Question

Let $f:R\to R$ be a function such that $\underset{x\to \infty }{lim}f\left(x\right)=L$, where $L$ is a finite real number. Then which of the following is/are true?

• A. $\underset{x\to \infty }{lim}x\sin (\frac{1}{x})\cdot f\left(x\right)=L$
• B. $\underset{x\to \infty }{lim}\sin \left(f\right(x\left)\right)=\sin L$
• C. $\underset{x\to \infty }{lim}x\sin \left(e^{-x}\right)\cdot f\left(x\right)=L$
• D. $\underset{x\to \infty }{lim}\frac{\sin x}{x}\cdot f\left(x\right)=L$

Answer 1

Answer: (A), (B)

The correct options are
A $\underset{x\to \infty }{lim}x\sin (\frac{1}{x})\cdot f\left(x\right)=L$
B $\underset{x\to \infty }{lim}\sin \left(f\right(x\left)\right)=\sin L$

$\underset{x\to \infty }{lim}x\sin (\frac{1}{x})\cdot f\left(x\right)=$
$=\underset{x\to \infty }{lim}\frac{\sin \left(1/x\right)}{\left(1/x\right)}\cdot f\left(x\right)$
$=1\times L=L$


$\underset{x\to \infty }{lim}\sin \left(f\right(x\left)\right)=\sin L$
$(\because \underset{x\to \infty }{lim}f\left(x\right)=L)$


$\underset{x\to \infty }{lim}x\sin \left(e^{-x}\right)\cdot f\left(x\right)$
$=\underset{x\to \infty }{lim}\frac{\sin \left(e^{-x}\right)}{e^{-x}}\cdot \frac{x}{e^x}\cdot f\left(x\right)$
$=1\times 0\times L=0$


$-1\le \sin x\le 1$
$-f\left(x\right)\le f\left(x\right)\cdot \sin x\le f\left(x\right)$
$\frac{-f\left(x\right)}{x}\le f\left(x\right)\cdot \frac{\sin x}{x}\le \frac{f\left(x\right)}{x}$
Thus, by sandwich theorm, we get:
$\underset{x\to \infty }{lim}\frac{-f\left(x\right)}{x}=\underset{x\to \infty }{lim}\frac{f\left(x\right)}{x}=0$
$\therefore \underset{x\to \infty }{lim}\frac{\sin x}{x}\cdot f\left(x\right)=0$