<s>[INST] $\lim_{x \to 0} \frac{x^n}{\cos\sin x -\cos x}=l$, 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.
 [/INST] 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 $$\lim_{x \to 0}\frac{\sin x - x}{x^{3}} = \lim_{x \to 0}\frac{\cos x - 1}{3x^{2}} = -\frac{1}{3}\cdot\frac{1}{2} = -\frac{1}{6}$$
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