Determine the value of the constant k so that the function

Given:

$f(x)=\left\{\begin{array}{l}\frac{\sin 2 x}{5 x}, \text { if } x \neq 0 \\ k, \text { if } x=0\end{array}\right.$

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

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

$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 2 x}{5 x}=k$

$\Rightarrow \lim _{x \rightarrow 0} \frac{2 \sin 2 x}{5 \times 2 x}=k$

$\Rightarrow \frac{2}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}=k$

$\Rightarrow \frac{2}{5} \times 1=k$

$\Rightarrow k=\frac{2}{5}$