JEE Advance - Mathematics (2024 - Paper 2 Online - No. 3)
Let $k \in \mathbb{R}$. If $\lim \limits_{x \rightarrow 0+}(\sin (\sin k x)+\cos x+x)^{\frac{2}{x}}=e^6$, then the value of $k$ is
1
2
3
4
Explanation
$$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{2}{x}(\sin (\sin k x)+\cos x+x-1)=6 \\
& \lim _{x \rightarrow 0} \frac{\sin (\sin k x) \cdot \sin k x}{(\sin k x) k x} \cdot k+1-\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2} \cdot x=3 \\
& k+1=3 \Rightarrow k=2
\end{aligned}$$
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