Without superior math, can we evaluate this limit? We all knew, with $$\lim_{x\to 0}\frac{\sin x - x}{x(1-\cos x)}$$ we can use L'Hôpital's rule or Taylor series to eliminate undefined form. But without all tools, by only using high school knowledge, how can we evaluate this limit? It seems difficult to transform numerator, any idea?
Thank you!
 A: I want to thank everyone for your help, after some hard work, i found the answer, i post it here for all of you (in case you all need reference later):
$$\lim_{x\to 0}\frac{\sin x - x}{x(1-\cos x)}=\lim_{x\to 0}\frac{\sin x - x}{2x\sin^2{\frac{x}{2}}}=\lim_{x\to 0}\frac{2(\sin x - x)}{x^3}\frac{\frac{x^2}{4}}{\sin^2{\frac{x}{2}}}=\lim_{x\to 0}\frac{2(\sin x - x)}{x^3}$$because $$\lim_{x\to 0}\frac{\frac{x^2}{4}}{\sin^2{\frac{x}{2}}}=1$$ when x approach to 0. Now, let $$L=\lim_{x\to 0}\frac{2(\sin x - x)}{x^3}(*)$$ use this indentity $$\sin x = 3\sin \frac{x}{3} - 4\sin^3\frac{x}{3}$$ $$=>L=\lim_{x\to 0}\frac{2(3\sin \frac{x}{3} - 4\sin^3\frac{x}{3} - x)}{x^3}=\lim_{x\to 0}\left(\frac{6}{27}\left(\frac{\sin\frac{x}{3}-\frac{x}{3}}{(\frac{x}{3})^3}\right)-\frac{8}{27}\right)$$ Now we can replace $$\lim_{x\to 0}\frac{\sin\frac{x}{3}-\frac{x}{3}}{(\frac{x}{3})^3}=\frac{L}{2}$$by comparing to (*), finally we have $$L=\frac{6}{27}\frac{L}{2}-\frac{8}{27}$$ solve this easy equation we conclude $$\lim_{x\to 0}\frac{\sin x - x}{x(1-\cos x)}=L=\frac{-1}{3}$$
And we have solved this limit without using L'Hopital's rule or Taylor series.
