$\lim_{x \to 0} \frac{x^n}{\cos\sin x -\cos x}=l$, 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.
 A: $$\lim_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$$
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 $n=4$.
A: Hint: Do you know how to work with power series?  If so, a good start is to find the power series for the function $f(x)=\cos(\sin(x))-\cos(x)$ about $x=0$; the relevant piece of information is the smallest power of $x$ that has a non-zero coefficient.
A: $\newcommand\O{\mathcal{O}}$You have $\cos x = 1-x^2/2+x^4/24+\O(x^6)$ and $\sin x=x-x^3/6+\O(x^5)$. Therefore
$$\cos\sin x-\cos x = \Bigl(1-\frac{\sin^2 x}{2} +\frac{\sin^4 x}{24}+\O(\sin^6 x)\Bigr) - \Bigl(1-\frac{x^2}{2}+\frac{x^4}{24}+\O(x^6)\Bigr).$$
You can now substitute for $\sin x$ and correctly eliminate the $\O$'s left. At some places, you can substitute $\sin x=x+\O(x^3)$ to make things shorter, I think you can think out where it is.
Remember that in the end, you need to have something like $x^n+\O(x^m)$ where $m$ is strictly larger than $n$.
A: Clearly we have
$\displaystyle \begin{aligned}l &= -\frac{1}{2}\lim_{x \to 0}\dfrac{x^{n}}{{\displaystyle \sin\left(\dfrac{\sin x + x}{2}\right)\sin\left(\dfrac{\sin x - x}{2}\right)}}\\
&= -\frac{1}{2}\lim_{x \to 0}\dfrac{x^{n}}{\left(\dfrac{\sin x + x}{2}\right)\left(\dfrac{\sin x - x}{2}\right)}\cdot\dfrac{\dfrac{\sin x + x}{2}}{\sin\left(\dfrac{\sin x + x}{2}\right)}\cdot\dfrac{\dfrac{\sin x - x}{2}}{\sin\left(\dfrac{\sin x - x}{2}\right)}\\
&= -\frac{1}{2}\lim_{x \to 0}\dfrac{x^{n}}{\left(\dfrac{\sin x + x}{2}\right)\left(\dfrac{\sin x - x}{2}\right)}\cdot 1\cdot 1\\
&= -2\lim_{x \to 0}\dfrac{x^{n}}{(\sin x + x)(\sin x - x)}\\
&= -2\lim_{x \to 0}\dfrac{x^{n}}{x\left(\dfrac{\sin x}{x} + \dfrac{x}{x}\right)x^{3}\left(\dfrac{\sin x - x}{x^{3}}\right)}\\
&= -2\lim_{x \to 0}\dfrac{x^{n - 4}}{(1 + 1)(-1/6)}\\
&= 6\lim_{x \to 0}x^{n - 4}\end{aligned}$
Clearly we can see that if $n > 4$ then $l = 0$ and if $n < 4$ then the above limit does not exist. Hence $n = 4$ and $l = 6$.
We have made use of $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$ and by L'Hospital's Rule $$\lim_{x \to 0}\frac{\sin x - x}{x^{3}} = \lim_{x \to 0}\frac{\cos x - 1}{3x^{2}} = -\frac{1}{3}\cdot\frac{1}{2} = -\frac{1}{6}$$
