Question

If $f\left(x\right)$ is continuous at $x=0$, where
$f\left(x\right)=\frac{1-\cos kx}{x^2}:$ $x\ne 0$
$=\frac{1}{2}:x=0$.

Answer 1

Given, the function $f\left(x\right)$ is continuous at $x=0$.

Then,

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{1-\cos kx}{x^2}=\frac{1}{2}$..........(1)

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

$=\underset{x\to 0}{lim}\frac{2{\sin }^2\left(\frac{kx}{2}\right)}{x^2}$

$=\underset{x\to 0}{lim}\frac{2}{x^2}\frac{{\sin }^2\left(\frac{kx}{2}\right)}{{\left(\frac{kx}{2}\right)}^2}{\left(\frac{kx}{2}\right)}^2$

$=\frac{2}{x^2}\times {\left(\frac{kx}{2}\right)}^2$

$=\frac{k^2}{2}$.......(2)

comparing (1) and (2),

we get, $k^2=1$

or, $k=\pm 1$.