Question

If f(x) is continuous at $x=\pi$, where $f\left(x\right)=\frac{\sqrt{2+cosx-1}}{(\pi -x^2)}$ for $x\ne \pi$ , find $f\left(\pi \right)$ .

Answer 1

Since f(x) is continuous at $x=\pi$
$f\left(\pi \right)={lim}_{x\to \pi }f\left(x\right)$

${lim}_{x\to \pi }\left[\frac{2+cosx-1}{\pi -x^2}\right]$$\times \frac{\sqrt{2+cosx+1}}{2+cosx+1}$

$f\left(\pi \right)={lim}_{x\to \pi }f\left(x\right)$

${lim}_{x\to \pi }\left[\frac{2+cosx-1}{\pi -x^2}\right]$

$f\left(\pi \right)={lim}_{x\to \pi }f\left(x\right)$

$\left[\frac{1}{\sqrt{2+cosx+1}}\right]$