Chapter 9 - Power Series - 9.4 Working with Taylor Series - 9.4 Exercises - Page 702: 20

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\[\begin{align} & \underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{\ln x} \\ & \text{Evaluating the limit directly} \\ & \underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{\ln x}=\frac{1-1}{\ln 1}=\frac{0}{0} \\ & \\ & \text{The limit has the indeterminate form }\frac{0}{0},\text{ then we can apply} \\ & \text{the L }\!\!'\!\!\text{ Hopital }\!\!'\!\!\text{ s Rule} \\ & \text{The Maclaurin series for }\ln x\text{ is: }\left( \text{go to page 694}\text{, table 9}\text{.5} \right) \\ & \ln x=\left( x-1 \right)-\frac{{{\left( x-1 \right)}^{2}}}{2}+\frac{{{\left( x-1 \right)}^{3}}}{3}-\frac{{{\left( x-1 \right)}^{4}}}{4}+\cdots \\ & \\ & \text{Substituting into the given limit} \\ & \underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{\ln x} \\ & \underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{\ln x}=\underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{\left( x-1 \right)-\frac{{{\left( x-1 \right)}^{2}}}{2}+\frac{{{\left( x-1 \right)}^{3}}}{3}-\frac{{{\left( x-1 \right)}^{4}}}{4}+\cdots } \\ & \text{Factoring the denominator} \\ & \underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{\ln x}=\underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{\left( x-1 \right)\left[ 1-\frac{\left( x-1 \right)}{2}+\frac{{{\left( x-1 \right)}^{2}}}{3}-\frac{{{\left( x-1 \right)}^{3}}}{4}+\cdots \right]} \\ & \text{Simplifying} \\ & \underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{\ln x}=\underset{x\to 1}{\mathop{\lim }}\,\frac{1}{1-\frac{\left( x-1 \right)}{2}+\frac{{{\left( x-1 \right)}^{2}}}{3}-\frac{{{\left( x-1 \right)}^{3}}}{4}+\cdots } \\ & \text{Evaluating the limit} \\ & \underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{\ln x}=\frac{1}{1-\frac{\left( 1-1 \right)}{2}+\frac{{{\left( 1-1 \right)}^{2}}}{3}-\frac{{{\left( 1-1 \right)}^{3}}}{4}+\cdots } \\ & \underset{x\to 1}{\mathop{\lim }}\,\frac{x-1}{\ln x}=1 \\ \end{align}\]