Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given $f\left(x\right)=\underset{n\to \infty }{\lim }\frac{\cos 2\pi x-x^{2n}\sin \left(x-1\right)}{1+x^{2n+1}-x^{2n}}$

$=\underset{n\to \infty }{\lim }\frac{\cos 2\pi x-{\left(x^2\right)}^n\sin \left(x-1\right)}{1+x{\left(x^2\right)}^n-{\left(x^2\right)}^n}$

Now for $-1<x<1,$ as $0<x^2<1\Rightarrow \underset{n\to \infty }{\lim }{x}^{2n}\to 0$

i.e. $f\left(x\right)=\cos 2\pi x$

Now again rewriting $f\left(x\right)=\underset{n\to \infty }{\lim }\frac{{\left(x^2\right)}^{-n}\cos 2\pi x-\sin \left(x-1\right)}{{\left(x^2\right)}^{-n}+x-1}$

For $x>1 \text{or} x<-1,$ $\underset{n\to \infty }{\lim }{x}^{-2n}\to 0$

i.e. $f\left(x\right)=-\frac{\sin \left(x-1\right)}{x-1}$

For $x=\pm 1, f\left(x\right)=\left\{\begin{array}{cc}1& \mathrm{if} x=1\\ \frac{\left(1+\sin 2\right)}{-1}& \mathrm{if} x=-1\end{array}\right.$

i.e. $\underset{x\to 1^{+}}{\lim }f\left(x\right)=-1,\underset{x\to 1^{-}}{\lim }f\left(x\right)=1$

So $f$ is discontinuous at $x=1$

$\underset{x\to -1^{+}}{\lim }f\left(x\right)=1,\underset{x\to -1^{-}}{\lim }f\left(x\right)=-\frac{\sin 2}{2}$

So $f\left(x\right)$ is discontinuous at $x=-1$