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If $f(x)=\left\{\begin{array}{cl}\frac{x}{\sin 3 x}, & x \neq 0 \\ k & , x=0\end{array}\right.$ is continuous at $x=0$, then write the value of $k$.


If $f(x)$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0} \frac{x}{\sin 3 x}=k$

$\Rightarrow \lim _{x \rightarrow 0} \frac{1}{\frac{\sin 3 x}{x}}=k$

$\Rightarrow \lim _{x \rightarrow 0} \frac{1}{\frac{3 \sin 3 x}{3 x}}=k$

$\Rightarrow \frac{1}{3}\left(\frac{1}{\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}}\right)=k$

$\Rightarrow k=\frac{1}{3}$

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