Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.7 Exercises - Page 564: 55

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