The value of $$\lim_{x \to 0}\frac{\log_e\!\left(\sec(ex)\cdot \sec(e^{2}x)\cdots \sec(e^{10}x)\right)}{e^{2}-e^{2\cos x}}$$ is equal to

To solve this limit, we can use logarithmic properties and standard Taylor expansions as $$x \to 0$$.\

Numerator

Using the property $$\ln(a \cdot b) = \ln a + \ln b$$ and $$\ln(\sec \theta) = -\ln(\cos \theta)$$:

$$L_N = \sum_{n=1}^{10} \ln(\sec(e^n x)) = -\sum_{n=1}^{10} \ln(\cos(e^n x))$$

Using the expansion $$\cos \theta \approx 1 - \frac{\theta^2}{2}$$ and $$\ln(1+u) \approx u$$:

$$\ln(\cos(e^n x)) \approx \ln\left(1 - \frac{(e^n x)^2}{2}\right) \approx -\frac{e^{2n}x^2}{2}$$

Thus, the numerator simplifies to:

$$L_N \approx -\sum_{n=1}^{10} \left( -\frac{e^{2n}x^2}{2} \right) = \frac{x^2}{2} \sum_{n=1}^{10} e^{2n}$$

Denominator

Factor out $$e^2$$:

$$L_D = e^2 - e^{2\cos x} = e^2(1 - e^{2(\cos x - 1)})$$

Using $$e^u \approx 1+u$$ as $$u \to 0$$ and $$(\cos x - 1) \approx -\frac{x^2}{2}$$:

$$L_D \approx e^2 \left( 1 - (1 + 2(-\frac{x^2}{2})) \right) = e^2(x^2)$$

Combine the simplified parts:

$$\lim_{x \to 0} \frac{\frac{x^2}{2} \sum_{n=1}^{10} (e^2)^n}{e^2 x^2} = \frac{1}{2e^2} \sum_{n=1}^{10} (e^2)^n$$

The sum is a Geometric Progression with $$a = e^2$$, $$r = e^2$$, and $$n = 10$$:

$$\text{Sum} = e^2 \frac{(e^2)^{10} - 1}{e^2 - 1} = \frac{e^2(e^{20}-1)}{e^2-1}$$

Substitute this back:

$$\text{Limit} = \frac{1}{2e^2} \cdot \frac{e^2(e^{20}-1)}{e^2-1} = \frac{e^{20}-1}{2(e^2-1)}$$

Correct Option: A