Question

The value of \[ \lim_{x \to 0} 2 \left( \frac{1 - \cos x \sqrt{\cos 2x} \, \sqrt[3]{\cos 3x} \ldots \sqrt[10]{\cos 10x}}{x^2} \right) \] is _____.

Answer 1

Correct Answer: 55

To resolve the problem, determine the value of:
\[ \lim_{x \to 0} 2 \left( \frac{1 - \cos x \sqrt{\cos 2x} \, \sqrt[3]{\cos 3x} \ldots \sqrt[10]{\cos 10x}}{x^2} \right) \]
Analyze the product of cosines:
Expression: \(\cos x \sqrt{\cos 2x} \sqrt[3]{\cos 3x} \ldots \sqrt[10]{\cos 10x}\)
Let this product be \(P(x)\). Using the exponential form of cosine, we get:
\(P(x) = \cos x \cdot (\cos 2x)^{1/2} \cdot (\cos 3x)^{1/3} \cdot \ldots \cdot (\cos 10x)^{1/10}\)
For small angles \(x\), the approximation \(\cos kx \approx 1 - \frac{(kx)^2}{2}\) applies. Applying this:
\(P(x) \approx 1 - \frac{x^2}{2} - \frac{(2x)^2}{4} - \frac{(3x)^2}{6} - \ldots - \frac{(10x)^2}{20}\)
Summing these terms yields:
\(P(x) \approx 1 - x^2 \left(\frac{1}{2} + \frac{4}{4} + \frac{9}{6} + \ldots + \frac{100}{20}\right)\)
This simplifies to a sum of \(P_k = \frac{k^2}{2k}\), which further simplifies to:
\(P(x) \approx 1 - x^2 \left( \frac{1}{2} + 1 + 1.5 + \ldots + 5 \right)\)
The arithmetic series sum is:
\(S = \frac{1}{2} + 1 + 1.5 + \ldots + 5 = 27.5\)
Therefore, \(P(x) \approx 1 - 27.5x^2\). Substituting this back into the limit expression:
\[ \lim_{x \to 0} 2 \left(\frac{1 - (1 - 27.5x^2)}{x^2}\right) \]
This simplifies to:
\[ \lim_{x \to 0} 2 \times \frac{27.5x^2}{x^2} \]
Further simplification results in:
\[ \lim_{x \to 0} 2 \times 27.5 = 55 \]
The computed value of 55 is within the expected range of 55 to 55.