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

Feb 26, 2016

This limit 'creates' the $\infty - \infty$ indeterminate form so the first step should be finding a common denominator.

#### Explanation:

${\lim}_{x \to 1} \left(\frac{1}{\ln} \left(x\right) - \frac{1}{x - 1}\right) = {\lim}_{x \to 1} \frac{x - 1 - \ln \left(x\right)}{\ln \left(x\right) \left(x - 1\right)} = \left[\frac{0}{0}\right]$

And now to get rid of $\frac{0}{0}$ you can use the de L'Hôspital's Rule which states that when evaluating $\frac{0}{0}$ or $\frac{\infty}{\infty}$ indeterminate forms the limit of the quotient stays the same if derivatives of the numerator and denominator (evaluated seperately, not using the formula for the derivative of a quotient) are substituted for the numerator and denominator, respectively (as long as the 'new' limit exists).

So (in this case using the de L'Hôpital's Rule twice):

${\lim}_{x \to 1} \frac{1 - \frac{1}{x}}{\frac{x - 1}{x} + \ln \left(x\right)} = \left[\frac{0}{0}\right] = {\lim}_{x \to 1} \frac{\frac{1}{x} ^ 2}{\frac{1}{x} ^ 2 + \frac{1}{x}} = \left[\frac{1}{1 + 1}\right] = \frac{1}{2}$