# How do you evaluate ( ( x / (x-1 ) - (1 / lnx ) ) as x approaches 1+?

Nov 16, 2017

$\frac{1}{2}$

#### Explanation:

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

Plugging in 1 gives the indeterminate form $\frac{0}{0}$

Using L'Hospital's Rule:

Differentiate numerator and denominator.

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

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

We still have the form $\frac{0}{0}$, so differentiate again.

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

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

$\therefore$

$\frac{\frac{1}{x}}{\frac{1}{x} + \frac{1}{x} ^ 2} = \frac{\frac{x}{x}}{\frac{x}{x} + \frac{x}{x} ^ 2} = \frac{1}{1 + \frac{1}{x}}$

Plugging in 1:

$\frac{1}{1 + \frac{1}{1}} = \frac{1}{1 + 1} = \frac{1}{2}$

$\therefore$

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