How to calculate $\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$?

Ok, one way to intimidate students is to use large numbers like 6000 in questions. A simple tool to beat this strategy of examiner is to replace the big number by a generic symbol say $n$. We thus calculate the limit $$f(n) = \lim_{x \to 0}\frac{x^{n} - \sin^{n}x}{x^{2}\sin^{n}x}$$ where $n$ is a positive integer. The answer for the question is $f(6000)$.

We have \begin{align} f(n) &= \lim_{x \to 0}\frac{x^{n} - \sin^{n}x}{x^{2}\sin^{n}x}\notag\\ &= \lim_{x \to 0}\dfrac{x - \sin x}{x^{3}\cdot\dfrac{\sin x}{x}}\cdot\dfrac{{\displaystyle \sum_{i = 0}^{n - 1}x^{i}\sin^{n - 1 - i}x}}{\sin^{n - 1}x}\notag\\ &= \lim_{x \to 0}\frac{x - \sin x}{x^{3}}\sum_{i = 0}^{n - 1}\left(\frac{x}{\sin x}\right)^{i}\notag\\ &= \sum_{i = 0}^{n - 1} 1\cdot \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{n}{6}\notag \end{align} and hence the desired answer is $f(6000) = 1000$.

You have $$ \sin x=x-x^3/6+o(x^3) $$ so $$ (\sin x)^{6000}=x^{6000}-6000\frac{x^{6002}}{6}+o(x^{6002}) $$ Hence your limit is $$ \lim_{x\to0}\frac{1000x^{6002}+o(x^{6002})}{x^2(\sin x)^{6000}} $$

Since $\sin(n+1)x=2\sin nx\cos x-\sin(n-1)x$, we have $$\begin{align}\sin3x&=2\sin2x\cos x-\sin x=2(2\sin x\cos x)\cos x-\sin x\\ &=4\sin x(1-\sin^2 x)-\sin x=3\sin x-4\sin^3x\end{align}$$ Then $\sin x=\sin(3(x/3))=3\sin(x/3)-4\sin^3(x/3)$. Then $$\begin{align}\frac{x^n-\sin^nx}{x^2\sin^nx}&=\frac{(3(x/3))^n-(3(\sin(x/3)-(4/3)\sin^3(x/3)))^n}{x^2\sin^nx}\\ &=\frac{(3(x/3))^n-3^n\sin^n(x/3)+n\cdot3^n(4/3)\sin^{n+2}(x/3)+O(\sin^{n+4}(x/3))}{x^2\sin^nx}\\ &=\frac{(3(x/3))^n-3^n\sin^n(x/3)}{x^2\sin^nx}+\frac{n\cdot3^n(4/3)\sin^{n+2}(x/3)}{x^2\sin^nx}+\frac{O(\sin^{n+4}(x/3)}{x^2\sin^nx}\\ &=\frac{(3(x/3))^n-3^n\sin^n(x/3)}{(3(x/3))^2(3^n\sin^n(x/3))+O(\sin^{n+2}(x/3)))}\\ &+\frac{n\cdot3^n(4/3)\sin^{n+2}(x/3)}{(3(x/3))^2(3^n\sin^n(x/3))+O(\sin^{n+2}(x/3)))}+\frac{O(\sin^{n+4}(x/3)}{x^2\sin^nx}\end{align}$$ So $$\begin{align}\lim_{x\rightarrow0}\frac{x^n-\sin^nx}{x^2\sin^nx}&=\lim_{x\rightarrow0}\frac{3^n}{3^{n+2}}\frac{(x/3)^n-\sin^n(x/3)}{(x/3)^2\sin^n(x/3)}\frac{1}{\left(1+\frac{O(\sin^{n+2}(x/3))}{3^n\sin^n(x/3)}\right)}\\&+\lim_{x\rightarrow0}\frac{4n\cdot3^{n-1}}{3^{n+2}}\frac{\sin^{n+2}(x/3)}{(x/3)^2\sin^n(x/3)}\frac{1}{\left(1+\frac{O(\sin^{n+2}(x/3))}{3^n\sin^n(x/3)}\right)}+\lim_{x\rightarrow0}\frac{O(\sin^{n+4}(x/3)}{x^2\sin^nx}\\ &=\frac19\lim_{x\rightarrow0}\frac{x^n-\sin^nx}{x^2\sin^nx}\frac1{1+0}+\frac{4n}{27}\frac1{1+0}+0=\frac98\frac{4n}{27}=\frac n6=\frac{6000}6=1000\end{align}$$ Sort of like the Taylor series approach but using trigonometric identities instead. That $O(\sin^{n+4}(x/3))$ term represents further terms of the binomial expansion which have at least the factor $\sin^{n+4}(x/3)$, so their limits when divided merely by $\sin^{n+2}(x/3)$ are all zero. We are, of course, using $$\lim_{x\rightarrow0}\frac{\sin x}x=\lim_{x\rightarrow0}\frac{\sin(x3)}{(x/3)}=1$$