Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$

Famously, a sector with a small angle proves $\lim_{x\to0}\tfrac{\sin x}{x}=1$ and hence$$\lim_{x\to0}\tfrac{1-\cos x}{x^2}=\lim_{x\to0}\tfrac{\sin^2\tfrac{x}{2}}{2(x/2)^2}=\tfrac12.$$We can improve the former result: if you approximate a sector's arc as a parabola, you can prove$$\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\tfrac16.$$Finally,$$\lim_{x\to0}\frac{\tfrac{\tan x}{x}-1}{x^2}=\lim_{x\to0}\frac{\tfrac{\sin x}{x}-\cos x}{x^2}=\lim_{x\to0}\frac{1-\cos x}{x^2}-\lim_{x\to0}\frac{1-\tfrac{\sin x}{x}}{x^2}=\tfrac12-\tfrac16=\tfrac13.$$