Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

$\lim\limits_{x \rightarrow 0}\left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right)^{\frac{2 m}{x}}$ is

Options:

1

$e^2$

$e^{lm}$

log 6m

Correct Answer:

$e^2$

Explanation:

$y=\lim\limits_{x \rightarrow 0}\left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right)^{\frac{2 m}{x}}$

$\log y=\lim\limits_{x \rightarrow 0}\frac{2 m}{x}\log \left(\sin \frac{x}{m}+\cos \frac{3 x}{m}\right) =\frac{0}{0}$ by L'hopital's rule

$\log y=\lim\limits_{x \rightarrow 0}\frac{\frac{1}{m}\frac{\cos x}{m}-\frac{3}{m}\frac{\sin x}{m}}{\left(\frac{\sin x}{m}+\frac{\cos x}{m}\right)×\frac{1}{2m}}=2$

$y=e^2$