For what value of k is the function
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

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

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

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

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

$\Rightarrow 2 \times 1=k$

$\Rightarrow k=2$