Question

${lim}_{x\to 3}(x^2-9)[\frac{1}{x+3}+\frac{1}{x-3}]$

Answer 1

$\begin{array}{c}Wehave\\ {lim}_{x\to 3}\left(x^2-9\right)\left[\frac{1}{x+3}+\frac{1}{x-3}\right]\\ ={lim}_{x\to 3}\left(x^2-9\right)\left[\frac{\left(x-3\right)+\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right]\\ ={lim}_{x\to 3}\left(x^2-9\right)\left[\frac{2x}{x^2-9}\right]\\ ={lim}_{x\to 3}\left(x^2-9\right)\times \frac{2x}{\left(x^2-9\right)}\\ ={lim}_{x\to 3}2x\\ =2\left(3\right)\\ =6\end{array}$

Hence, this is the answer.