JEE Main 2026 Jan 28 Shift 1, Mathematics Q14: Methods to Evaluate Limits

Numerator: ${\sum }_{k=1}^{10}\ln (\sec (e^{k}x))=-{\sum }_{k=1}^{10}\ln (\cos (e^{k}x))$. As $x\to 0$, $\cos (e^{k}x)\approx 1-\frac{(e^{k}x{)}^2}{2}$, so $-\ln \cos (e^{k}x)\sim \frac{(e^{k}x{)}^2}{2}=\frac{e^{2k}{x}^2}{2}$.

Sum: $\frac{x^2}{2}{\sum }_{k=1}^{10}{e}^{2k}=\frac{x^2}{2}\cdot e^2\cdot \frac{e^{20}-1}{e^2-1}$.

Denominator: $e^2-e^{2\cos x}=e^{2\cos x}(e^{2-2\cos x}-1)$. As $x\to 0$: $e^{2\cos x}\to e^2$ and $e^{2-2\cos x}-1\sim 2-2\cos x\sim x^2$.

So denominator $\sim e^2\cdot x^2$.

Limit: $\underset{x\to 0}{\lim }\frac{\frac{x^2}{2}{e}^2\cdot \frac{e^{20}-1}{e^2-1}}{e^2{x}^2}=\frac{e^{20}-1}{2(e^2-1)}$.

Option (3).