Question

If $\displaystyle\lim _{x \rightarrow 0} \frac{1-\sqrt{\cos 2 x} \cdot \sqrt[3]{\cos 3 x} \cdot \sqrt[4]{\cos 4 x} \ldots \cdot \sqrt[n]{\cos n x}}{x^{2}}$ has the value equal to $10$ , then the value of $n$ equals.

Answer 1

$L=\displaystyle\lim _{x \rightarrow 0}=-\displaystyle\lim _{x \rightarrow 0} \frac{D \displaystyle\prod_{r=2}^{n}(\cos r x)^{1 / r}}{2 x}$
(using L’Hospital’s rule)
let $y=\displaystyle\prod_{r=2}^{n}(\cos r x)^{1 / r}$
$\Rightarrow \ln y=\displaystyle\sum_{r=2}^{n}\left(\frac{1}{r} \ln \cos r x\right)$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=-\displaystyle\sum_{r=2}^{n} \tan (r x)$
$\Rightarrow -D y=y \displaystyle\sum_{r=2}^{n} \tan (r x)$
$D \displaystyle\prod_{ r =2}^{ n }(\cos r x )^{1 / r }$
$=- y \displaystyle\sum_{ r =2}^{ n } \tan ( r x )$
$L=\displaystyle\lim _{x \rightarrow 0} \frac{y \cdot \displaystyle\sum_{r=2}^{n} \tan (r x)}{2 x}$
$=\frac{1}{2}[2+3+4+\ldots .+ n ]$
$=\frac{1}{2}\left[\frac{ n ( n +1)}{2}-1\right]$
$=\frac{ n ^{2}+ n -2}{4}$
$\Rightarrow \frac{ n ^{2}+ n -2}{4}=10$
$\Rightarrow n ^{2}+ n -42=0$
$\Rightarrow ( n +7)( n -6)=0$
$\Rightarrow n =6$