SOLUTION: lim 4x-1/8X^2+3X+1 ->OO what is the trick to this

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Question 198498: lim 4x-1/8X^2+3X+1
->OO
what is the trick to this

Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
lim 4x-1/8X^2+3X+1
->OO
what is the trick to this
------------------------------
Since x is getting larger without bound,
it certainly is not zero.
---
So divide numerator and denominator by x^2 to get:
{[4x-1]/x^2}/{[8x^2+3x+1]/x^2
-----------
Expand that to get:
{(4/x - 1/x^2)] / [8 + 3/x + 1/x^2]
---------------
Take the limit of that as x goes to infinity.
Notice that the fractions with x in the denominator
approach zero as x approaches infinity.
--------------------
The limit is (0 - 0)/(8 + 0 + 0) = 0/8 = 0
====================================================
Hope that helps.
-------------------
what you are really finding is the horizontal asymptote.
=========================================================
Cheers,
Stan H.

Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!


Edwin's solution.
Stanbon is right. You are finding the horizontal asymptote.  Here's
a little more detailed way to look at it. 

 lim 
x->OO 


Finding the limit of a rational function as x approaches infinity depends
on this theorem that we accept on faith:

limx->oo = 0, if n>0 and k is any constant.

Method: Divide every term on top and bottom of 
by the largest power of x that appears in there.

So,

limx->oo 

becomes

limx->oo 

and upon simplifying the fractions, you have:

limx->oo

Now every term on top approaches 0 as x approaches infinity.
So the whole numerator approaches 0.  

And the two terms of the denominator other than the 8 both
approack 0, and so, the fraction approaches  

So the limit is 0.

Note:
[Note here that the denominator did not approach 
0 too, and we were saved from having to face that dilemma 
of having both a numerator and a denominator approach 0.
But that dilemma occurs in other problems you'll be studying.]

Edwin


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