# How do you determine the limit of (x² - 3x - 2)/(x² - 5) as x approaches infinity?

Apr 28, 2016

${\lim}_{x \rightarrow \infty} \frac{{x}^{2} - 3 x - 2}{{x}^{2} - 5} = 1$

#### Explanation:

Given,

${\lim}_{x \rightarrow \infty} \frac{{x}^{2} - 3 x - 2}{{x}^{2} - 5}$

Divide every term by the term with the highest degree in the denominator.

$= {\lim}_{x \rightarrow \infty} \frac{{x}^{2} / \textcolor{b l u e}{{x}^{2}} - \frac{3 x}{\textcolor{b l u e}{{x}^{2}}} - \frac{2}{\textcolor{b l u e}{{x}^{2}}}}{\frac{\textcolor{b l u e}{{x}^{2}}}{\textcolor{b l u e}{{x}^{2}}} - \frac{5}{\textcolor{b l u e}{{x}^{2}}}}$

$= {\lim}_{x \rightarrow \infty} \frac{1 - \frac{3}{x} - \frac{2}{x} ^ 2}{1 - \frac{5}{x} ^ 2}$

Substitute $x = \infty$.

$= {\lim}_{x \rightarrow \infty} \frac{1 - \frac{3}{\infty} - \frac{2}{\infty} ^ 2}{1 - \frac{5}{\infty} ^ 2}$

Any constant divided by $\infty$ has a limit of $0$ since the quotient would approach $0$.

$= {\lim}_{x \rightarrow \infty} \frac{1 - 0 - 0}{1 - 0}$

$= {\lim}_{x \rightarrow \infty} \frac{1}{1}$

$= 1$