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 $1000$.
 A: 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
$$\lim_{x\rightarrow0}\frac{\sin x}x=\lim_{x\rightarrow0}\frac{\sin(x/3)}{(x/3)}=1$$
A: 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 $f(6000) = 1000$.
A: HINT
Divide denominator and numerator by $x^{6000}$ then use $\lim_{x \to 0} \frac {\sin(x)} x =1$
You get: $$\lim_{x \to 0} \frac{1 -  (\frac {\sin x} x)^{6000}}{x^2}  $$
Now apply L'Hospital:
$$
\lim_{x \to 0} \frac{- 6000 (\frac {\sin x} x)^{5999} (\frac {\sin x} x )'} {2x} = \lim_{x \to 0} \frac{- 6000 ( (\frac {\sin x} x )'} {2x}
$$ then again L'Hospital after calculating $(\frac {\sin x} x )'$ etc.
A: 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:

A: $$\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}=\lim_{x \to 0} \frac{\left(\frac{x}{\sin x}\right)^{6000} - 1}{x^2}$$
Lets call $u(x)=\frac{x}{\sin x}$ and apply L'Hospital twice:
$$\lim_{x \to 0} \frac{\left(\frac{x}{\sin x}\right)^{6000} - 1}{x^2}=3000\lim_{x \to 0}\left[ u'^2+u\cdot u''(x)\right]$$
Evaluate the last limit and note that $\lim_{x\to 0}\frac{x}{\sin x}=1$.
A: $$\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}} =\lim_{x \to 0} \frac{x-\sin x}{x^3}\cdot\lim_{x \to 0}\dfrac{\sum_{r=0}^{6000-1-1}x^r\sin^{6000-r}}{(\sin x)^{6000}}$$
Now,
$$\lim_{x \to 0}\dfrac{\sum_{r=0}^{6000-1}x^r\sin^{6000-1-r}}{(\sin x)^{6000}}=\sum_{r=0}^{6000-1}\left(\dfrac x{\sin x}\right)^r$$
For the rest, see http://www.enotes.com/homework-help/solve-lim-x-gt-0-x-sin-x-x-3-211979
A: 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
$$
\lim_{x\to0}\frac{1000x^{6002}+o(x^{6002})}{x^2(\sin x)^{6000}}
$$
A: A lovely cheat that complex analysis offers:
Recall that the Laurent expansion of $\csc(z)$ is $\frac{1}{z}(1+\frac{z^2}{6}+o(z^4))$.
Apply this mindlessly to the question to obtain
$\frac{z^{5998}}{(\sin(x))^{6000}}-\frac{1}{z^2}=\frac{z^{5998}}{z^{6000}}(1+\frac{6000}{6}z^{2}+o(z^4))-\frac{1}{z^2}$ and the answer pops right out.
