Question

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

Answer 1

(b) not continuous at x = − 2

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{\left|x+2\right|}{{\tan }^{-1}(x+2)},x\ne -2\\ 2,x=-2\end{array}\right.$
$\Rightarrow f\left(x\right)=\left\{\begin{array}{l}\frac{-(x+2)}{{\tan }^{-1}(x+2)},x<-2\\ \begin{array}{l}\frac{(x+2)}{{\tan }^{-1}(x+2)},x>-2\\ 2,x=-2\end{array}\end{array}\right.$
Continuity at x = − 2.
(LHL at x= − 2) = $\underset{x\to -2^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f(-2-h)=\underset{h\to 0}{\lim }\frac{-(-2-h+2)}{{\tan }^{-1}(-2-h+2)}=\underset{h\to 0}{\lim }\frac{h}{{\tan }^{-1}(-h)}=-1.$

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

Also $f(-2)=2$
Thus, $\underset{x\to -2^{-}}{\lim }f\left(x\right)\ne \underset{x\to -2^{+}}{\lim }f\left(x\right)$≠ $f(-2).$
Therefore, given function is not continuous at x = − 2