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 181104 by cortano1 last updated on 21/Nov/22

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln (1+\frac{4}{\mathrm{x}})}{\pi -\arctan \left(2\mathrm{x}\right)}=?$$

Commented by Frix last updated on 21/Nov/22

$$\mathrm{wrong}.$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln (1+\frac{4}{x})}{\pi -\arctan 2x}=\frac{0}{\frac{\pi }{2}}=0$$$${\mathrm{l}}^{\prime }\mathrm{H}\widehat{\mathrm{o}pital}\mathrm{is}\mathrm{not}\mathrm{allowed}\left(\mathrm{and}\mathrm{not}\mathrm{necessary}\right)$$$$\mathrm{in}\mathrm{this}\mathrm{case}$$

Answered by a.lgnaoui last updated on 21/Nov/22

$$f\left(x\right)=\frac{\ln (1+\frac{4}{\mathrm{x}})}{\pi -\arctan \left(2\mathrm{x}\right)}$$$$\mathrm{x}\to \mathrm{\infty }\arctan \left(2\mathrm{x}\right)\to \frac{\pi }{2}$$$${\lim }_{\mathrm{x}\to \mathrm{\infty }}f\left(x\right)=\frac{2}{\pi }{\lim }_{\mathrm{x}\to \mathrm{\infty }}\left[\ln (1+\frac{4}{\mathrm{x}})\right]=\frac{2}{\pi }\times \ln \left(1\right)$$$${\lim }_{\mathrm{x}\to \mathrm{\infty }}\frac{\ln (1+\frac{4}{\mathrm{x}})}{\pi -\arctan \left(2\mathrm{x}\right)}=0$$$$$$

Answered by Gamil last updated on 21/Nov/22

Commented by Frix last updated on 21/Nov/22

$$\mathrm{wrong}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com