Q. If $\lim_{x \rightarrow 0}{\frac{\left( e^{kx} - 1 \right)\sin{kx}}{x^{2}} = 4}$, then $k$ is equal to
A) 2
B) -2
C) $\pm 2$
D) $\pm 4$
Explanation
(c)
$\lim_{x \rightarrow 0}{\frac{\left( e^{kx} - 1 \right)\sin{kx}}{x^{2}} = 4}$
$\Rightarrow \ \ \ \ \ \ \lim_{x \rightarrow 0}{\frac{e^{kx} - 1}{kx} \times k \times \frac{\sin{kx}}{kx} \times k = 4}$
$\Rightarrow \ \ \ \ \ \ \ \ \ k^{2} = 4\$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ k = \pm 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.