Question
For what value of k is the function
$\text{f}\text{(x)}=\begin{cases}\frac{\sin5\text{x}}{3\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$ is continuous at x = 0?

Answer

Given,
$\text{f}\text{(x)}=\begin{cases}\frac{\sin5\text{x}}{3\text{x}}, &\text{if}\text{ x}\neq0\\\text{k}, &\text{if}\text{ x}=0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{\sin5\text{x}}{3\text{x}}=\text{k}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\frac{5\sin5\text{x}}{3\times5\text{x}}=\text{k}$
$\Rightarrow\frac{5}{3}\lim\limits_{\text{x} \rightarrow 0}\frac{\sin5\text{x}}{5\text{x}}=\text{k}$
$\Rightarrow\frac{5}{3}\times1=\text{k}$
$\Rightarrow\text{k}=\frac{5}{3}$

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