Question

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

Answer 1

Given, $f\left(x\right)$ is continious at $x=0$

$\therefore \underset{x\to 0}{lim}f\left(x\right)=f\left(U\right)$

$=\underset{x\to 0}{lim}\frac{1-\cos kx}{x\sin x}=\frac{1}{2}$

$=\underset{x\to 0}{lim}\frac{2{\sin }^2\frac{kx}{2}}{x\times x}=\frac{1}{2}$

$=\underset{x\to 0}{lim}\frac{2\times k^2{x}^2}{4x^2}=\frac{1}{2}$

$\therefore k=\pm 1$