Question 198498
<pre><font size = 4 color = "indigo"><b>

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

 lim {{{(4x-1)/(8x^2+3x+1)}}}
x->OO 


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

lim<sub>x->oo</sub>{{{k/x^n}}} = 0, if n>0 and k is any constant.

Method: Divide every term on top and bottom of {{{(4x-1)/(8X^2+3X+1)}}}
by the largest power of x that appears in there.

So,

lim<sub>x->oo</sub>{{{(4x-1)/(8x^2+3x+1))}}} 

becomes

lim<sub>x->oo</sub>{{{(4x/x^2-1/x^2)/(8x^2/x^2+3x/x^2+1/x^2))}}} 

and upon simplifying the fractions, you have:

lim<sub>x->oo</sub>{{{(4/x-1/x^2)/(8+3/x+1/x^2))}}}

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 {{{(0-0)/(8+0+0)=0/8=0}}} 

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</pre>