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#### Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 40 Maths Textbook Solution.

Discontinuous

Hint:

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

Given:

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

Solution:

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

$f(x)=\left\{\begin{array}{l} 0, \text { if } x>0 \\ 2, \text { if } x<0 \\ 2, \text { if } x=0 \end{array}\right.$

We have

(LHL at  $x=0$ )

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0} 2=2$

(RHL at  $x=0$ )

$\begin{gathered} \lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0} 0=0 \\ \lim _{x \rightarrow 0^{+}} f(x) \neq \lim _{x \rightarrow 0^{-}} f(x) \end{gathered}$

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