Using L Hospital Rule Find: $\mathop {\lim }\limits_{x \to \infty } (\frac{1}{{x - 1}} - \frac{1}{{\ln x}})$

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

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

Explanation:

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

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

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

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

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

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

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

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

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

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

$=\frac{1}{2}$

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