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  • Find n given that  <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{1-\sqrt{\cos 2 x} \cdot \sqrt[3]{\cos 3 x} \cdot \sqrt[4]{\cos 4 x} \ldots \sqrt[n]{\cos n x}}{x^2}=10}" src="https://en

Find n given that  \mathrm{\lim _{x \rightarrow 0} \frac{1-\sqrt{\cos 2 x} \cdot \sqrt[3]{\cos 3 x} \cdot \sqrt[4]{\cos 4 x} \ldots \sqrt[n]{\cos n x}}{x^2}=10}

Option: 1

6


Option: 2

0


Option: 3

4


Option: 4

1


Answers (1)

best_answer

Using the fact that:
                                                \mathrm{ \cos (n x) \sim 1-\frac{n^2}{2} x^2 }
we have:
\mathrm{ \frac{1-\sqrt{\cos 2 x} \cdot \sqrt[3]{\cos 3 x} \cdot \sqrt[4]{\cos 4 x} \cdots \sqrt[n]{\cos n x}}{x^2} \sim \frac{1-\left(\left(1-2 x^2\right)^{\frac{1}{2}} \cdots\left(1-\frac{n^2}{2} x^2\right)^{\frac{1}{n}}\right)}{x^2} }
Now, remember that:

                                                    \mathrm{ (1-f(x))^\alpha \sim 1-\alpha f(x) }
when \mathrm{f(x) \rightarrow 0}, so:

\mathrm{ \begin{gathered} \frac{1-\left(\left(1-2 x^2\right)^{\frac{1}{2}} \cdots\left(1-\frac{n^2}{2} x^2\right)^{\frac{1}{n}}\right)}{x^2} \sim \frac{\left(\sum_{i=2}^n \frac{1}{i} \cdot \frac{i^2}{2}\right) x^2}{x^2}=\sum_{i=2}^n \frac{i}{2}=\frac{1}{2} \cdot \sum_{i=2}^n i= \\ =\frac{n \cdot(n+1)}{4}-\frac{1}{2} \end{gathered} }
Thus, we have:

                                                \mathrm{\frac{n \cdot(n+1)}{4}-\frac{1}{2}=10 \leftrightarrow n=6}

Posted by

Anam Khan

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