# What is the limit of (7x^2+x-100)/(2x^2-5x)  as x approaches infinity?

Aug 11, 2015

${\lim}_{x \to \infty} f \left(x\right) = \frac{7}{2}$

#### Explanation:

Notice that you're dealing with the ratio of two polynomials that have the same degree, which means that the limit when $x \to \infty$ will be equal to the ratio of the coefficients that belong to the terms that have the highest degree.

$\frac{7 {x}^{\textcolor{red}{2}} + x - 100}{2 {x}^{\textcolor{red}{2}} - 5 x}$

The highest degree terms are $7 {x}^{2}$ and $2 {x}^{2}$, which means that the limit will be

${\lim}_{x \to \infty} \frac{\textcolor{b l u e}{7} {x}^{2} + x - 100}{\textcolor{b l u e}{2} {x}^{2} - 5 x} = \frac{\textcolor{b l u e}{7}}{\textcolor{b l u e}{2}}$

Alternatively, you can prove this by doing some algebraic manipulation of the fraction's numerator and denominator. More specifically, you can write

$7 {x}^{2} + x - 100 = {x}^{2} \cdot \left(7 + \frac{1}{x} - \frac{100}{x} ^ 2\right)$

and

$2 {x}^{2} - 5 x = {x}^{2} \cdot \left(2 - \frac{5}{x}\right)$

The limit now becomes

${\lim}_{x \to \infty} \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} \cdot \left(7 + \frac{1}{x} - \frac{100}{x} ^ 2\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} \cdot \left(2 - \frac{5}{x}\right)} = {\lim}_{x \to \infty} \frac{7 + \frac{1}{x} - \frac{100}{x} ^ 2}{2 - \frac{5}{x}}$

This is equal to

${\lim}_{x \to \infty} \frac{7 + \frac{1}{x} - \frac{100}{x} ^ 2}{2 - \frac{5}{x}} = \frac{7 + 0 + 0}{2 + 0} = \frac{7}{2}$