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Mathematics 16 Online
OpenStudy (anonymous):

find the limit as x approaches infinity of x^2+x/3-x thanks

OpenStudy (anonymous):

An explanation of how to solve limits approaching infinity would be helpful. Thanks :)

OpenStudy (anonymous):

l'hopital's rule

OpenStudy (anonymous):

so the answer would just be 2x

OpenStudy (jdoe0001):

ahemm, something tells me joshy12 is not doing l'hopital yet if at all

OpenStudy (anonymous):

Do you know l'hopital's rule joselin?

OpenStudy (anonymous):

jdoe, actaully i do

OpenStudy (dumbcow):

there is no horizontal asymptote but there is a limit , its just that it goes to neg infinity

OpenStudy (jdoe0001):

@joselin12 ohh, okie dokie

OpenStudy (anonymous):

Activate l'hopital's rule and plug in infinity and tell me what you get

OpenStudy (anonymous):

\[\frac{( x^2-x)' }{(3-x)' }=\frac{ 2x-1 }{ -1 }=-2\infty-1= -\infty\]

OpenStudy (anonymous):

When I do l'hopital's rule i get 2x..

OpenStudy (anonymous):

You can't divide by parts! Look at how I did my operations

OpenStudy (anonymous):

Isn't it suppose to be 2x+1 ?

OpenStudy (anonymous):

I guess so, but it's irrelevant

OpenStudy (anonymous):

-2x+1

OpenStudy (anonymous):

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