JEE Main 2022 — Mathematics Calculus

Given $f(x)=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{\mathrm{cos}2\pi x-{x}^{2n}\mathrm{sin}(x-1)}{1+{x}^{2n+1}-{x}^{2n}}$f(x)=n→∞lim​1+x2n+1−x2ncos2πx−x2nsin(x−1)​

$$=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{\mathrm{cos}2\pi x-{({x}^{2})}^{n}\mathrm{sin}(x-1)}{1+x{({x}^{2})}^{n}-{({x}^{2})}^{n}}$$=n→∞lim​1+x(x2)n−(x2)ncos2πx−(x2)nsin(x−1)​

Now for $-1<x<1,$−1<x<1, as $0<{x}^{2}<1\Rightarrow \underset{n\rightarrow \infty }{\mathrm{lim}}{x}^{2n}\rightarrow 0$0<x2<1⇒n→∞lim​x2n→0

i.e. $f(x)=\mathrm{cos}2\pi x$f(x)=cos2πx

Now again rewriting $f(x)=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{{({x}^{2})}^{-n}\mathrm{cos}2\pi x-\mathrm{sin}(x-1)}{{({x}^{2})}^{-n}+x-1}$f(x)=n→∞lim​(x2)−n+x−1(x2)−ncos2πx−sin(x−1)​

For $x>1\text{or}x<-1,$x>1orx<−1, $\underset{n\rightarrow \infty }{\mathrm{lim}}{x}^{-2n}\rightarrow 0$n→∞lim​x−2n→0

i.e. $f(x)=-\frac{\mathrm{sin}(x-1)}{x-1}$f(x)=−x−1sin(x−1)​

For x=\pm 1,f(x)={\begin{matrix}1 & \mathrm{if}x=1 \\ \frac{(1+\mathrm{sin}2)}{-1} & \mathrm{if}x=-1\end{matrix}

i.e. $\underset{x\rightarrow {1}^{+}}{\mathrm{lim}}f(x)=-1,\underset{x\rightarrow {1}^{-}}{\mathrm{lim}}f(x)=1$x→1+lim​f(x)=−1,x→1−lim​f(x)=1

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

$$\underset{x\rightarrow -{1}^{+}}{\mathrm{lim}}f(x)=1,\underset{x\rightarrow -{1}^{-}}{\mathrm{lim}}f(x)=-\frac{\mathrm{sin}2}{2}$$x→−1+lim​f(x)=1,x→−1−lim​f(x)=−2sin2​

So $f(x)$f(x) is discontinuous at $x=-1$x=−1