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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

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

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$\lim \limits_{x \to -2} f(x) =\lim \limits _{x \to {-2}} \large\frac{|x+2|}{\tan ^{-1} (x+2)}$
$\qquad= \lim \limits_{x \to {-2^-}} \large\frac{-(x+2)}{\tan^{-1}(x+2)}$$=-1$
$\qquad= \lim \limits_{x \to {-2^+}} f(x)=\lim \limits_{x \to {2^+}} \large\frac{|x+2|}{\tan^{-1}(x+2)}$
$\qquad= \lim \limits_{x \to {-2^+}} \large\frac{x+2}{\tan^{-1}(x+2)}$$=1$
Hence it continuous but not differentiable $x=-2$
Hence D is the correct answer.
answered Apr 22, 2014 by meena.p

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