Chapter 4 - Section 4.4 - Indeterminate Forms and l''Hospital''s Rule - 4.4 Exercises - Page 317: 66

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$$\eqalign{ & \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 - \cos x} \right)^{\sin x}} \cr & {\text{Evaluating the limit}} \cr & \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 - \cos \left( 0 \right)} \right)^{\sin 0}} = {\left( {1 - 1} \right)^0} = {0^0} \cr & {\text{This limit has the form }}{0^0},{\text{ then}} \cr & {\text{Let }}y = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 - \cos x} \right)^{\sin x}} \cr & {\text{Take natural log on both sides of the equation}} \cr & \ln y = \ln \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 - \cos x} \right)^{\sin x}} \cr & {\text{Using the continuity}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\ln {{\left( {1 - \cos x} \right)}^{\sin x}}} \right) \cr & {\text{Use the power property of logarithms}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\sin x\ln \left( {1 - \cos x} \right)} \right) \cr & {\text{Rewrite}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{\ln \left( {1 - \cos x} \right)}}{{\frac{1}{{\sin x}}}}} \right) \cr & {\text{Evaluating}} \cr & \ln y = \frac{\infty }{\infty } \cr & {\text{Using the L'Hopital's rule}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{{\frac{d}{{dx}}\left[ {\ln \left( {1 - \cos x} \right)} \right]}}{{\frac{d}{{dx}}\left[ {\frac{1}{{\sin x}}} \right]}}} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\frac{{\sin x}}{{1 - \cos x}}}}{{\frac{{\cos x}}{{{{\sin }^2}x}}}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{{{\sin }^3}x}}{{\cos x\left( {1 - \cos x} \right)}} = \frac{0}{0} \cr & {\text{Using the L'Hopital's rule}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{3{{\sin }^2}x\cos x}}{{\cos x\left( {\sin x} \right) + \left( {1 - \cos x} \right)\sin x}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{3\sin x\cos x}}{{\cos x + \left( {1 - \cos x} \right)}} \cr & {\text{Evaluating the limit}} \cr & \ln y = \frac{{3\left( 0 \right)}}{{1 - 0}} \cr & \ln y = 0 \cr & y = 1 \cr & {\text{Therefore}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 - \cos x} \right)^{\sin x}} \cr & \mathop {\lim }\limits_{x \to {0^ + }} {\left( {1 - \cos x} \right)^{\sin x}} = 1 \cr} $$