Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 148328 by liberty last updated on 27/Jul/21

$$\underset{x\to 0}{\lim }\frac{\sin 2\mathrm{x}+2\sin {}^2\mathrm{x}-2\sin \mathrm{x}}{\cos \mathrm{x}-\cos {}^2\mathrm{x}}=?$$

Answered by EDWIN88 last updated on 27/Jul/21

Answered by mathmax by abdo last updated on 27/Jul/21

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com