# Find the following limit. lim_{nrightarrowinfty}frac{2n^2+2}{5n^3-3n-1}

Lipossig 2020-11-10 Answered
Find the following limit.
$\underset{n\to \mathrm{\infty }}{lim}\frac{2{n}^{2}+2}{5{n}^{3}-3n-1}$
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## Expert Answer

Caren
Answered 2020-11-11 Author has 96 answers
Solution:
$\underset{n\to \mathrm{\infty }}{lim}\frac{2{n}^{2}+2}{5{n}^{3}-3n-1}=\underset{n\to \mathrm{\infty }}{lim}\frac{{n}^{2}\left(2+\frac{2}{{n}^{2}}\right)}{{n}^{3}\left(5-\frac{3}{{n}^{2}}-\frac{1}{{n}^{3}}\right)}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{\left(2+\frac{2}{{n}^{2}}\right)}{n\left(5-\frac{3}{{n}^{2}}-\frac{1}{{n}^{3}}\right)}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}×\underset{n\to \mathrm{\infty }}{lim}\frac{\left(2+\frac{2}{{n}^{2}}\right)}{\left(5-\frac{3}{{n}^{2}}-\frac{1}{{n}^{3}}\right)}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}×\frac{\left(2+0\right)}{\left(5-0-0\right)}$
$=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}×\frac{2}{5}$
$=0×\frac{2}{5}$
$=0$
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Jeffrey Jordon
Answered 2022-04-01 Author has 2047 answers

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