# How do you find the limit of (ln x^2) / (x^2-1) as x approaches 1?

Jul 1, 2016

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

#### Explanation:

$\frac{{\log}_{e} {x}^{2}}{{x}^{2} - 1} = {\log}_{e} \frac{x}{x - 1} - {\log}_{e} \frac{x}{x + 1}$
but the ${\log}_{e} x$ series expansion in the $x = 1$ vicinity is
${\log}_{e} x = {\sum}_{k = 1}^{\infty} \frac{{\left(x - 1\right)}^{k}}{k}$ so

${\log}_{e} \frac{x}{x - 1} = 1 + {\sum}_{k = 2}^{\infty} \frac{{\left(x - 1\right)}^{k - 1}}{k}$

so

${\lim}_{x \to 1} {\log}_{e} \frac{x}{x - 1} = 1$

Finally

${\lim}_{x \to 1} \frac{{\log}_{e} {x}^{2}}{{x}^{2} - 1} = {\lim}_{x \to 1} {\log}_{e} \frac{x}{x - 1} - {\lim}_{x \to 1} {\log}_{e} \frac{x}{x + 1} = 1 + 0 = 1$