Question

$f\left(x\right)=\frac{\sqrt{2+\cos \left(\pi x\right)}-1}{\left(1-x^2\right)}x\ne 1$
$\frac{\pi }{3},x=1$
Discuss continuity at $x=1$.

Answer 1

For continuity at $x=1$ we just need to check whether the value of function around $x=1$ matches to the value given at $x=1$.

$f\left(x\right)=\frac{\sqrt{2+\cos \left(\pi x\right)}-1}{1-x^2}$

at $x=1$ it is in the form of $\frac{0}{0}$ so apply L'Hoptell Rule;

$f\left(x\right)=\frac{\frac{-\pi \sin \left(\pi x\right)}{2\sqrt{2+\cos \left(\pi x\right)}}}{-2x}=\frac{\pi \sin \left(\pi x\right)}{4x\sqrt{2+\cos \left(\pi x\right)}}$

$\underset{x\to 1}{lim}f\left(x\right)=0$

but the value at $x=1$ is $\frac{\pi }{3}$

both are not matching therefore the function is discontinuous at $x=1$