Question

$\underset{x\to \infty }{lim}\left[\frac{2+2x+\sin 2x}{(2x+\sin 2x)e^{\sin x}}\right]$ is equal to

• A. 0
• B. 1
• C. -1
• D. Does not exist

Answer 1

The correct option is D Does not exist

$\underset{x\to \infty }{lim}\left[\frac{2+2x+sin2x}{(2x+sin2x)e^{sinx}}\right]$
$=\underset{x\to \infty }{lim}\frac{\frac{2}{x}+2+\frac{\sin 2x}{x}}{(2+\frac{\sin 2x}{x})e^{\sin x}}$
$=\frac{0+2+0}{(2+0)\times (avaluebetween\frac{1}{e}ande)}[\because \underset{x\to \infty }{lim}\sin x\in (-1,1)]$
Hence, limit does not exist.