x0 g  x!x0 g x
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since as x ! x0 ,  ! x0 . Modi cation of the above procedure can be used to establish the result if x ! x0 , x ! x0 , x ! 1, x ! 1. (b) We suppose that f x and g x are di erentiable in a < x < b, and lim f x 1, lim g x 1 x!x0 x!x0 where a < x0 < b. Assume x1 is such that a < x0 < x < x1 < b. By Cauchy s generalized mean value theorem, f x f x1 f 0  g x g x1 g 0  Hence f x f x1 f x 1 f x1 =f x f 0  g x g x1 g x 1 g x1 =g x g 0  x <  < x1
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CHAP. 4]
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from which we see that f x f 0  1 g x1 =g x g x g 0  1 f x1 =f x Let us now suppose that lim f 0 x L and write (1) as x!x0 g 0 x  0     f x f  1 g x1 =g x 1 g x1 =g x L 0  L g x g 1 f x1 =f x 1 f x1 =f x 1
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We can choose x1 so close to x0 that j f 0  =g 0  Lj < . Keeping x1 xed, we see that   1 g x1 =g x lim 1 since lim f x 1 and lim g x 1 x!x0 1 f x1 =f x x!x0 x!x0 Then taking the limit as x ! x0 on both sides of (2), we see that, as required,
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f x f 0 x L lim 0 x!x0 g x g x
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Appropriate modi cations of the above procedure establish the result if x ! x0 , x ! x0 , x ! 1, x ! 1.
4.27. Evaluate
(a) lim
e2x 1 x!0 x
b lim
x!1 x2
1 cos x 2x 1
All of these have the indeterminate form 0/0. a e2x 1 2e2x lim 2 x!0 x!0 1 x lim 1 cos x  sin x 2 cos x 2 lim lim x!1 x2 2x 1 x!1 2x 2 x!1 2 2 lim Note: Here L Hospital s rule is applied twice, since the rst application again yields the indeterminate form 0/0 and the conditions for L Hospital s rule are satis ed once more.
4.28. Evaluate (a) lim
3x2 x 5 5x2 6x 3
b lim x2 e x
All of these have or can be arranged to have the indeterminate form 1=1. a 3x2 x 5 6x 1 6 3 lim lim x!1 5x2 6x 3 x!1 10x 6 x!1 10 5 lim lim x2 e x lim x2 lim 2x lim 2 0
x!1 ex
x!1 ex
x!1 ex
4.29. Evaluate lim x2 ln x.
x!0
x!0
lim x2 ln x lim
ln x
x!0 1=x2
lim
x!0 2=x3
lim
x!0
x2 0 2
The given limit has the indeterminate form 0 1. In the second step the form is altered so as to give the indeterminate form 1=1 and L Hospital s rule is then applied.
4.30. Find lim cos x 1=x .
Since lim cos x 1 and lim 1=x2 1, the limit takes the indeterminate form 11 .
x!0 x!0
Let F x cos x 1=x . have lim
DERIVATIVES
[CHAP. 4
Then ln F x ln cos x =x2 to which L Hospital s rule can be applied.
ln cos x sin x = cos x sin x cos x 1 lim lim lim x!0 x!0 2x cos x x!0 2x sin x 2 cos x 2x 2 x2
Thus, lim ln F x 1. But since the logarithm is a continuous function, lim ln F x ln lim F x . Then 2
x!0 x!0
ln lim F x 1 2
lim F x lim cos x 1=x e 1=2
4.31. If F x e3x 5x 1=x , nd
(a) lim F x and
(b) lim F x .
The respective indeterminate forms in (a) and (b) are 10 and 11 . ln e3x 5x . Then lim G x and lim G x assume the indeterminate forms 1=1 Let G x ln F x x!1 x!0 x and 0/0 respectively, and L Hospital s rule applies. We have a ln e3x 5x 3e3x 5 9e3x 27e3x lim 3x 3 lim 3x lim x!1 x!1 e 5x x!1 9e3x x!0 3e 5 x lim Then, as in Problem 4.30, lim e3x 5x 1=x e3 .
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