Limit of a sequence I

Calculus Level 3

$\large \lim _{n\to\infty }\left(\frac{n^2+1}{n^3+2n+1}+\frac{1}{5^n}+\frac{\cos (3n+4)}{n^2+1}\right)= \ ?$

1 solution

Dimitris Kouroupos
Dec 23, 2016

First examines the limit in pieces

$\large \lim _{n\to\infty }\left(\frac{1}{5^n}\right)= \ \large \lim _{n\to\infty }\left(\frac{1}{5}\right)^n= \ 0$
$\large \lim _{n\to\infty }\left(\frac{n^2+1}{n^3+2n+1}\right)= \large \lim _{n\to\infty }\left(\frac{n^2(1+1/n^2)}{n^3(1+2/n^2+1/n^3)}\right)= \ \large \lim _{n\to\infty }\left(\frac{(1+1/n^2)}{(1+2/n^2+1/n^3)}*1/n\right)= \ 0$
$\large \lim _{n\to\infty }\left(\frac{cos(3n+4)}{n^2+1}\right)= \ 0$

* BECAUSE *

- $\frac{1}{n^2+1}$ ≤ $\frac{cos (3n+4)}{n^2+1}$ ≤ $\frac{1}{n^2+1}$

and

$\large \lim _{n\to\infty }\left(\frac{n^2+1}{n^3+2n+1}+\frac{1}{5^n}+\frac{cos (3n+4)}{n^2+1}\right)= \ 0+0+0 =0$

Be careful of the small typo for the Squeeze Theorem where you imposed greater than instead of "less than".

Michael Huang - 4 years, 5 months ago

I would like to apologize for my mistake, thank you very much for your comment

dimitris kouroupos - 4 years, 5 months ago