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 83690 by jagoll last updated on 05/Mar/20

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

Commented by mathmax by abdo last updated on 05/Mar/20

$$letf\left(x\right)=\frac{tan\left(\frac{\pi x+1}{2x+2}\right)}{x+1}\Rightarrow f\left(x\right)=\frac{tan\left(\frac{\pi }{2}\times \frac{x+\frac{1}{\pi }}{x+1}\right)}{x+1}$$$$=\frac{tan\left(\frac{\pi }{2}\times \left(\frac{x+1+\frac{1}{\pi }-1}{x+1}\right)\right)}{x+1}=\frac{tan(\frac{\pi }{2}(1+\frac{1-\pi }{\pi (x+1)})}{x+1}$$$$=\frac{tan(\frac{\pi }{2}+\frac{(1-\pi )}{2(x+1)})}{x+1}=-\frac{1}{tan\left(\frac{1-\pi }{2(x+1)}\right)(x+1)}$$$$\Rightarrow f\left(x\right)\sim -\frac{1}{\frac{1-\pi }{2(x+1)}\times (x+1)}=\frac{-2}{1-\pi }=\frac{2}{\pi -1}\Rightarrow$$$${lim}_{x\to +\mathrm{\infty }}f\left(x\right)=\frac{2}{\pi -1}$$

Answered by john santu last updated on 05/Mar/20

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{x+1}\cot (\frac{\pi }{2}-\frac{\pi x+1}{2x+2})=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\cot \left(\frac{2(\pi -1)}{4(x+1)}\right)}{x+1}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{(x+1)\tan \left(\frac{\pi -1}{2(x+1)}\right)}\times \frac{\left(\frac{\pi -1}{2(x+1)}\right)}{\left(\frac{\pi -1}{2(x+1)}\right)}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{2(x+1)}{(\pi -1)(x+1)}=\frac{2}{\pi -1}\Leftarrow \mathrm{the}\mathrm{answer}$$

Commented by jagoll last updated on 05/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{mister}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com