Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If $\lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \sin k x}{x^{2}}=4$, then $k$ is equal to
MathematicsLimitsMHT CETMHT CET 2009
Options:
  • A 2
  • B $-2$
  • C $\pm 2$
  • D $\pm 4$
Solution:
1544 Upvotes Verified Answer
The correct answer is: $\pm 2$
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \sin k x}{x^{2}}=4 \\ & \Rightarrow \quad \lim _{x \rightarrow 0}\left[\frac{e^{k x} \sin k x}{x^{2}}-\frac{\sin k x}{x^{2}}\right]=4 \\ & \Rightarrow \lim _{x \rightarrow 0}\left[\frac{e^{k x} \cos k x \cdot k+k e^{k x} \sin k x}{2 x}-\frac{k \cos k x}{2 x}\right]=4 \\ & \Rightarrow \lim _{x \rightarrow 0}\left[\frac{k^{2} e^{k x} \cos k x-k^{2} e^{k x} \sin k x+k^{2} e^{k x} \sin k x}{+k^{2} e^{k x} \cos k x}{2}\right.\\ & \Rightarrow \frac{k^{2}-0+0+k^{2}}{2}+0=4 \\ & \Rightarrow \quad k^{2}=4 \\ & \Rightarrow \quad k=\pm 2 \end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.