Question

The function $f:R\to R$ defined by $f\left(x\right)=\underset{n\to \infty }{lim}\frac{\cos \left(2\pi x\right)-x^{2n}\sin (x-1)}{1+x^{2n+1}-x^{2n}}$ is continuous for all $x$ in

• A. $R$ - $\{-1,1\}$
• B. $R$ - $\{-1\}$
• C. $R$ - $\left\{0\right\}$
• D. $R$ - $\left\{1\right\}$

Answer 1

The correct option is A $R$ - $\{-1,1\}$

$f\left(x\right)=\underset{n\to \infty }{lim}\frac{\cos \left(2\pi x\right)-x^{2n}\sin (x-1)}{1+x^{2n+1}-x^{2n}}$
For $|x|<1,f\left(x\right)=\cos 2\pi x$, continuous function
$|x|>1,f\left(x\right)=\underset{n\to \infty }{lim}\frac{\frac{1}{x^{2n}}\cos 2\pi x-\sin (x-1)}{\frac{1}{x^{2n}}+x-1}$
$=\frac{-\sin (x-1)}{x-1}$ , continuous
For $|x|=1,f\left(x\right)=\{\begin{array}{cc}1& \text{if}x=1\\ -(1+\sin 2)& \text{if}x=-1\end{array}$
Now,
$\underset{x\to 1^{+}}{lim}f\left(x\right)=-1,\underset{x\to 1^{-}}{lim}f\left(x\right)=1,$ so
discontinuous at x = 1
$\underset{x\to -1^{+}}{lim}f\left(x\right)=1,\underset{x\to -1^{-}}{lim}=-\frac{\sin 2}{2}$, so
discontnuous at x = -1
$\therefore f\left(x\right)$ is continuous for all $x\in R-\{-1,1\}$