Question

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

• A. $f$ is continuous at $x=1$
• B. $\underset{x\to 1^{+}}{\lim }f\left(x\right)=log3$
• C. $\underset{x\to 1^{+}}{\lim }f(x)=-sin1$
• D. $\underset{x\to 1^{-}}{\lim }f(x)$ does not exist

Answer 1

Answer: (C)

For $\left|x\right|<1,x^{2n}\to 0$ as $n\to \infty$, and for $\left|x\right|>1,1/x^{2n}\to 0$ as
$n\to \infty$ So,
$f(x)=\{\begin{array}{ll}log(2+x),& |x|<1\\ \underset{n\to \infty }{\lim }\frac{x^{-2n}log\left(2+x\right)-sinx}{x^{-2n}+1}=-sinx,& |x|>1\\ \frac{1}{2}[log(2+x)-sinx],& |x|=1\end{array}$
Thus,
$\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1}{\lim }\left(-sinx\right)=-sin1$
and
$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1}{\lim }log\left(2+x\right)=log3$