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### 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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