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

Dec 4, 2016

Combine, using a common denominator, then use L'Hôpital's rule , twice.

#### Explanation:

lim_(xto1)[1/ln(x) - 1/(x - 1)] = ?

Make a common denominator:

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

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

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

The above evaluates at the limit to an indeterminate form, $\frac{0}{0}$. Therefore, the use of L'Hôpital's rule is warranted:

Compute the first derivative of the numerator:

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

Compute the first derivative of the denominator:

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

Make a new fraction out of the new numerator and new denominator:

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

Multiply by $\frac{x}{x}$

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

It is still the indeterminate form $\frac{0}{0}$, therefore, we apply the rule, again:

Numerator:

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

Denominator:

$\frac{d \left(x - 1 + x \ln \left(x\right)\right)}{\mathrm{dx}} = 1 + \ln \left(x\right) + \frac{x}{x} = 2 + \ln \left(x\right)$

Here is the new expression:

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

The above can be evaluated at the limit:

$\frac{1}{2 + \ln \left(1\right)} = \frac{1}{2}$

Therefore, the original expression has the same limit:

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