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

विकल्प

• continuous at x = − 2

• not continuous at x = − 2

• differentiable at x = − 2

• continuous but not derivable at x = − 2

Answer 1

(b) not continuous at x = − 2 

Given: 

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

 

⇒ `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

\[f( - 2) = 2\]

Thus,  

\[\lim_{x \to - 2^-} f(x) \neq \lim_{x \to - 2^+} f(x)\]

\[f( - 2) .\]

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