Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.7 Exercises - Page 564: 56

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$$\eqalign{ & \mathop {\lim }\limits_{x \to {0^ + }} {\left[ {\cos \left( {\frac{\pi }{2} - x} \right)} \right]^x} \cr & {\text{Use the Cofunction identity cos}}\left( {\frac{\pi }{2} - x} \right) = \sin x \cr & = \mathop {\lim }\limits_{x \to {0^ + }} {\left[ {\sin x} \right]^x} \cr & {\text{Evaluating the limit}} \cr & \mathop {\lim }\limits_{x \to {0^ + }} {\left[ {\sin x} \right]^x} = {\left[ {\sin 0} \right]^0} = {0^0} \cr & {\text{This limit has the form }}{0^0}{\text{ }}\left( {{\text{See example 6, page 562}}} \right) \cr & {\text{Let }}y = \mathop {\lim }\limits_{x \to {0^ + }} {\left[ {\sin x} \right]^x} \cr & {\text{Take the natural log of each side}} \cr & \ln y = \ln \mathop {\lim }\limits_{x \to {0^ + }} {\left[ {\sin x} \right]^x} \cr & {\text{Continuity}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} {\left[ {\ln \sin x} \right]^x} \cr & {\text{Product property of logarithms}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {x\ln \sin x} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{{\ln \sin x}}{{\frac{1}{x}}}} \right] \cr & {\text{Evaluating}} \cr & \ln y = \frac{\infty }{\infty } \cr & {\text{Using L'Hopital's rule}} \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{{\frac{d}{{dx}}\left[ {\ln \sin x} \right]}}{{\frac{d}{{dx}}\left[ {\frac{1}{x}} \right]}}} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{{\frac{{\cos x}}{{\sin x}}}}{{ - \frac{1}{{{x^2}}}}}} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{{\cot x}}{{ - \frac{1}{{{x^2}}}}}} \right] = - \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{{{x^2}}}{{\tan x}}} \right] \cr & {\text{Evaluating the limit}} \cr & \ln y = \frac{0}{0} \cr & {\text{Using L'Hopital's rule}} \cr & \ln y = - \mathop {\lim }\limits_{x \to {0^ + }} \left[ {\frac{{2x}}{{{{\sec }^2}x}}} \right] \cr & \ln y = - \frac{{2\left( 0 \right)}}{{{{\sec }^2}\left( 0 \right)}} = 0 \cr & \ln y = 0 \cr & y = 1 \cr & {\text{Therefore}} \cr & y = \mathop {\lim }\limits_{x \to {0^ + }} {\left[ {\sin x} \right]^x} \cr & \mathop {\lim }\limits_{x \to {0^ + }} {\left[ {\sin x} \right]^x} = 1 \cr & \mathop {\lim }\limits_{x \to {0^ + }} {\left[ {\cos \left( {\frac{\pi }{2} - x} \right)} \right]^x} = 1 \cr} $$