106 4.4 L’Hôpital’s Rule ■ Example 4.4.3 If the derivatives of the functions f and g themselves satisfy the assumptions of Theorem 4.4.1 we may apply L’Hôpital’s rule to determine frst the limit of f ′ (x)/g ′ (x) and then apply the rule again to determine the original limit. Consider the limit 2x lim . x→0 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, x ≠ 0. Also, f ′ (0)= 0 = g ′ (0) and g ′′ (x)= cosx ≠ 0 for x near 0. Moreover, ′′ (x) f 2 lim = lim = 2. x→0 g ′′ (x) x→0 cosx By L’Hôpital’s rule we get ′′ (x) f ′ (x) f 2 lim = lim = lim = 2. x→0 g ′ (x) x→0 g ′′ (x) x→0 cosx Applying L’Hôpital’s rule one more time we get 2x f (x) f ′ (x) lim = lim = lim = 2. x→0 1 − cosx x→0 g(x) x→0 g ′ (x) ■ Example 4.4.4 Let g(x)= x + 3x 2 and let f : R → R be given by x2 sin 1 , if x ≠ 0; f (x)= x 0, if x = 0. Now consider the limit f (x) x 2 sin 1 x lim = lim x→0 g(x) x→0 x + 3x2 . Using the derivative rules at x ̸ = 0 and the defnition of derivative at x = 0 we can see that f is differentiable and 1 1 2xsin − cos , if x ≠ 0; f ′ (x)= x x 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. x2 sin 1 x On the other hand limx→0 does exist as we can see from x + 3x2 x2 sin 1 xsin 1 limx→0(xsin 1 ) x x x lim = lim = = 0. x→0 x + 3x2 x →0 1 + 3x limx →0(1+ 3x)
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