Question:
For what value of k is the function
$f(x)=\left\{\begin{aligned} \frac{\sin 5 x}{3 x}, & \text { if } \quad x \neq 0 \\ k & \text {, if } x=0 \end{aligned}\right.$ is continuous at $x=0 ?$
Solution:
Given:
$f(x)=\left\{\begin{array}{l}\frac{\sin 5 x}{3 x}, \text { if } x \neq 0 \\ k, \text { if } x=0\end{array}\right.$
If $f(x)$ is continuous at $x=0$, then
$\lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{\mathrm{x} \rightarrow 0} \frac{\sin 5 x}{3 x}=k$
$\Rightarrow \lim _{\mathrm{x} \rightarrow 0} \frac{5 \sin 5 x}{3 \times 5 x}=k$
$\Rightarrow \frac{5}{3} \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x}=k$
$\Rightarrow \frac{5}{3} \times 1=k$
$\Rightarrow k=\frac{5}{3}$