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

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Answer 1 · 2017-06-16T02:02:15

The limit equals #5/3#

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 · 2017-06-16T02:08:41

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

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#