Question
For what value of k is the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{\text{x}}, & \text{x} \neq 0\\\text{k}, &\text{x} = 0\end{cases}$ continuous at x = 0.

Answer

Given, $\text{f}\text{(x)}=\begin{cases}\frac{\sin2\text{x}}{\text{x}}, & \text{x} \neq 0\\\text{k}, &\text{x} = 0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{\sin2\text{x}}{\text{x}}=\text{k}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{2\sin2\text{x}}{2\text{x}}=\text{k}$
$\Rightarrow2\lim\limits_{\text{x} \rightarrow 0}\frac{\sin2\text{x}}{2\text{x}}=\text{k}$
$\Rightarrow2\times1=\text{k}$
$\Rightarrow\text{k} =2$

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