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
(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
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