Find the limits without L'Hôpital:$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=? $

Question

Find the limits without L'Hôpital's rule $$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=? $$ My Try: $$\lim_{ x \to 0 }\frac{\sin(\pi-x)-\sin x}{\tan(\pi+x)-\tan x}=?\\\lim_{ x \to 0 }\frac{2\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2})}{\frac{\sin(\frac{π}{2}-x)}{\cos(\pi+x)\cos(x)}}=\lim_{ x \to 0 }\frac{(2\cos x)(-\cos x)(\cos(\frac{\pi}{2}))}{\cos x}=0$$ but: $$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=-1/2$$ Where is my mistake?

Answer 1

You can write the Sin and the Tan as exponential series, and then you get

$$x-\sin(x)=\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040}-\frac{x^9}{362880}+O\left(x^{11}\right )$$

and

$$-\frac12 (x-\tan(x))=\frac{x^3}{6}+\frac{x^5}{15}+\frac{17 x^7}{630}+\frac{31 x^9}{2835}+O\left(x^{11}\right)$$

If you divide the two sums you get $$ 1-\frac{9 x^2}{20}+\frac{27 x^4}{1400}-\frac{27 x^6}{56000}+\frac{201 x^8}{43120000}+O\left(x^{10}\right)$$ which goes clearly to 1.