Tazmin Horton

2020-11-07

Find the limits
$\underset{x\to 0}{lim}\frac{\frac{1}{x-1}+\frac{1}{x+1}}{x}$

Khribechy

Given that $\underset{x\to 0}{lim}\frac{\frac{1}{x-1}+\frac{1}{x+1}}{x}$
Compute the limit as follows.
$\underset{x\to 0}{lim}\frac{\frac{1}{x-1}+\frac{1}{x+1}}{x}=\underset{x\to 0}{lim}\frac{x+1+x-1}{x\left(x-1\right)\left(x+1\right)}$
$=\underset{x\to 0}{lim}\frac{2x}{x\left({x}^{2}-1\right)}$
$=\underset{x\to 0}{lim}\frac{2}{\left({x}^{2}-1\right)}$
$=\frac{2}{0-1}$
$=-2$
Thus, $\underset{x\to 0}{lim}\frac{\frac{1}{x-1}+\frac{1}{x+1}}{x}=-2$

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