+0  
 
0
543
5
avatar+564 

What is the limit of (x^2)/(x^2 - 4) as x approaches infinity?

 

I have been having trouble with this one because of the x^2 at the top. I factored out the bottom but I don't know where to go from there. Is there a thereom that I'm not following/should know for a problem like this?

chilledz3non  Sep 5, 2014

Best Answer 

 #3
avatar+90180 
+15

This IS a seemingly confusing situation.......here's a page that may help explain Honga's answer.....

http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx

 

CPhill  Sep 8, 2014
 #1
avatar+169 
+10

$$f(x)=\frac{x^2}{x^2-4}$$ 

 

This function is quite similar to the function:

 

$$\forall\,{x},\,x\neq0:f(x)=\frac{x^2}{x^2}=1$$

 

The original function is different in the fact that the denominator is smaller than the numerator:

 

$$\frac{n}{d}=\frac{n}{n-4}$$

 

The fact that the denominator 'd' is four less than the numerator 'n', becomes more and more insignificant as 'n' becomes larger. Hence when 'n' becomes infinite the four becomes entirely meaningless. This means:

 

$$\lim_{x\to\infty}:f(x)=\frac{\infty}{\infty-4}=\frac{\infty}{\infty}=1$$

Honga  Sep 5, 2014
 #2
avatar+564 
0

Thanks so much for answering Honga.  

 

In my notes though, my professor told us that infinity divided by infinity is indeterminant. Are there exception to when it can be divided and the answer will be 1? 

Sorry for asking further into this on here. My professor isn't a very approachable guy...

chilledz3non  Sep 8, 2014
 #3
avatar+90180 
+15
Best Answer

This IS a seemingly confusing situation.......here's a page that may help explain Honga's answer.....

http://tutorial.math.lamar.edu/Classes/CalcI/LimitsAtInfinityI.aspx

 

CPhill  Sep 8, 2014
 #4
avatar+93691 
+5

I really should look at some of these tutorials as well. !  thanks Chris.    

Melody  Sep 9, 2014
 #5
avatar+564 
0

Thanks for the link CPhill I'll check it out. Also sorry for the late reply >_<

chilledz3non  Sep 11, 2014

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