<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] 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 $\ne0, \infty$.
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