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.
In order to simplify matters we are going to look at the reciprocal of the given expression. The denominator can be developed into a series as follows: From $$\cos x=1-{1\over2}x^2+{1\over24}x^4+?x^6$$ and $$\eqalign{\cos(\sin x) &=1-{1\over2}x^2\left(1-{1\over 6}x^2+?x^4\right)^2+{1\over24}x^4(1+?x^2)^4+?x^6\cr &=1-{1\over2}x^2+\left({1\over6}+{1\over24}\right)x^4+?x^6\cr}$$ it follows that $$\cos(\sin x)-\cos x={1\over6}x^4+?x^6\ .$$ This implies $$\lim_{x\to0}{\cos(\sin x)-\cos x\over x^4}={1\over6}\ ,$$ respectively: $$\lim_{x\to0}{x^4\over \cos(\sin x)-\cos x}={6}\ ,$$ and $n=4$ is the only exponent leading to a limit $\ne0, \infty$.
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 $$\frac{1}{\color{blue}{\frac{(-1)^1}{2}}\color{red}{\frac{-1}{3}}}$$
