Question

The value of $\underset{x\to 0}{lim}{(\sin \frac{x}{m}+\cos \frac{3x}{m})}^{\frac{2m}{x}}$ is equal to:

• A. $e$
• B. $1$
• C. $e^{-1}$
• D. $e^2$

Answer 1

The correct option is D $e^2$

Let $y={(\sin \frac{x}{m}+\cos \frac{3x}{m})}^{\frac{2m}{x}}$
$\log y=\frac{2m}{x}\log (\sin \frac{x}{m}+\cos \frac{3x}{m})$
$y=e^{\frac{2m}{x}\log (\sin \frac{x}{m}+\cos \frac{3x}{m})}$
$\Rightarrow \underset{x\to 0}{lim}y=\underset{x\to 0}{lim}{e}^{\frac{2m}{x}\log (\sin \frac{x}{m}+\cos \frac{3x}{m})}$
$=e^{\underset{x\to 0}{lim}\frac{2m}{x}\log (\sin \frac{x}{m}+\cos \frac{3x}{m})}$
$=e^{\underset{x\to 0}{lim}\frac{2m\log (\sin \frac{x}{m}+\cos \frac{3x}{m})}{x}}$
$=e^{\underset{x\to 0}{lim}\frac{2m\log (1+(\sin \frac{x}{m}+\cos \frac{3x}{m}-1))}{x}}$
$=e^{\underset{x\to 0}{lim}2m\frac{\frac{\log (1+(\sin \frac{x}{m}+\cos \frac{3x}{m}-1))(\sin \frac{x}{m}+\cos \frac{3x}{m}-1)}{(\sin \frac{x}{m}+\cos \frac{3x}{m}-1)}}{x}}$
$=e^{\underset{x\to 0}{lim}2m\frac{\sin \frac{x}{m}+\cos \frac{3x}{m}-1}{x}}$..............$(\because \underset{x\to 0}{lim}\frac{log(1+x)}{x}=1)$
$=e^{2m\underset{x\to 0}{lim}\frac{\frac{\frac{x}{m}\cdot \sin \frac{x}{m}}{\frac{x}{m}}-\frac{\frac{3x}{m}\cdot (1-\cos \frac{3x}{m})}{\frac{3x}{m}}}{x}}$.......................$(\because \underset{x\to 0}{lim}\frac{sinx}{x}=1\&\underset{x\to 0}{lim}\frac{1-cosx}{x}=0)$
$=e^2$