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

##### 2 Answers

The limit equals

#### Explanation:

Divide by the highest power. Call the limit

#L = lim_(y-> -oo) ((5y^3 + 3y^2 + 2)/y^3)/((3y^3 - 6y + 1)/y^3)#

#L = lim_(y-> -oo) (5 + 3/y + 2/y^3)/(3 - 6/y^2 + 1/y^3)#

Now consider the graph of

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

You should be able to see that the limit as x approaches negative infinity is

#L = (5 + 0 + 0)/(3 - 0 + 0)#

#L = 5/3#

Hopefully this helps!

#### Explanation:

Divide by the highest denominator power:

Take the limit for both the numerator and denominator:

**Recall**:

**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