Chapter 12 - Sequences and Series - 12.7 L'Hospital's Rule - 12.7 Exercises - Page 660: 45

$$\frac{1}{2}$$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 1} \left( {\frac{x}{{x - 1}} - \frac{1}{{\ln x}}} \right) \cr & {\text{The common denominator is }}\left( {x - 1} \right)\ln x,{\text{ then}} \cr & \mathop {\lim }\limits_{x \to 1} \left( {\frac{{x\ln x - x + 1}}{{\left( {x - 1} \right)\ln x}}} \right) \cr & {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr & {\text{evaluate the limit substituting }}1{\text{ for }}x{\text{ into the numerator and }} \cr & {\text{denominator}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to 1} \left( {\frac{{x\ln x - x + 1}}{{\left( {x - 1} \right)\ln x}}} \right) = \frac{{1\ln 1 - 1 + 1}}{{\left( {1 - 1} \right)\ln 1}} = \frac{0}{0} \cr & \cr & {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{. then differentiating the numerator and}} \cr & {\text{denominator}}{\text{.}} \cr & {\text{for }}x\ln x - x + 1 \to {D_x}\left( {x\ln x - x + 1} \right) = 1 + \ln x - 1 \cr & {\text{for }}\left( {x - 1} \right)\ln x \to {D_x}\left( {\left( {x - 1} \right)\ln x} \right) = 1 - \frac{1}{x} + \ln x \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 1} \left( {\frac{{1 + \ln x - 1}}{{1 - \frac{1}{x} + \ln x}}} \right) = \mathop {\lim }\limits_{x \to 1} \frac{{1 + \ln 1 - 1}}{{1 - \frac{1}{1} + \ln 1}} = \frac{0}{0} \cr & {\text{Since the limits of both numerator and denominator are 0}}{\text{, }} \cr & {\text{l'Hospital's rule applies}}{\text{.}} \cr & {\text{for }}1 + \ln x - 1 \to {D_x}\left( {1 + \ln x - 1} \right) = 1/x \cr & {\text{for }}1 - \frac{1}{x} + \ln x \to {D_x}\left( {1 - \frac{1}{x} + \ln x} \right) = {x^{ - 2}} + \frac{1}{x} \cr & {\text{then}} \cr & \mathop {\lim }\limits_{x \to 1} \left( {\frac{{1 + \ln x - 1}}{{1 - \frac{1}{x} + \ln x}}} \right) = \mathop {\lim }\limits_{x \to 1} \left( {\frac{{1/x}}{{{x^{ - 2}} + \frac{1}{x}}}} \right) \cr & {\text{Find the limit of the quotient of the derivatives}} \cr & = \frac{1}{{{1^{ - 2}} + \frac{1}{1}}} = \frac{1}{2} \cr & {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr & \mathop {\lim }\limits_{x \to 1} \left( {\frac{x}{{x - 1}} - \frac{1}{{\ln x}}} \right) = \frac{1}{2} \cr} $$