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Question Number 140905 by bramlexs22 last updated on 14/May/21

$$\underset{x\to 0}{\lim }\frac{1-\cos (1-\cos \mathrm{x})}{\mathrm{x}(\tan \mathrm{x}-\mathrm{x})}=?$$

Answered by bramlexs22 last updated on 14/May/21

Answered by mathmax by abdo last updated on 14/May/21

$$\mathrm{let}\mathrm{L}\left(\mathrm{x}\right)=\frac{1-\cos (1-\mathrm{cosx})}{\mathrm{x}(\mathrm{tanx}-\mathrm{x})}$$$$\mathrm{we}\mathrm{have}1-\mathrm{cosx}\sim \frac{{\mathrm{x}}^2}{2}\Rightarrow \cos (1-\mathrm{cosx})\sim \cos \left(\frac{{\mathrm{x}}^2}{2}\right)\sim 1-\frac{{\mathrm{x}}^4}{8}\Rightarrow$$$$1-\cos (1-\mathrm{cosx})\sim \frac{{\mathrm{x}}^4}{8}$$$$\mathrm{tanx}\sim \mathrm{x}+\frac{{\mathrm{x}}^3}{3}\Rightarrow \mathrm{tanx}-\mathrm{x}\sim \frac{{\mathrm{x}}^3}{3}\Rightarrow \mathrm{x}(\mathrm{tanx}-\mathrm{x})\sim \frac{{\mathrm{x}}^4}{3}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{{\mathrm{x}}^4}{8}.\frac{3}{{\mathrm{x}}^4}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\frac{3}{8}$$

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