Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{|x+2|}{ta{n}^{-1}(x+2)}& x\ne -2\\ 2,& x=-2\end{array}$ 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 differentiable at $x=-2$

Answer 1

Answer: (B)

not continuous at $x=-2$

$LHL={lim}_{x\to {-2}^{-}}f\left(x\right)$
$={lim}_{x\to {-2}^{-}}\frac{|x+2|}{ta{n}^{-1}(x+2)}$
$={lim}_{x\to {-2}^{-}}\frac{-(x+2)}{ta{n}^{-1}(x+2)}=-1$
$RHL={lim}_{x\to {-2}^{+}}f\left(x\right)$
$={lim}_{x\to {-2}^{+}}\frac{|x+2|}{ta{n}^{-1}(x+2)}$
$={lim}_{x\to {-2}^{+}}\frac{(x+2)}{ta{n}^{-1}(x+2)}=1$
Since, $RHL\ne LHL$
$f\left(x\right)$ is not continuous at $x=-2$