Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{|x+2|}{{\tan }^{-1}(x+2)},& x\ne -2\\ 2,& x=-2\end{array}$, then $f\left(x\right)$ is

• A. Continuous at $x=-2$
• B. Not continuous at $x=-2$
• C. Differentiable at $x=-2$
• D. Continuous but not derivable at $x=-2$

Answer 1

Answer: (B)

Not continuous at $x=-2$

Given, $f\left(x\right)=\{\begin{array}{cc}\frac{|x+2|}{{\tan }^{-1}(x+2)},& x\ne -2\\ 2,& x=-2\end{array}$
$\therefore \underset{x\to -2}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(-2-h)$
$=\underset{h\to 0}{lim}\frac{|-2-h+2|}{(-2-h+2)}$
$=\underset{h\to 0}{lim}\frac{-h}{{\tan }^{-1}h}=-1$
and $\underset{x\to -2^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(-2+h)$
$=\underset{h\to 0}{lim}\left[\frac{|-2+h+2|}{{\tan }^{-1}(-2+h+2)}\right]$
$=\underset{h\to 0}{lim}\frac{h}{{\tan }^{-1}h}=1$
$\because \underset{x\to -2^{-}}{lim}f\left(x\right)\ne \underset{x\to -2^{+}}{lim}f\left(x\right)$
So, $f\left(x\right)$ is not continuous as well as not differentiable at $x=-2$.