Without superior math, can we evaluate this limit?

Solution 1:

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.