Question

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{1-\cos kx}{x\sin x},& x\ne 0\\ \frac{1}{2},& x=0\end{array}\right.\mathrm{is}\mathrm{continuous}\mathrm{at}x=0,\mathrm{find}k.$

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{l}\frac{1-\cos kx}{x\sin x},x\ne 0\\ \frac{1}{2},x=0\end{array}\right.$
If $f\left(x\right)$ is continuous at x = 0, then
$\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$ ...(1)

Consider:
$\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{1-\cos kx}{x\sin x}\right)=\underset{x\to 0}{\lim }\left(\frac{2{\sin }^2{\displaystyle \frac{kx}{2}}}{x\sin x}\right)$
$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{2{\sin }^2{\displaystyle \frac{kx}{2}}}{x^2\left({\displaystyle \frac{\sin x}{x}}\right)}\right)$
$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left(\frac{{\displaystyle \frac{2k^2}{4}}{\left(\sin {\displaystyle \frac{kx}{2}}\right)}^2{\displaystyle }}{{\left({\displaystyle \frac{kx}{2}}\right)}^2\left({\displaystyle \frac{\sin x}{x}}\right)}\right)$
$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\frac{2k^2}{4}\underset{x\to 0}{\lim }\left(\frac{{\left({\displaystyle \mathrm{s}}\mathrm{in}{\displaystyle \frac{kx}{2}}\right)}^2}{{\left({\displaystyle \frac{kx}{2}}\right)}^2\left({\displaystyle \frac{\sin x}{x}}\right)}\right)$
$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\frac{2k^2}{4}\left(\frac{\underset{x\to 0}{\lim }{\displaystyle \frac{{\left(\sin {\displaystyle \frac{kx}{2}}\right)}^2}{{\left({\displaystyle \frac{kx}{2}}\right)}^2}}}{{\displaystyle \underset{x\to 0}{\lim }\frac{\sin x}{x}}}\right)$
$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\frac{2k^2}{4}\times 1=\frac{k^2}{2}$

From equation (1), we have
$\frac{k^2}{2}=f\left(0\right)$
$$\begin{gathered} \Rightarrow \frac{k^2}{2}=\frac{1}{2} \\ \Rightarrow k=\pm 1 \end{gathered}$$