# Solve this

Question:

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

Solution:

(b) not continuous at x = − 2

Given:

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

$\Rightarrow f(x)= \begin{cases}\frac{-(x+2)}{\tan ^{-1}(x+2)}, & x<-2 \\ \frac{(x+2)}{\tan ^{-1}(x+2)}, & x>-2 \\ 2, & x=-2\end{cases}$

Continuity at x = − 2.

$(\mathrm{LHL}$ at $x=-2)=\lim _{x \rightarrow-2^{-}} f(x)=\lim _{h \rightarrow 0} f(-2-h)=\lim _{h \rightarrow 0} \frac{-(-2-h+2)}{\tan ^{-1}(-2-h+2)}=\lim _{h \rightarrow 0} \frac{h}{\tan ^{-1}(-h)}=-1 .$

$(\mathrm{RHL}$ at $x=-2)=\lim _{x \rightarrow-2^{+}} f\left(x=\lim _{h \rightarrow 0} f\left(-2+h=\lim _{h \rightarrow 0} \frac{(-2+h+2)}{\tan ^{-1}(-2+h+2)}=\lim _{h \rightarrow 0} \frac{h}{\tan ^{-1}(h)}=1\right.\right.$

Also $f(-2)=2$

Thus, $\lim _{x \rightarrow-2^{-}} f(x) \neq \lim _{x \rightarrow-2^{+}} f(x) \neq f(-2)$.

Therefore, given function is not continuous at x = − 2