Question

If the value of $\underset{x\to 0}{lim}\left(\frac{x^n{\sin }^{n}x}{x^n-{\sin }^{n}x}\right)$ is non-zero finite, then $n$ is equal to

Answer 1

$\underset{x\to 0}{lim}\frac{x^n{\sin }^{n}x}{x^n-{\sin }^{n}x}=\underset{x\to 0}{lim}\frac{x^{2n}\frac{{\sin }^{n}x}{x^n}}{x^n-{\sin }^{n}x}$
$=\underset{x\to 0}{lim}\frac{x^{2n}}{x^n-{\sin }^{n}x}$

Now, using expansion we have:
$=\underset{x\to 0}{lim}\frac{x^{2n}}{x^n-{(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots )}^n}$
$=\underset{x\to 0}{lim}\frac{x^n}{1-{(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots )}^n}$

Now, using binomial expansion, we have:
$=\underset{x\to 0}{lim}\frac{x^n}{n(\frac{x^2}{3!}-\frac{x^4}{5!}+\cdots )-\frac{n(n-1)}{2}\cdot {(\frac{x^2}{3!}-\frac{x^4}{5!}+\cdots )}^2\cdots }$
$\therefore$ For $n=2$ limit exist.