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

Answers (1)

Answer:

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.

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