Limit to infinity question 
Possible Duplicate:
Finding the limit of $\frac{Q(n)}{P(n)}$ where $Q,P$ are polynomials 

Does $\lim_{x\rightarrow \infty} \frac{5x}{(1+x^2)} = 0$ or $\lim_{x\rightarrow \infty} \frac{5x}{(1+x^2)} = 1$?
I am asking because I was wondering if $\infty^2$ at the denominator is "considered" bigger than $5\infty$. Or do we just take the above as $\frac{\infty}{\infty}$?
 A: You have (dividing through by $x$)
$$\lim_{x\to \infty} \frac{5x}{1+x^2} = \lim_{x\to \infty} \frac{5}{x^{-1} + x} = 0.$$
Of if you know about L'Hopital's rule:
$$\lim_{x\to \infty} \frac{5x}{1+x^2} = \lim_{x\to \infty} \frac{5}{2x} = 0.$$
Note that in general we can't talk about $\frac{\infty}{\infty}$. $\infty$ is not a number so we can't really use it as such. Usually in calculus when we talk about infinity we think of it in terms of limits.
A: The limit is  $0$. 
In a vague (and somewhat dangerous) sense, you are correct concerning $\infty^2$ verses $5\infty$. To see that the limit is $0$ more rigorously, divide everything in the limit expression by the dominate term $x^2$ (that, is multiply both numerator and denominator by $1/x^2$):
$$
{5x\over 1+x^2}= { {5x\over x^2}\over {1\over x^2}+{x^2\over x^2}}= { {5 \over x }\over {1\over x^2}+1}\quad\buildrel{x\rightarrow\infty}\over\longrightarrow \quad{0\over0+1}=0.
$$
A: We claim that the limit is zero. To prove this, we need to show that given $\epsilon > 0$, there exists $M \in \mathbb{R}$ such that $\forall x >M$, we have $$\displaystyle \left \lvert \frac{5x}{1+x^2} \right \rvert < \epsilon$$ Take $\displaystyle M = \frac{5}{\epsilon}$. We then have that $\forall x > M$, $$\frac{5x}{1+x^2} \leq \frac{5x}{x^2} = \frac{5}{x} < \frac5{M} = \epsilon.$$ Hence, given $\epsilon > 0$, we have that $$\displaystyle \left \lvert \frac{5x}{1+x^2} \right \rvert < \epsilon, \,\,\,\,\, \forall x > \frac5{\epsilon}.$$ Hence, the limit $\displaystyle \lim_{x \rightarrow \infty} \frac{5x}{1+x^2} = 0$.
A: The answer is $0$ because the degree of $x$ ($n$) in the denominator is bigger than numerator. 
