# How do you find the limit of ((2x^2-6)/(5x-x^2)) as x approaches infinity?

Jan 27, 2016

${\lim}_{x \to \infty} \left(\frac{2 {x}^{2} - 6}{5 x - {x}^{2}}\right) = - 2$

#### Explanation:

This is one way to approach this problem :

${\lim}_{x \to \infty} \left(\frac{2 {x}^{2} - 6}{5 x - {x}^{2}}\right)$

If you notice the highest degree of both the denominator and numerator is 2. We can divide every term by ${x}^{2}$ to get

${\lim}_{x \to \infty} \left(\frac{\frac{2 {x}^{2}}{x} ^ 2 - \frac{6}{x} ^ 2}{\frac{5 x}{x} ^ 2 - \left({x}^{2} / {x}^{2}\right)}\right)$

${\lim}_{x \to \infty} \left(\frac{2 \frac{{\cancel{x}}^{2}}{\cancel{{x}^{2}}} - \frac{6}{x} ^ 2}{\frac{5 x}{x} ^ 2 - \cancel{{x}^{2} / {x}^{2}}}\right)$

${\lim}_{x \to \infty} \frac{2 - \frac{6}{x} ^ 2}{\frac{5}{x} - 1}$
${\lim}_{x \to \infty} \frac{2 - \frac{6}{\infty} ^ 2}{\frac{5}{\infty} - 1}$

Note: As the denominator get larger, the number will be smaller, and almost close to 0. We can stated as follow

$\frac{2 - 0}{0 - 1} = - 2$

${\lim}_{x \to \infty} \left(\frac{2 {x}^{2} - 6}{5 x - {x}^{2}}\right) = - 2$

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The other method is to use L'Hopitals' Rule

If we direct substitute we will get an intermediate form

Direct Sub: $\frac{2 {\left(\infty\right)}^{2} - 6}{5 \left(\infty\right) - {\left(\infty\right)}^{2}} = - \frac{\infty}{\infty}$
We can simply differentiate numerator and denominator separately like so

${\lim}_{x \to \infty} \left(\frac{2 {x}^{2} - 6}{5 x - {x}^{2}}\right)$

Note: Derivative of $2 {x}^{2} - 6 = 4 x$

Derivative of $5 x - {x}^{2} = 5 - 2 x$

We can rewrite it as : ${\lim}_{x \to \infty} \left(\frac{2 {x}^{2} - 6}{5 x - {x}^{2}}\right) = {\lim}_{x \to \infty} \left(\frac{4 x}{5 - 2 x}\right)$

by direct substitution, we get $\frac{- \infty}{\infty}$ = intermediate form

We can differentiate again to get

${\lim}_{x \to \infty} \left(\frac{2 {x}^{2} - 6}{5 x - {x}^{2}}\right) = {\lim}_{x \to \infty} \left(\frac{4 x}{5 - 2 x}\right) = {\lim}_{x \to \infty} \frac{4}{- 2} = - 2$