# How do you find the limit of x/(x-1) - 1/(ln(x)) as x approaches 1?

Apr 28, 2016

Gather terms into a single ratio and apply L'Hopital's rule twice to find

${\lim}_{x \to 1} \left(\frac{x}{x - 1} - \frac{1}{\ln} \left(x\right)\right) = \frac{1}{2}$

#### Explanation:

${\lim}_{x \to 1} \left(\frac{x}{x - 1} - \frac{1}{\ln} \left(x\right)\right) = {\lim}_{x \to 1} \left(1 + \frac{1}{x - 1} - \frac{1}{\ln} \left(x\right)\right)$

$= {\lim}_{x \to 1} \left(1 + \frac{\ln \left(x\right) - x + 1}{\left(x - 1\right) \ln \left(x\right)}\right)$

$= 1 + {\lim}_{x \to 1} \frac{\ln \left(x\right) - x + 1}{\left(x - 1\right) \ln \left(x\right)}$

As the above limit is a $\frac{0}{0}$ indeterminate form, we may apply L'Hopital's rule.

$= 1 + \lim \left(x \to 1\right) \frac{\frac{d}{\mathrm{dx}} \left(\ln \left(x\right) - x + 1\right)}{\frac{d}{\mathrm{dx}} \left(x - 1\right) \ln \left(x\right)}$

$= 1 + {\lim}_{x \to 1} \frac{\frac{1}{x} - 1}{1 + \ln \left(x\right) - \frac{1}{x}}$

This is another $\frac{0}{0}$ indeterminate form, so we apply L'Hopital's rule again.

$= 1 + {\lim}_{x \to 1} \frac{\frac{d}{\mathrm{dx}} \left(\frac{1}{x} - 1\right)}{\frac{d}{\mathrm{dx}} \left(1 + \ln \left(x\right) - \frac{1}{x}\right)}$

$= 1 + {\lim}_{x \to 1} \frac{- \frac{1}{x} ^ 2}{\frac{1}{x} + \frac{1}{x} ^ 2}$

$= 1 + \frac{- 1}{1 + 1}$

$= \frac{1}{2}$