JEE Advanced 2024 Paper 2, Mathematics Section 1 Q3: Methods to Evaluate Limits

This is a $1^{\infty }$ form. Taking logarithm and using the standard limit:

$\ln L=\underset{x\to 0^{+}}{\lim }\frac{2}{x}(\sin (\sin kx)+\cos x+x-1).$

Split the bracket into manageable pieces:

$\ln L=\underset{x\to 0^{+}}{\lim }2\left(\frac{\sin (\sin kx)}{\sin kx}\cdot \frac{\sin kx}{kx}\cdot k+1-\frac{1-\cos x}{x^2}\cdot x\right).$

As $x\to 0^{+}$, the first product tends to $k$, and the last term vanishes since $\frac{1-\cos x}{x^2}\to \frac{1}{2}$ while it is multiplied by $x$.

So $\ln L=2(k+1)$. Setting this equal to $6$ gives $k+1=3$, hence $k=2$. Option (B).