# Prove that $\lim \limits_{n\to\infty}\frac{n}{n^2+1} = 0$ from the definition

This is a homework question:

Prove, using the definition of a limit, that $$\lim_{n\to\infty}\frac{n}{n^2+1} = 0.$$

Now this is what I have so far but I'm not sure if it is correct:

Let $\epsilon$ be any number, so we need to find an $M$ such that:

$$\left|\frac{n}{n^2 + 1}\right| < \epsilon \text{ whenever }x \gt M.$$

$$n \lt \epsilon(n^2 + 1)$$

$$n \lt \epsilon n^2 + \epsilon$$

Now what?

I am completely clueless on how to do this!

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try completing the square –  badatmath Oct 24 '11 at 16:32
Sorry, but since you (a) have shown work, (b) were able to identify and apply the right definition to use, and (c) have done some valid algebraic rewritings, you're ineligible for calling yourself clueless. –  Henning Makholm Oct 24 '11 at 16:33
I mean that I am clueless on where to go from here, sorry! –  Logan Serman Oct 24 '11 at 16:35

You had the right start. We want to show that for any $\epsilon>0$, there exists an integer $N$ such that $\left|\frac{n}{n^2+1}-0\right| <\epsilon$ whenever $n>N$. Or alternately, we want to show that there is an $M$, not necessarily an integer, such that $\left|\frac{n}{n^2+1}-0\right| <\epsilon$ whenever $n>M$.

We solve the problem by producing a suitable $N$ (alternately, $M$).

So let $\epsilon>0$ be given. Note that if $n>0$ then
$$\left|\frac{n}{n^2+1}-0\right|=\frac{n}{n^2+1}<\frac{n}{n^2}=\frac{1}{n}.$$

So if we choose $n$ positive, it is enough to make sure that $\frac{1}{n}<\epsilon$. This inequality is easy to solve for $n$.

The inequality can be rewritten as $n >\frac{1}{\epsilon}$. It will hold if $n > \lceil \frac{1}{\epsilon}\rceil$. So we can choose, for example, $N=\lceil\frac{1}{\epsilon}\rceil$. Here by $\lceil x \rceil$ we mean the smallest integer which is $\ge x$.

Alternately and more simply, if $M=\frac{1}{\epsilon}$, then the desired inequality holds whenever $n>M$.

We have shown that given $\epsilon>0$, if we choose $N=\lceil\frac{1}{\epsilon}\rceil$, then $\left|\frac{n}{n^2+1}-0\right|<\epsilon$ whenever $n>N$. By definition we therefore have $$\lim_{n\to \infty}\frac{n}{n^2+1} =0.$$

Comment: There is no need to solve the inequality $\frac{n}{n^2+1}<\epsilon\:$ exactly. We are not being asked to find the cheapest $N$ such that if $n >N$, then the inequality holds. All we are asked to do is to show that there is such an $N$. It can be very helpful, in this case and elsewhere, to replace our expression $\frac{n}{n^2+1}$ by something which is guaranteed to be larger (but still approaches $0$), and is substantially simpler. In the solution we replaced $\frac{n}{n^2+1}$ by $\frac{1}{n}$. This saved us from the slightly painful task of solving a quadratic inequality.

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I don't think $N$ necessarily need not be an integer –  Ramana Venkata Oct 24 '11 at 16:41
I do not understand how to go from this to a final answer. How do I actually close out a problem like this? –  Logan Serman Oct 24 '11 at 16:55
@Ramana Venketa: About whether we find an $N$, not necessarily an integer, or an $N$ which is an integer depends on the conventions being used in the course. I agree that fussing about making the thing an integer is essentially pointless. However, I have seen books in which by definition the "$N$" must be an integer, and the formal solution iincludes the fiddly little stuff with the "ceiling" function. –  André Nicolas Oct 24 '11 at 16:58
@Logan Serman: I have added a small concluding paragraph. Your solution started by saying that given an $\epsilon>0$ we needed to show that there is an $M$ such that $\dots$. I used the symbol $N$ instead. But apart from that small change, the solution I wrote out produces a suitable $M$ (and shows that it works). Once we have that $M$, the problem is finished. –  André Nicolas Oct 24 '11 at 17:06
@Ramana Venkata: Thanks for the comment. I added an alternative for courses that do not ask that "N" be an integer. –  André Nicolas Oct 24 '11 at 17:29
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First, $\epsilon$ should not be "any number", it should be "any positive number."

Now, you are on the right track. What do you need in order for $\frac{n}{n^2+1}$ to be smaller than $\epsilon$? You need $n\lt \epsilon n^2 + \epsilon$. This is equivalent to requiring $$\epsilon n^2 - n + \epsilon \gt 0.$$

You want to find out for what values of $n$ this is true. This is a quadratic inequality: you first solve $$\epsilon n^2 - n + \epsilon = 0,$$ and then you use the solution to determine where the quadratic is positive, and where it is negative. The answer will, of course, depend on $\epsilon$.

Using the quadratic formula, we have that $$\epsilon n^2 - n + \epsilon = 0$$ has solutions $$n = \frac{1 + \sqrt{1-4\epsilon^2}}{2\epsilon}, \quad n= \frac{1-\sqrt{1-4\epsilon^2}}{2\epsilon}.$$ That is, $$\epsilon n^2 - n + \epsilon = \epsilon\left( n - \frac{1-\sqrt{1-4\epsilon^2}}{2\epsilon}\right)\left(n - \frac{1+\sqrt{4\epsilon^2}}{2\epsilon}\right).$$

Now, we can assume that $\epsilon\lt \frac{1}{2}$, so that $4\epsilon^2\lt 1$ (if it works for all small enough $\epsilon$, then it works for all $\epsilon$. Since $\epsilon\gt 0$, then the quadratic is positive if $n$ is smaller than the smallest of the roots, or if $n$ is larger than the larger of the two roots. The larger root is $\displaystyle \frac{1 + \sqrt{1-4\epsilon^2}}{2\epsilon}$. So if $$n \gt \frac{1+\sqrt{1-4\epsilon^2}}{2\epsilon},$$ then $$\epsilon n^2 -n + \epsilon \gt 0.$$

Can you finish it up from here?

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P.S. I pushed forward your attempt, so you could see that it was a beginning that can be used to finish off the result. However, André Nicolas's answer is better because it is much simpler, and yet uses the definition and works. –  Arturo Magidin Oct 24 '11 at 16:54
Interesting! I never thought of solving these types of problems in such a way. –  Pedro Tamaroff May 1 '12 at 21:47

$n<\epsilon n^2 +\epsilon$ if and only if $n^2 - \frac{n}{\epsilon} + 1 > 0$, if and only if $(n - \frac{1}{2\epsilon})^2 + (1 - \frac{1}{4 \epsilon^2}) > 0$ (complete the square)

You should be able to do it from here! Good luck.

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$\frac{n}{n^2+1} = \frac{1}{n+\frac{1}{n}}$ for any $n$ in $\mathbb{R}^\star$, and for any $\varepsilon>0,n>0$ then $\frac{1}{n+\frac{1}{n}}<\varepsilon \Leftrightarrow n + \frac{1}{n} > \frac{1}{\varepsilon}$.
On the range $\left[1;+\infty\right[$ then $\frac{1}{n}$ is upper-bounded by $1$.