Question

The function $\frac{\left|x\right|}{x^2+2x},x\ne 0$ and $f\left(0\right)=0$ is not continuous at $x=0$ because-

• A. $\underset{x\to 0}{lim}f\left(x\right)\ne f\left(0\right)$
• B. $\underset{x\to 0^{+}}{lim}f\left(x\right)$ does not exist
• C. $\underset{x\to 0^{-}}{lim}f\left(x\right)$ does not exist
• D. $\underset{x\to 0}{lim}f\left(x\right)$ does not exist.

Answer 1

Answer: (D)

$\underset{x\to 0}{lim}f\left(x\right)$ does not exist.

$f\left(x\right)=\left\{\begin{array}{c}\frac{-x}{x^2+2x}\text{if}x<0\\ \frac{x}{x^2+2x}\text{if}x>0\end{array}\right\}$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{c}\frac{-1}{x+2}\text{if}x<0\\ \frac{1}{x+2}\text{if}x>0\end{array}\right\}$

Clearly L.H.L $=-1\ne$ R.H.S $=1$

Hence $\underset{x\to 0}{lim}f\left(x\right)$ does not exist.