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 106173 by john santu last updated on 03/Aug/20

$$\underset{x\to \pi /2}{\lim }\left(\frac{1-\sin \mathrm{x}}{{\mathrm{x}}^2\cot {}^2\mathrm{x}}\right)?$$

Answered by bemath last updated on 03/Aug/20

$$\mathrm{set}\mathrm{x}=\frac{\pi }{2}+\mathrm{♭}\to \frac{4}{{\pi }^2}\times \underset{\mathrm{♭}\to 0}{\lim }\frac{1-\sin (\frac{\pi }{2}+\mathrm{♭})}{\cot {}^2(\frac{\pi }{2}+\mathrm{♭})}=$$$$\frac{4}{{\pi }^2}\times \underset{\mathrm{♭}\to 0}{\lim }\frac{1-\cos \mathrm{♭}}{\tan {}^2\mathrm{♭}}=\frac{4}{{\pi }^2}\times \underset{\mathrm{♭}\to 0}{\lim }\frac{2\sin {}^2\left(\frac{\mathrm{♭}}{2}\right)}{\tan {}^2\mathrm{♭}}$$$$=\frac{8}{{\pi }^2}\times {\left[\underset{\mathrm{♭}\to 0}{\lim }\frac{\sin \left(\frac{\mathrm{♭}}{2}\right)}{\tan \mathrm{♭}}\right]}^2$$$$=\frac{8}{{\pi }^2}\times \frac{1}{4}=\frac{2}{{\pi }^2}★$$

Answered by mathmax by abdo last updated on 03/Aug/20

$$\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)=\frac{1-\mathrm{sinx}}{{\mathrm{x}}^2{\mathrm{cotan}}^2\mathrm{x}}\Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\frac{{\sin }^2\mathrm{x}(1-\mathrm{sinx})}{{\mathrm{x}}^2{\cos }^2\mathrm{x}}$$$$\mathrm{changement}\mathrm{x}=\frac{\pi }{2}-\mathrm{t}\mathrm{give}\mathrm{g}\left(\mathrm{x}\right)=\mathrm{h}\left(\mathrm{t}\right)=\frac{{\cos }^2\mathrm{t}(1-\mathrm{cost})}{{(\frac{\pi }{2}-\mathrm{t})}^2{\sin }^2\mathrm{t}}$$$$(\mathrm{x}\to \frac{\pi }{2}\Rightarrow \mathrm{t}\to 0)\Rightarrow \mathrm{h}\left(\mathrm{t}\right)\sim \frac{{(1-\frac{{\mathrm{t}}^2}{2})}^2.\frac{{\mathrm{t}}^2}{2}}{{(\frac{\pi }{2}-\mathrm{t})}^2{\mathrm{t}}^2}=\frac{1}{2}{\left(\frac{1-\frac{{\mathrm{t}}^2}{2}}{\frac{\pi }{2}-\mathrm{t}}\right)}^2$$$$\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{h}\left(\mathrm{t}\right)=\frac{1}{2}\left(\frac{4}{{\pi }^2}\right)=\frac{2}{{\pi }^2}={\lim }_{\mathrm{x}\to \frac{\pi }{2}}\mathrm{g}\left(\mathrm{x}\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com