Question

The value of \(\lim_{x \to 0} \frac{\ln(\sec(ex) \cdot \sec(e^2x) \dots \sec(e^{10}x))}{e^2 - e^{2\cos x}}\) is :

• A. \(\frac{1}{2} \frac{(e^2 - 1)}{(e^{20} - 1)}\)
• B. \(\frac{1}{2} \frac{(e^{20} - 1)}{(e^2 - 1)}\)
• C. \(\frac{1}{2} \frac{(e - 1)}{(e^{20} - 1)}\)
• D. None of these

Hint:

Using Taylor expansions like \( \cos x \approx 1 - x^2/2 \) is often much faster for limits involving transcendental functions than L'Hopital's rule.

Answer 1

The Correct Option is B

To find the value of \(\lim_{x \to 0} \frac{\ln(\sec(ex) \cdot \sec(e^2x) \cdots \sec(e^{10}x))}{e^2 - e^{2\cos x}}\), we start by analyzing both the numerator and the denominator separately.

Step 1: Simplifying the Numerator 

The numerator is \(\ln(\sec(ex) \cdot \sec(e^2x) \cdots \sec(e^{10}x))\). This can be rewritten using the properties of logarithms:

\(\ln(\sec(ex)) + \ln(\sec(e^2x)) + \cdots + \ln(\sec(e^{10}x))\)

For small values of \(x\), \(\sec(\theta) \approx 1 + \frac{\theta^2}{2}\) since \(\sec(\theta) = \frac{1}{\cos(\theta)}\) and \(\cos(\theta) \approx 1 - \frac{\theta^2}{2}\).

Thus, each term becomes:

\(\ln(1 + \frac{(ex)^2}{2})\), \ln(1 + \frac{(e^2x)^2}{2})\), \ldots, \ln(1 + \frac{(e^{10}x)^2}{2})\)

Using \(\ln(1 + u) \approx u\) for small \(u\), the numerator approximates to:

\(\frac{(ex)^2}{2} + \frac{(e^2x)^2}{2} + \cdots + \frac{(e^{10}x)^2}{2}\)

Factoring \(\frac{x^2}{2}\), the expression simplifies to:

\(\frac{x^2}{2}(e^2 + e^4 + \cdots + e^{20})\)

This is a geometric series with first term \(e^2\) and common ratio \(e^2\). Thus, the sum is:

\(\frac{x^2}{2} \times \frac{e^2(e^{20} - 1)}{e^2 - 1}\)

Step 2: Simplifying the Denominator

The denominator is \(e^2 - e^{2\cos x}\). Using the approximation \(\cos x \approx 1 - \frac{x^2}{2}\) for small \(x\), the expression becomes:

\(e^2 - e^{2(1 - \frac{x^2}{2})} = e^2 - e^{2}e^{-x^2} = e^2(1 - e^{-x^2})\)

For small \(x\), \(e^{x} \approx 1 + x\), so \(e^{-x^2} \approx 1 - x^2\). Therefore, the denominator approximates to:

\(e^2 \times x^2\)

Step 3: Evaluating the Limit

Substitute the expressions into the limit:

\(\lim_{x \to 0} \frac{\frac{x^2}{2} \times \frac{e^2(e^{20} - 1)}{e^2 - 1}}{e^2 \times x^2}\)

The \(x^2\) terms cancel out, leaving:

\(\frac{1}{2} \times \frac{e^2(e^{20} - 1)}{e^2(e^2 - 1)}\)

This further simplifies to:

\(\frac{1}{2} \times \frac{(e^{20} - 1)}{(e^2 - 1)}\)

Hence, the value of the limit is \(\frac{1}{2} \times \frac{(e^{20} - 1)}{(e^2 - 1)}\), which corresponds to the correct answer:

\(\frac{1}{2} \frac{(e^{20} - 1)}{(e^2 - 1)}\)