How to calculate $\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$? Evaluate this limit
$$\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$$
What's the method?
The answer is $1000$.

This answer is pretty useless for the results per se, much shorter proofs have been given. Honestly, it took me some time to write it, and I could non just abandon it in the forest.
[EDIT] Its intent is to play with standard inequalities, that are useful to keep in mind, and keep them as long as possible, without using de l'Hôpital rule, which I revere for the insight. However it requires some care, and probably should mention Johan Bernoulli in its name. Keeping inequalities may provide you with convergence rate estimates.  
The function (call it $f(x)$) is even. Borrowing from @Paramanand Singh notation, it can be factorized as:
$$f(x) = \frac{1 - \left(\frac{\sin x }{x}\right)^n}{x^2 \left(\frac{\sin x }{x}\right)^n}\,.$$
We use  standard inequalities close enough to $0^+$:
$$ 1-ny \le (1-y)^n \le 1-ny + \frac{n(n+1)}{2} y^2$$
for $n>0$ and 
$$ x - \frac{x^3}{6} \le \sin x \le x - \frac{x^3}{6} +  \frac{x^5}{5!} $$
hence
$$ 1 - \frac{x^2}{6} \le \frac{\sin x }{x} \le 1 - \frac{x^2}{6} +  \frac{x^4}{5!}\,.$$
From the last one, we see that $\left(\frac{\sin x }{x}\right)^n \to 1$, so we forget about it and study $g(x) = f(x)\left(\frac{\sin x }{x}\right)^n$.
By replacing above $y$ by $ \frac{x^2}{6} $ or  $ \frac{x^2}{6} - \frac{x^4}{5!}$ , we have:
$$ n x^2 \left(\frac{1}{6} - \frac{x^2}{5!}\right) - \frac{n(n+1)x^4}{2}\left(\frac{1}{6} - \frac{x^2}{5!}\right) ^2\le 1 - \left(\frac{\sin x }{x}\right)^n \le n\left(\frac{x^2}{6} \right)$$
hence
$$\frac{n}{6}+a_nx^2+b_n x^4+c_nx^6\le g(x) \le \frac{n}{6}$$
with constants $a_n,b_n,c_n $ which you can compute explicitly, especially if you want some rate of convergence, for instance I got $a_n = -\frac{n(5n+8)}{360}$. And for your question of course the limit is $1000 = 6000/6$.
Here is a visual simulation of the bounds for function $g(x)$, only for $n=4$ for numerical reasons:

