Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises - Page 580: 78

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$$\eqalign{ & \mathop {\lim }\limits_{x \to {1^ + }} {\left( {x - 1} \right)^{\ln x}} \cr & {\text{Evaluating the limit}} \cr & \mathop {\lim }\limits_{x \to {1^ + }} {\left( {x - 1} \right)^{\ln x}} = {\left( {1 - 1} \right)^{\ln 1}} = {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 {1^ + }} {\left( {x - 1} \right)^{\ln x}} \cr & {\text{Take the natural log of each side}} \cr & \ln y = \ln \left[ {\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {x - 1} \right)}^{\ln x}}} \right] \cr & {\text{Continuity}} \cr & \ln y = \left[ {\mathop {\lim }\limits_{x \to {1^ + }} \ln {{\left( {x - 1} \right)}^{\ln x}}} \right] \cr & {\text{Power property of logarithms}} \cr & \ln y = \mathop {\lim }\limits_{x \to {1^ + }} \ln x\ln \left( {x - 1} \right) \cr & \ln y = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\ln \left( {x - 1} \right)}}{{\frac{1}{{\ln x}}}} \cr & {\text{Evaluating when }}x \to {1^ + } \cr & \ln y = \frac{{\ln \infty }}{{\frac{1}{0}}} = \frac{\infty }{\infty } \cr & {\text{Using L'Hopital's rule}} \cr & \ln y = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {\frac{{\frac{d}{{dx}}\left[ {\ln \left( {x - 1} \right)} \right]}}{{\frac{d}{{dx}}\left[ {\frac{1}{{\ln x}}} \right]}}} \right] \cr & \ln y = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {\frac{{\frac{1}{{x - 1}}}}{{ - \frac{{1/x}}{{{{\left( {\ln x} \right)}^2}}}}}} \right] \cr & \ln y = - \mathop {\lim }\limits_{x \to {1^ + }} \left[ {\frac{{x{{\left( {\ln x} \right)}^2}}}{{x - 1}}} \right] \cr & {\text{Evaluating the limit}} \cr & \ln y = - \mathop {\lim }\limits_{x \to {1^ + }} \left[ {\frac{{x{{\left( {\ln x} \right)}^2}}}{{x - 1}}} \right] = \frac{0}{0} \cr & {\text{Using L'Hopital's rule}} \cr & \ln y = - \mathop {\lim }\limits_{x \to {1^ + }} \left[ {\frac{{2x\left( {\ln x} \right)\left( {1/x} \right) + {{\left( {\ln x} \right)}^2}}}{1}} \right] \cr & \ln y = - 2\left( 1 \right)\left( {\ln 1} \right)\left( {1/1} \right) + {\left( {\ln 1} \right)^2} \cr & \ln y = 0 \cr & y = 1 \cr & {\text{Therefore}} \cr & y = \mathop {\lim }\limits_{x \to {1^ + }} {\left( {x - 1} \right)^{\ln x}} \cr & \mathop {\lim }\limits_{x \to {1^ + }} {\left( {x - 1} \right)^{\ln x}} = 1 \cr} $$