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

Question

$\begin{array}{1 1}(A)\;continuous \;at\; x=2 \\(B)\;not \;continuous\;at\;x=-2 \\(C)\;differentiable\;at\;x=-2 \\(D)\;continuous\;but\;not\;differentiable\;x=-2 \end{array}$

Comments

f(x) is not continuous at x=-2 because both the limit are different.
So option B is correct answer.