Question

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

• A. $1$
• B. $\frac{1}{2}$
• C. $0$
• D. $2$

Answer 1

The correct option is A $1$

$\frac{1}{4n}\to 0⟹n\to \infty$

$\therefore f\left(0\right)={lim}_{n\to \infty }\frac{\sin e^n}{e^{n^2}}+{lim}_{n\to \infty }\frac{n^2}{n^2+1}$

$\sin e^n\in [0,1]$ and $e^{\infty }\approx \infty$

$\Rightarrow f\left(0\right)=0+{lim}_{n\to \infty }\frac{2n}{2n}=1$ (Using L'hopital's rule)