<s>[INST] $\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of n such that l is non zero finite real number. Find l Problem : 
$\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of $n$ such that $l$ is non zero finite real number. Find value of $l$. 
My approach : 
$$\lim_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$$
$$ \Rightarrow \lim_{x \to 0} \frac{x^n}{-2\sin(\frac{\sin x+x}{2})\sin(\frac{\sin x-x}{2})} =l $$
Please suggest how to move further , will be of great help thanks. 
 [/INST] We have $cos x= \sum_{j=0}^{\infty}\frac{(-1)^j}{(2j)!}x^{2j}$ hence \begin{align}
\frac{x^n}{\cos \sin x -\cos x}&= \frac{x^n}{\sum_{j=0}^{\infty}\frac{(-1)^j}{(2j)!}(\sin^{2j}x-x^{2j})}\\
&= \frac{x^n}{\sum_{j=1}^{\infty}\frac{(-1)^j}{(2j)!}(\sin^{2j}x-x^{2j})}\\
&=\frac{1}{\color{blue}{\frac{(-1)^1}{2!}}\frac{\sin^{2}x-x^{2}}{x^n}+\frac{1}{4!}\frac{\sin^{4}x-x^{4}}{x^n}+...}\\
\end{align}
Now note that $$\sin^2x-x^2=\color{red}{-\frac13}x^4+\frac{2 x^6}{45}+O\left(x^8\right)$$ and $$\sin^4x-x^4=-\frac{2 x^6}{3}+\frac{x^8}{5}+O\left(x^{10}\right)$$ This indicates that choosing $n=4$ we will have a leading non-zero term in denumerator (coming from $(\sin^2x-x^2)/x^4$) making for a limit of $6$. Choosing $x>4$ say $x=6$, endows the second term a nonzero limit but the first term will be diverging making for a zero limit. Hence your choice is only $n=4$ and in this case the limit is $$\frac{1}{\color{blue}{\frac{(-1)^1}{2}}\color{red}{\frac{-1}{3}}}$$
</s>