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  • If    <img alt="f(x)=\left\{\begin{array}{ll} \frac{|x+2|}{\tan ^{-1}(x+2)}, & x \neq-2 \\ 2, , x=-2 \end{array} \text {, then } f(x)\right. \text { is }" src="https://entranceco

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

Option: 1

continuous at x=-2


Option: 2

not continuous at x=-2


Option: 3

differentiable at x=-2


Option: 4

continuous but not derivable at x=-2


Answers (1)

best_answer

We have,
\mathrm{\lim _{x \rightarrow 2^{-}} f(x) =\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} \frac{|-2-h+2|}{\tan ^{-1}(-2-h+2)} }
\mathrm{=\lim _{h \rightarrow 0} \frac{h}{\tan ^{-1}(-h)}=\lim _{h \rightarrow 0} \frac{-h}{\tan ^{-1} h}=-1 }

\mathrm{ \text { and, } \lim _{x \rightarrow-2^{+}} f(x) =\lim _{h \rightarrow 0} f(-2+h)=\lim _{h \rightarrow 0} \frac{|-2+h+2|}{\tan ^{-1}(-2+h+2)} }
\mathrm{ =\lim _{h \rightarrow 0} \frac{h}{\tan ^{-1} h}=1 } 
\mathrm{ \therefore \lim _{x \rightarrow-2^{-}} f(x) \neq \lim _{h \rightarrow-2^{+}} f(x) . }

So, f(x) is neither continuous nor differentiable at x=-2.

Posted by

sudhir.kumar

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