Question:
For what value of k is the function
$f(x)= \begin{cases}\frac{\sin 2 x}{x}, & x \neq 0 \\ k & , x=0\end{cases}$
continuous at x = 0?
Solution:
Given: $f(x)=\left\{\begin{array}{l}\frac{\sin 2 x}{x}, x \neq 0 \\ k, x=0\end{array}\right.$
If $f(x)$ is continuous at $x=0$, then
$\lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin 2 x}{x}=k$
$\Rightarrow \lim _{x \rightarrow 0} \frac{2 \sin 2 x}{2 x}=k$
$\Rightarrow 2 \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}=k$
$\Rightarrow 2 \times 1=k$
$\Rightarrow k=2$