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# 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

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

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