Question

If $\underset{x\to 0}{lim}\frac{1-\sqrt{\cos 2x}.\sqrt[3]{3\cos 3x}.\sqrt[4]{4\cos 4x}......\sqrt[n]{n\cos nx}}{x^2}$ has the value equal to $14$, then the value of $n$ equals

Answer 1

We have,

$\cos nx=1-\frac{(nx{)}^2}{2!}+\frac{(nx{)}^4}{4!}-.......$.

Now, $(\cos nx{)}^{\frac{1}{n}}=1-n\frac{x^2}{2!}+\text{higer power of x}$.

$\therefore \sqrt{\cos 2x}.\sqrt[3]{3cos3x}.......\sqrt[n]{n\cos nx}=1-(1+2+3+.....+n)\frac{x^2}{2}+\text{higher power of x}$

or, $1-\sqrt{\cos 2x}.\sqrt[3]{3cos3x}.......\sqrt[n]{n\cos nx}=(1+2+3+.....+n)\frac{x^2}{2}-\text{higher power of x}$

or, $\frac{1-\sqrt{\cos 2x}.\sqrt[3]{3cos3x}.......\sqrt[n]{n\cos nx}}{x^2}=\frac{(1+2+3+...+n)}{2}-\text{higer power of x}$

$\therefore \underset{x\to 0}{lim}\frac{1-\sqrt{\cos 2x}.\sqrt[3]{3cos3x}.......\sqrt[n]{n\cos nx}}{x^2}=\frac{(1+2+3+...+n)}{2}=\frac{n(n+1)}{4}$

According to the problem,

$\frac{n(n+1)}{4}=14\Rightarrow n=7$