# What is the limit of (3x^2+20x)/(4x^2+9) as x goes to infinity?

Oct 10, 2015

${\lim}_{x \rightarrow \infty} \frac{3 {x}^{2} + 20 x}{4 {x}^{2} + 9} = \frac{3}{4}$

#### Explanation:

$\frac{3 {x}^{2} + 20 x}{4 {x}^{2} + 9} = \frac{{x}^{2} \left(3 + \frac{20}{x}\right)}{{x}^{2} \left(4 + \frac{9}{x} ^ 2\right)}$ for $x \ne 0$

$= \frac{3 + \frac{20}{x}}{4 + \frac{9}{x} ^ 2}$ for $x \ne 0$.

As $x \rightarrow \infty$, we get $\frac{20}{x} \rightarrow 0$ and $\frac{9}{x} ^ 2 \rightarrow 0$, so we have

${\lim}_{x \rightarrow \infty} \frac{3 {x}^{2} + 20 x}{4 {x}^{2} + 9} = {\lim}_{x \rightarrow \infty} \frac{3 + \frac{20}{x}}{4 + \frac{9}{x} ^ 2} = \frac{3}{4}$

Oct 10, 2015

$\frac{3}{4}$

#### Explanation:

${\lim}_{x \to \infty} \frac{3 {x}^{2} + 20 x}{4 {x}^{2} + 9} = {\lim}_{x \to \infty} \frac{\frac{3 {x}^{2}}{x} ^ 2 + \frac{20 x}{x} ^ 2}{\frac{4 {x}^{2}}{x} ^ 2 + \frac{9}{x} ^ 2} = {\lim}_{x \to \infty} \frac{3 + \frac{20}{x}}{4 + \frac{9}{x} ^ 2} = \frac{3}{4}$

Note:

$\frac{20}{x} \to 0$ when $x \to \infty$

$\frac{9}{x} ^ 2 \to 0$ when $x \to \infty$