Question

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

Answer 1

The limit equals `5/3`

Explanation:

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

`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 `y = 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 = (5 + 0 + 0)/(3 - 0 + 0)`

`L = 5/3`

Hopefully this helps!

Answer 2

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

Explanation:

Divide by the highest denominator power: `(y^3)`

`lim_(y->-oo)(5y^3+3y^2+2)/(3y^3-6y+1)* (1/y^3)/(1/y^3)`

`lim_(y->-oo)((5y^3)/y^3+(3y^2)/y^3+2/y^3)/((3y^3)/y^3-(6y)/y^3+1/y^3)`

`lim_(y->-oo) (5+3/y+2/y^3)/(3-6/y^2+1/y^3)`

Take the limit for both the numerator and denominator:

`(lim_(y->-oo) 5+3/y+2/y^3)/(lim_(x->-oo) 3-6/y^2+1/y^3)`

Recall: `lim_(y->-oo)(c/y^a)=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`

`:.lim_(y->-oo) (5+0+0)/(3-0+0)=5/3`