Limits – Evaluation of Limits – JEE Main 28 Jan 2026 Shift 1

Solution:

The given limit is of the form $\frac{0}{0}$. We can apply L’Hopital’s Rule.
$$L = \lim_{x\rightarrow0}\frac{\sum_{k=1}^{10} \ln(\sec(e^k x))}{e^2 – e^{2\cos x}}$$

Differentiating numerator and denominator with respect to $x$:
Numerator derivative: $\frac{d}{dx} \sum \ln(\sec(e^k x)) = \sum \frac{1}{\sec(e^k x)} \cdot \sec(e^k x)\tan(e^k x) \cdot e^k = \sum e^k \tan(e^k x)$
Denominator derivative: $\frac{d}{dx} (e^2 – e^{2\cos x}) = -e^{2\cos x} \cdot (-2\sin x) = 2\sin x \cdot e^{2\cos x}$

Substituting these back into the limit:
$$L = \lim_{x\rightarrow0} \frac{\sum_{k=1}^{10} e^k \tan(e^k x)}{2\sin x \cdot e^{2\cos x}}$$

As $x \to 0$, $e^{2\cos x} \to e^2$. We can rewrite the limit using standard limits $\lim_{\theta\to0}\frac{\tan\theta}{\theta}=1$ and $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$:
$$L = \frac{1}{2e^2} \sum_{k=1}^{10} e^k \cdot \lim_{x\to 0} \left( \frac{\tan(e^k x)}{e^k x} \cdot \frac{e^k x}{\sin x} \right)$$
$$L = \frac{1}{2e^2} \sum_{k=1}^{10} e^k \cdot (1) \cdot \frac{e^k}{1}$$
$$L = \frac{1}{2e^2} \sum_{k=1}^{10} e^{2k}$$

The sum is a Geometric Progression (G.P.) with first term $a=e^2$, common ratio $r=e^2$, and number of terms $n=10$:
$$Sum = \frac{e^2((e^2)^{10} – 1)}{e^2 – 1} = \frac{e^2(e^{20} – 1)}{e^2 – 1}$$

Substituting the sum back into the expression for $L$:
$$L = \frac{1}{2e^2} \cdot \frac{e^2(e^{20} – 1)}{e^2 – 1}$$
$$L = \frac{e^{20} – 1}{2(e^2 – 1)}$$

Ans. (3)