$\lim_{x \to 0} \frac{x^n}{\cos\sin x -\cos x}=l$, find $n$ such that $l$ is non zero finite real Problem: Consider $\lim_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$. The value of $n$ such that $l$ is non zero finite real is:
(a) $3$, (b) $4$, (c) $5$, (d) $6$
Solution: The given limit can be written as $$-\frac{1}{2}\lim_{x \to 0} \frac{x^n}{\sin\bigl(\frac{\sin x+x}{2}\bigr) \sin\bigl(\frac{sinx-x}{2}\bigr)}.$$
Please suggest how to proceed further ... getting no clue on this. Thanks.

$$\lim_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$$
By the clue
$$l = \frac{-1}{2} \lim_{x \to 0}\frac{x^n}{  \sin(\frac{\sin x + x}{2}  )\sin(\frac{\sin x − x}{2})}$$ 
I would use the approximation $$\sin(x) \approx x - \frac{x^3}{3!} = x - \frac{x^3}{6}$$ since $x \to 0$, then
$$l = \frac{-1}{2} \lim_{x \to 0}\frac{x^n}{  \sin(x + x^3/12  )\sin(-x^3/12)} = $$ 
$$=\frac{-1}{2} \lim_{x \to 0}\frac{x^n}{  \sin(x)\sin(-x^3/12)} = $$ 
$$=\frac{-1}{2} \lim_{x \to 0}\frac{x^n}{  x \cdot (-x^3/12)} = $$ 
$$=\frac{1}{2} \lim_{x \to 0}\frac{x^n}{ (x^4/12)} = 6 \lim_{x \to 0}{(x^{n-4})}$$ 
This limit is different than 0 or $\infty$ only for $n=4$.
