Question

Let $f$ be a continuous function on $R$ such that $f\left(\frac{1}{4n}\right)=\left(\sin e^n\right)e^{-n^2}+\frac{n^2}{n^2+1}$. Then, the value of $f\left(0\right)$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $0$
• D. none of these

Answer 1

The correct option is A $1$

As $f$ is continuous on $R,$ so $\underset{x\to 0}{lim}f\left(x\right)$ $=\underset{n\to \infty }{lim}f\left(x_n\right)$

Thus, $f\left(0\right)=\underset{n\to \infty }{lim}f\left(\frac{1}{4n}\right)$

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

$=\underset{n\to \infty }{lim}\left(\sin e^n\right)e^{{-n}^2}+1=1$ as $\left|\left(\sin e^n\right)e^{{-n}^2}\right|\le e^{{-n}^2}$ and $e^{{-n}^2}\to 0$ as $n\to \infty )$