JEE Advanced 2024 · Paper 2, Mathematics question 3

The base tends to $1$ and the exponent to $\infty$ (a $1^\infty$ form), so
$$L=\exp\left(\lim_{x\to0^+}\frac{2}{x}\big(\sin(\sin kx)+\cos x-1+x\big)\right).$$

Each term behaves as follows as $x\to0$:

• $\dfrac{\sin(\sin kx)}{x}\to k$
• $\dfrac{\cos x-1}{x}\to0$
• $\dfrac{x}{x}=1$

So $L=e^{2(k+1)}$. Setting $2(k+1)=6$ gives $k=2$.