Introduction to Mathematical Analysis I - 3rd Edition

106 4.4 L’Hôpital’s Rule ■ Example 4.4.3 If the derivatives of the functions f andgthemselves satisfy the assumptions of Theorem 4.4.1 we may apply L’Hôpital’s rule to determine first the limit of f ′(x)/g′(x) and then apply the rule again to determine the original limit. Consider the limit lim x→0 x2 1−cosx . Here f (x) =x2 and g(x) =1−cosx so both functions and all its derivatives are continuous. Now g′(x) =sinx and, so, g′(x)̸ =0 for x near zero, x̸ =0. Also, f ′(0) =0=g′(0) and g′′(x) =cosx̸ =0 for x near 0. Moreover, lim x→0 f ′′(x) g′′(x) =lim x→0 2 cosx =2. By L’Hôpital’s rule we get lim x→0 f ′(x) g′(x) =lim x→0 f ′′(x) g′′(x) =lim x→0 2 cosx =2. Applying L’Hôpital’s rule one more time we get lim x→0 x2 1−cosx =lim x→0 f (x) g(x) =lim x→0 f ′(x) g′(x) =2. ■ Example 4.4.4 Let g(x) =x+3x 2 and let f : R→Rbe given by f (x) =  x2sin 1 x , if x̸ =0; 0, if x =0. Now consider the limit lim x→0 f (x) g(x) =lim x→0 x2sin1 x x+3x2 . Using the derivative rules at x̸ =0 and the definition of derivative at x =0 we can see that f is differentiable and f ′(x) =  2xsin 1 x − cos 1 x , if x̸ =0; 0, if x =0, However, f ′ is not continuous at 0 (since limx→0 f ′(x) does not exist) and, hence, L’Hôpital’s rule cannot be applied in this case. On the other hand limx→0 x2sin1 x x+3x2 does exist as we can see from lim x→0 x2sin1 x x+3x2 =lim x→0 xsin1 x 1+3x = limx→0(xsin 1 x) limx→0(1+3x) =0.

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