Question

If $f\left(x\right)=\underset{n\to \infty }{lim}\frac{\log (x+2)-x^{2n}\sin x}{x^{2n}+1}$, examine the continuity of $f\left(x\right)$ at $x=1$.

Write $1$ if continuous and $0$ if not.

Answer 1

If $x>1\underset{n\to \infty }{lim}{x}^{2n}=\infty$

$x<1\underset{n\to \infty }{lim}{x}^{2n}=\frac{1}{\infty }=0$

$x>1$

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

$=\frac{0-\sin x}{1+0}$

$f\left(x\right)=-\sin x$

For $x<1$

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

$=\frac{\log (x+2)-0}{0+1}$

$f\left(x\right)=\log (x+2)$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}\left[\log (x+2)\right]=3$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}[-\sin x]=-\sin 1$

LHL$\ne$RHL $\therefore$ Discontinuous at $x=1$