Q. The value of

\[ \lim_{x \to 0} \frac{\log_e \left( \sec(ex)\cdot \sec(e^2x)\cdots \sec(e^{10}x) \right)} {e^2 - e^{2\cos x}} \]

is equal to

Correct Answer: \(\dfrac{e^{10}-1}{2(e^2-1)}\)

Explanation

Step 1:

For small x, \(\sec(ax) \approx 1 + \frac{a^2x^2}{2}\). Hence, \[ \log(\sec(ax)) \approx \frac{a^2x^2}{2} \]

Step 2:

So numerator becomes: \[ \log\left(\prod_{k=1}^{10}\sec(e^k x)\right) = \sum_{k=1}^{10} \frac{e^{2k}x^2}{2} \]

Step 3:

This is a geometric series: \[ \sum_{k=1}^{10} e^{2k} = \frac{e^{2}(e^{20}-1)}{e^{2}-1} \]

Step 4:

Denominator: \[ e^2 - e^{2\cos x} \approx e^2 - e^{2(1-\frac{x^2}{2})} = e^2 x^2 \]

Step 5:

Taking ratio and simplifying:

\(\dfrac{e^{10}-1}{2(e^2-1)}\)

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