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) andg′′(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 givenby 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(xsin1 x) limx →0(1+3x) =0.
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