# How do you find lim (5y^3+3y^2+2)/(3y^3-6y+1) as y->-oo?

##### 2 Answers
Jun 16, 2017

The limit equals $\frac{5}{3}$

#### Explanation:

Divide by the highest power. Call the limit $L$.

$L = {\lim}_{y \to - \infty} \frac{\frac{5 {y}^{3} + 3 {y}^{2} + 2}{y} ^ 3}{\frac{3 {y}^{3} - 6 y + 1}{y} ^ 3}$

$L = {\lim}_{y \to - \infty} \frac{5 + \frac{3}{y} + \frac{2}{y} ^ 3}{3 - \frac{6}{y} ^ 2 + \frac{1}{y} ^ 3}$

Now consider the graph of $y = \frac{1}{x}$.

graph{y = 1/x [-10, 10, -5, 5]}

You should be able to see that the limit as x approaches negative infinity is $0$. Hence:

$L = \frac{5 + 0 + 0}{3 - 0 + 0}$

$L = \frac{5}{3}$

Hopefully this helps!

Jun 16, 2017

${\lim}_{y \to - \infty} \frac{5 {y}^{3} + 3 {y}^{2} + 2}{3 {y}^{3} - 6 y + 1} = \frac{5}{3}$

#### Explanation:

Divide by the highest denominator power: $\left({y}^{3}\right)$

${\lim}_{y \to - \infty} \frac{5 {y}^{3} + 3 {y}^{2} + 2}{3 {y}^{3} - 6 y + 1} \cdot \frac{\frac{1}{y} ^ 3}{\frac{1}{y} ^ 3}$

${\lim}_{y \to - \infty} \frac{\frac{5 {y}^{3}}{y} ^ 3 + \frac{3 {y}^{2}}{y} ^ 3 + \frac{2}{y} ^ 3}{\frac{3 {y}^{3}}{y} ^ 3 - \frac{6 y}{y} ^ 3 + \frac{1}{y} ^ 3}$

${\lim}_{y \to - \infty} \frac{5 + \frac{3}{y} + \frac{2}{y} ^ 3}{3 - \frac{6}{y} ^ 2 + \frac{1}{y} ^ 3}$

Take the limit for both the numerator and denominator:

$\frac{{\lim}_{y \to - \infty} 5 + \frac{3}{y} + \frac{2}{y} ^ 3}{{\lim}_{x \to - \infty} 3 - \frac{6}{y} ^ 2 + \frac{1}{y} ^ 3}$

Recall: ${\lim}_{y \to - \infty} \left(\frac{c}{y} ^ a\right) = 0$

Also: The limit of a constant is the constant itself!

Using this property you'll know that every fraction you see with a variable in the denominator, its limit is $0$

$\therefore {\lim}_{y \to - \infty} \frac{5 + 0 + 0}{3 - 0 + 0} = \frac{5}{3}$