# Limit to infinity question [duplicate]

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 $\frac{\infty}{\infty}$?

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## marked as duplicate by Marvis, Mariano Suárez-Alvarez♦May 7 '12 at 1:57

Have a look at this question and answers given there. – Martin Sleziak May 6 '12 at 16:11
It is best to avoid algebraic manipulation of the symbol $\infty$, that can easily lead to disaster. But if $M$ is humongous, then $5M$ is negligibly small compared to $1+M^2$, the $1+M^2$ in the denominator crushes poor $5M$. – André Nicolas May 6 '12 at 16:51
Thanks, most appreciated. – Gary May 6 '12 at 22:11
Now that this has answers (and even an accepted one) crafted for this special case, I'll mark it as a duplicate of the question mentioned by Martin in the first comment and a few others in flags. – Mariano Suárez-Alvarez May 7 '12 at 1:57

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.

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Thanks, I did not think of it that we could use L'Hopital's rule! – Gary May 6 '12 at 15:23

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 $1/x^2$): $${5x\over 1+x^2}= { {5x\over x^2}\over {1\over x^2}+{x^2\over x^2}}= { {5 \over x }\over {1\over x^2}+1}\quad\buildrel{x\rightarrow\infty}\over\longrightarrow \quad{0\over0+1}=0.$$

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Thanks for your help, nice clear way to see it! – Gary May 6 '12 at 15:24

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 $\displaystyle \lim_{x \rightarrow \infty} \frac{5x}{1+x^2} = 0$.

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Thanks @Marvis, this is quite similar to the epsilon-delta continuity definition! – Gary May 6 '12 at 22:10

The answer is $0$ because the degree of $x$ ($n$) in the denominator is bigger than numerator.

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Thanks @Gigili ,nice pointer! – Gary May 6 '12 at 22:10