Question

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

Answer 1

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