Question

If $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{x^2-1}{x+1}& x\ne -1\\ -2& x=-1\end{array}\end{array}$ so verify whether the function $f\left(x\right)$ is continuous at $x=-1$.

Answer 1

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

$=\underset{x\to -1^{-}}{lim}\frac{(x-1)(x+1)}{(x+1)}$

$=\underset{x\to -1^{-}}{lim}(x-1)$

$=(-1-1)$

$=-2$

and $\underset{x\to -1^{+}}{lim}f\left(x\right)=\underset{x\to -1^{+}}{lim}\frac{x^2-1}{x+1}$

$=\underset{x\to -1^{+}}{lim}\frac{(x-1)(x+1)}{(x+1)}$

$=\underset{x\to -1^{+}}{lim}(x-1)$

$=-2$

At $x=-1,f\left(x\right)=-2$

$\Rightarrow f(-1)=-2$

$\therefore \underset{x\to -1^{-}}{lim}f\left(x\right)=\underset{x\to -1^{+}}{lim}f\left(x\right)=f(-1)$

Hence, $f\left(x\right)$ is continuous at $x=-1$.