#### Question

If \[f\left( x \right) = \begin{cases}\frac{\left| x + 2 \right|}{\tan^{- 1} \left( x + 2 \right)} & , x \neq - 2 \\ 2 & , x = - 2\end{cases}\] then f (x) is

##### Options

continuous at x = − 2

not continuous at x = − 2

differentiable at x = − 2

continuous but not derivable at x = − 2

#### Solution

(b) not continuous at *x* = − 2

Given:

⇒ `f(x) = {((-|x+2|)/(tan^(-1)(x+2)), x< -2),((|x+2|)/(tan^(-1)(x+2)), x> -2),(2, x = -2):}`

Continuity at *x* = − 2.

(LHL at *x*= − 2) =

\[\lim_{x \to - 2^-} f(x) = \lim_{h \to 0} f( - 2 - h) = \lim_{h \to 0} \frac{- ( - 2 - h + 2)}{\tan^{- 1} ( - 2 - h + 2)} = \lim_{h \to 0} \frac{h}{\tan^{- 1} ( - h)} = - 1 . \]

(RHL at x = −2) =

\[\lim_{x \to - 2^+} f(x = \lim_{h \to 0} f( - 2 + h = \lim_{h \to 0} \frac{( - 2 + h + 2)}{\tan^{- 1} ( - 2 + h + 2)} = \lim_{h \to 0} \frac{h}{\tan^{- 1} (h)} = 1 .\]

Also

Therefore, given function is not continuous at x = − 2