Question

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

• A. f is continuous at x=-2
• B. f is not derivable at x=-2
• C. f is not continuous at x=-2
• D. f is continuous at x=-2 but not derivable

Answer 1

Answer: (B), (C)

f is not derivable at x=-2
f is not continuous at x=-2

Given, $\underset{x\to -2+}{lim}f\left(x\right)=\underset{x\to -2+}{lim}\frac{|x+2|}{{\tan }^{-1}(x+2)}$

$\underset{x\to -2+}{lim}\frac{x+2}{{\tan }^{-1}(x+2)}=\underset{y\to 0+}{lim}\frac{y}{{\tan }^{-1}y}=1$

where $y=x+2$, and
$\underset{x\to -2-}{lim}f\left(x\right)=\underset{x\to -2-}{lim}\frac{|x+2|}{{\tan }^{-1}(x+2)}=\underset{y\to 0-}{lim}\frac{-y}{{\tan }^{-1}y}=-1$

Hence f is neither continuous nor derivable at $x=-2.$