Question

evaluate.

lim 2x^5+3x/2x^3-4x^2
x->inf

Answer 1

This is pretty much the same as all the other ones you've seen us do.. Did you attempt it yourself?

Answer 2

yea aren't i supposed to divide the everything by the highest power of x

Answer 3

Yes.

Answer 4

There is a calculus rule that says that you can cancel all but the highest power in a fraction when taking a limit to infinity

Answer 5

Really what you are doing is factoring out powers of x from the top and the bottom and cancelling them.

Answer 6

If you want a calculus proof
invert the fraction and you get
1/(2x^3-4x^2)/1/(2x^5+3x) where x goes to inf
thus you get 0/0
So you can take L'Hopital's rule
And eventually you get 120x^2 where x goes to inf
and that is equal to infinity

Answer 7

it should equal infinity

Answer 8

if you use the process of dividing everthing by x^5 and then setting x-> inf, then it would become 2/0
which is infinity

Answer 9

okay i get it now

Answer 10

No need for l'hopital..
Like I said, factor out powers of x.
\[{ 2x^5+3x\over 2x^3-4x^2} = {x^3(2x^2 + {3\over x^2}) \over x^3(2 - {4\over x})}\]
Cancel the \( x^3\) and you have:
\[{2x^2 + {3\over x^2} \over 2 - {4\over x}}\]

Answer 11

Which you can see will go to infinity because there's still an x squared in the numerator that goes to infinity.