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 54992 by Mikael_Marshall last updated on 15/Feb/19

$$\underset{n\to \mathrm{\infty }}{lim}(\sqrt{n^3+n^2+n+1}-n)$$

Answered by kaivan.ahmadi last updated on 15/Feb/19

$$\times \frac{\sqrt{n^3+n^2+n+1}+n}{\sqrt{n^3+n^2+n+1}+n}=li\underset{n\to \mathrm{\infty }}{m}\frac{n^3+n+1}{\sqrt{n^3+n^2+n+1}+n}\approx$$$$li\underset{n\to \mathrm{\infty }}{m}\frac{n^3}{n^{\frac{3}{2}}}=\mathrm{\infty }$$$$$$

Answered by tm888 last updated on 16/Feb/19

$$=\mathrm{\infty }$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com