Question

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

• A. $f$ is continuous at $x=1$
• B. $\underset{x\to 1^{+}}{lim}f\left(x\right)\ne \underset{x\to 1^{-}}{lim}f\left(x\right)$
• C. $\underset{x\to 1^{+}}{lim}f\left(x\right)=\sin 1$
• D. $\underset{x\to 1^{-}}{lim}f\left(x\right)$ doesn't exist

Answer 1

The correct option is B $\underset{x\to 1^{+}}{lim}f\left(x\right)\ne \underset{x\to 1^{-}}{lim}f\left(x\right)$

For $\left|x\right|<1,x^{2n}\to 0$ as $n\to \infty$ and for $\left|x\right|>1,1/2^{2n}\to 0$ as $n\to \infty$. So
$f\left(x\right)=\{\begin{array}{cc}log(2+x)& \text{if}\left|x\right|<1\\ {lim}_{n\to \infty }\frac{x^{-2n}log(2+x)-\sin x}{x^{-2n}+1}=-\sin x& \text{if}\left|x\right|>1\\ \frac{1}{2}\left[log\right(2+x)-\sin x]& \left|x\right|=1\end{array}$
Thus $\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}(-\sin x)=-\sin 1$
and $\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}log(2+x)=log3$