# 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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berggansS
The degree of both numerator and denominator is=2
So we divide numerator and denominator by ${x}^{2}$
Calculation:
$\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{8{x}^{2}-3}{2{x}^{2}+x}}$
$=\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{\frac{\left(8{x}^{2}-3\right)}{{x}^{2}}}{\frac{\left(2{x}^{2}+x\right)}{{x}^{2}}}}$
$=\underset{x\to \mathrm{\infty }}{lim}\sqrt{\frac{\left(8-\frac{3}{{x}^{2}}\right)}{\left(2+\frac{1}{x}\right)}}$
$=\sqrt{\frac{\left(8-0\right)}{\left(2+0\right)}}$
$=\sqrt{\frac{8}{2}}$
$=\sqrt{4}$
$=2$