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  • Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 5

Need solution for RD Sharma Maths Class 12 Chapter 8 Continuity Excercise 8.1 Question 5

Answers (1)

Answer:

                x=0  (Discontinuous)

Hint:

For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Solution:

Given,

f(x)=\left(\begin{array}{cc} \frac{\sin 3 x}{x}, \text { if } x \neq 0 \\\\ 1 & , \text { if } x=0 \end{array}\right)   

We observe,

[LHL at x=0 ]

\begin{aligned} &\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h) \\\\ &\lim _{h \rightarrow 0} \frac{\sin (-3 h)}{-h}=\lim _{h \rightarrow 0} \frac{-\sin (3 h)}{-h}=\lim _{h \rightarrow 0} \frac{3 \sin (3 h)}{3 h}=3 \lim _{h \rightarrow 0} \frac{\sin (3 h)}{3 h}=3 \times 1=3 \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\begin{aligned} &{[\because \sin (-x)=-\sin x]} \\ &{\left[\because \frac{\sin x}{x}=1\right]} \end{aligned}\end{aligned}                                         

 [RHL at x=0 ]

\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h) \\\\ &\lim _{h \rightarrow 0} \frac{\sin (3 h)}{h}=\lim _{h \rightarrow 0} \frac{3 \sin (3 h)}{3 h}=3 \lim _{h \rightarrow 0} \frac{\sin (3 h)}{3 h}=3 \times 1=3 \end{aligned}

Given

f\left ( 0 \right )=1

It is known that for a function f\left ( x \right ) is to be continuous at x=a ,

\lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow a^{-}} f(x) \neq f(a)

But here

\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x) \neq f(0)

Hence, f\left ( x \right ) is discontinuous at x=0 .

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