$\lim\limits_{x \rightarrow 0}\left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right)^{\frac{2 m}{x}}$ is
Answer & explanation
Correct answer: option 2
$y=\lim\limits_{x \rightarrow 0}\left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right)^{\frac{2 m}{x}}$
$\log y=\lim\limits_{x \rightarrow 0}\frac{2 m}{x}\log \left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right) =\frac{0}{0}$ by L'hopital's rule
$\log y=\lim\limits_{x \rightarrow 0}\frac{\frac{1}{m}\frac{\cos x}{m}-\frac{3}{m}\frac{\sin x}{m}}{\left(\frac{\sin x}{m}+\frac{\cos x}{m}\right)×\frac{1}{2m}}=2$
$y=e^2$