As limits approach infinity

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

$$f(x)=\frac{x^2}{x^2-4}$$ 

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:

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

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...

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

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