I'm trying to solve this rather interesting problem. We have been given that $\lim\limits_{x\rightarrow0} \frac{1-\sqrt{\cos2x}.\sqrt[3]{\cos3x}.\sqrt[4]{\cos4x}…\sqrt[n]{\cos{nx}}}{x^2}$ = 10 and we are required to find n.
This is the $\frac{0}{0}$ form, so we can use L'Hôpital's rule. After taking the derivatives of the numerator and the denominator separately, the problem becomes-
$\lim\limits_{x\rightarrow0} \frac{-\frac{d}{dx}\sqrt{\cos2x}.\sqrt[3]{\cos3x}.\sqrt[4]{\cos4x}…\sqrt[n]{\cos{nx}}}{2x}$ = 10
$\lim\limits_{x\rightarrow0} \frac{-\frac{d}{dx}\prod_{i = 2}^{n} (\cos ix)^\frac{1}{i}}{2x}$ = 10
Now, the numerator looks like a pretty difficult expression to differentiate. Here, I decided to simplify the expression in the numerator first.
Let y= $\prod_{i = 2}^{n} (\cos ix)^\frac{1}{i}$
Now, we can simplify the expression by taking the natural logarithm of both sides.
$\log(y)$= $\log(\prod_{i = 2}^{n} (\cos ix)^\frac{1}{i})$
Now we can use the property of logarithms to simplify the expression, $\log(xy)=\log(x)+\log(y)$
$\log(y)$= $\sum_{i = 2}^{n} \frac{1}{i}.\log(\cos ix)$
This is where I ran out of ideas to simplify this expression any further. Any ideas would be appreciated.
Best Answer
You proceed further by implicitly differentiating: $\dfrac{dy}{dx}(\log y)=\dfrac1y \dfrac{dy}{dx}$.
$$\dfrac1y \dfrac{dy}{dx}=\sum_{i=2}^n\frac1i\frac{1}{\cos ix}(-i\sin ix)=-\sum_{i=2}^n\tan ix.$$ Therefore, $\dfrac{dy}{dx}=y'=-y\sum_{i=2}^n\tan ix.$ Now, as $x\to 0$, $\sum_{i=2}^n\tan ix\to 0$ and $y\to 1$, so $y'\to 0$.
Thus, we have again got a $\dfrac00$ form, $\displaystyle \frac{y\sum_{i=2}^n\tan ix}{2x}$, on our hands. Differentiating again, we get $$\lim_{x\to 0}\frac{y'\sum_{i=2}^n\tan ix+y\tfrac{d}{dx}\left(\sum_{i=2}^n\tan ix\right)}{2}=\lim_{x\to 0}\frac{-y\left(\sum_{i=2}^n\tan ix\right)^2+y\left(\sum_{i=2}^ni\sec^2 ix\right)}{2}$$$$=\dfrac{\sum_{i=2}^ni}{2}=\dfrac{n^2+n-2}{4}=10 \ (given).$$ Therefore, we get $n=6, -7.$ But since $n$ is the index of a sum, it must be a natural number. Thus, $n=6$.
EDIT: As per user @Vishu's comment, $\displaystyle \lim_{x\to 0}\frac{y\sum_{i=2}^n\tan ix}{2x}$ can be simplified more easily as $\displaystyle \lim_{x\to 0}\dfrac y2\sum_{i=2}^n\frac{\tan ix}{x}=\displaystyle \lim_{x\to 0}\dfrac y2\sum_{i=2}^ni\frac{\tan ix}{ix}=\displaystyle \lim_{x\to 0}\dfrac y2\sum_{i=2}^n i$ by using the fact that $\lim_{u\to 0}\dfrac{\tan u}{u}=1.$