Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 24

Since,$\lim _{x \rightarrow 0^{-}} f(x)$   and $\lim _{x \rightarrow 0^{+}} f(x)$  are not equal,  $f\left ( x \right )$  is discontinuous.

Thus,$f\left ( x \right )$  is discontinuous at  $x=0$

Hint:

$f\left ( x \right )$  must be defined. The limit of the  $f\left ( x \right )$ approaches the value $x$  must exist.

Given:

$f(x)=\left\{\begin{array}{c} \frac{x}{|x|+2 x^{2}}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right.$

Solution:

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

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

We observe

(LHL at $x=0$ )

$\lim _{h \rightarrow 0^{-}} \frac{1}{-2 h-1}=-1$

(RHL at $x=0$ )

$\lim _{h \rightarrow 0^{+}} \frac{1}{2 h+1}=1$

Since, $\lim _{x \rightarrow 0^{-}} f(x)$  and  $\lim _{x \rightarrow 0^{+}} f(x)$ are not equal, $f\left ( x \right )$ is discontinuous.

Thus, $f\left ( x \right )$ is discontinuous at $x=0$ , regardless of choice of  $k$