# Find the limits: \lim_{x\to\infty}\sqrt{\frac{8x^2-3}{2x^2+x}}

Find the limits:
$\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{8{x}^{2}-3}{2{x}^{2}+x}}$
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FieniChoonin
Given:
Consider rational function inside the radical. The highest power of x in the denominator is 2.
So divide numerator and denominator by x^2,
$\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{8{x}^{2}-3}{2{x}^{2}+x}}=\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{\frac{8{x}^{2}}{{x}^{2}}-\frac{3}{{x}^{2}}}{\frac{2{x}^{2}}{{x}^{2}}+\frac{x}{{x}^{2}}}}$
$=\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{8-\frac{3}{{x}^{2}}}{2+\frac{1}{x}}}$
Use the limit laws,
$\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{8{x}^{2}-3}{2{x}^{2}+x}}=\sqrt{\underset{x\to \mathrm{\infty }}{lim}\frac{\left(8-\frac{3}{{x}^{2}}\right)}{\left(2+\frac{1}{x}\right)}}$
$=\sqrt{\frac{\underset{x\to \mathrm{\infty }}{lim}\left(8-\frac{3}{{x}^{2}}\right)}{\underset{x\to \mathrm{\infty }}{lim}\left(2+\frac{1}{x}\right)}}$
$=\sqrt{\frac{8-0}{2+0}}$
$=\sqrt{4}$
$=2$
Therefore, $\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{8{x}^{2}-3}{2{x}^{2}+x}}=2$
Jeffrey Jordon