Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3\text{x}^2-4\text{x}^2+6\text{x}-1}{2\text{x}^3+\text{x}^2-5\text{x}+7}$

Answer

$\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3\text{x}^2-4\text{x}^2+6\text{x}-1}{2\text{x}^3+\text{x}^2-5\text{x}+7}$
$=\lim\limits_{\text{x}\rightarrow{\infty}}\frac{3-\frac{4}{\text{x}}+\frac{6}{\text{x}^2}-\frac{1}{\text{x}^3}}{2+\frac{1}{\text{x}}-\frac{5}{\text{x}^2}+\frac{7}{\text{x}^3}}$
$=\frac{3-0+0-0}{2+0-0+0}$
$=\frac32$

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