# $\epsilon$-$\delta$ limit proof, $\lim_{x \to 2} \frac{x^{2}-2x+9}{x+1}$

Prove that $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ using an epsilon delta proof.

So I have most of the work done. I choose $\delta = min{\frac{1}{2}, y}$,
$f(x)$ factors out to $\frac{|x-3||x-2|}{|x+1|}$ But $|x-3| \lt \frac{3}{2}$ for $\delta = \frac{1}{2}$ and also $|x+1| > 5/2$ (I'll spare you the details).

I'm not sure how to choose my y here. If I take $\lim\limits_{x \to 2} \frac{x^{2}-2x+9}{x+1}$ < $(3/5) \delta$ How do I choose my epsilon here (replace y with this) to satisfy this properly?

Thanks

• $x^2-2x+9\neq (x-3)(x-2)$. – TCL Nov 25 '10 at 20:19
• You can simplify life a little if you divide $x+1$ into $x^2 - 2x + 9$ and prove your statement for what comes out. Do you have a candidate for a limit? – Gunnar Þór Magnússon Nov 25 '10 at 20:23
• @TCL. Try subtracting 3 first. – Aryabhata Nov 25 '10 at 20:26

## 2 Answers

First: You don't choose $\epsilon$!

In an $\epsilon$-$\delta$ proof, you begin by taking an arbitrary $\epsilon$; you then need to show that it is possible to choose a $\delta$ that "works" for that $\epsilon$. So you are going at things exactly backwards, by first trying to work with $\delta$.

Second: You cannot "prove $\lim\limits_{x\to 2}\frac{x^2-2x+9}{x+1}$" using $\epsilon$-$\delta$ proofs. What you can try to prove is that the limit is equal to something (in this case, to $3$).

Third: $f(x)$ does not factor as $\frac{|x-2||x-3|}{|x+1|}$; what factors that way is $|f(x)-3|$.

So. Let's start from scratch, shall we?

An $\epsilon$-$\delta$ proof that $\lim\limits_{x\to 2}\frac{x^2-2x+9}{x+1}=3$ would work as follows: given some arbitrary $\epsilon\gt 0$, we must find a $\delta\gt 0$ (which may depend on $\epsilon$), with the property that if $0\lt |x-2|\lt \delta$, then we will have $\left|\frac{x^2-2x+9}{x+1} - 3\right|\lt \epsilon$.

So, let $\epsilon\gt 0$ be given. You want to make sure that $$\left|\frac{x^2-2x+9}{x+1} - 3\right| = \left|\frac{x^2-2x+9-3x-3}{x+1}\right| = \left|\frac{x^2-5x+6}{x+1}\right| = \left|\frac{(x-3)(x-2)}{x+1}\right|$$ is small, by making sure that $|x-2|$ is small (that is, that $0\lt |x-2|\lt\delta$ for some chosen $\delta$).

Clearly, you also want to make sure that $\delta\lt \frac{|x+1|}{|x-3|}\epsilon$, because then we will have: $$\left|\frac{x^2-x+0}{x+1}-3\right| = \frac{|x-3||x-2|}{|x+1|} \lt \delta\frac{|x-3|}{|x+1|} \lt \frac{|x+1|\epsilon}{|x-3|}\frac{|x-3|}{|x+1|} = \epsilon.$$

You can make sure that $\frac{1}{|x+1|}$ is no larger than some constant by making sure that $x$ is close enough to $2$. You can do the same thing for $|x-3|$. So then you can pick a $\delta$ that is simultaneously small enough to ensure that $\frac{1}{|x+1|}$ is smaller than some $C$, that $|x-3|$ is smaller than some $D$, and that $|x-2|$ is smaller than $CD\epsilon$. Try that.

• Don't you mean $\delta < |x+1|/|x-3|$? +1 for a well written answer, anyway... – Aryabhata Nov 26 '10 at 1:53
• Thanks for a reply. I was in a rush when I posted this earlier I so sorry it was sloppy. I was trying to mention that by taking $\delta = \frac{1}{2}$ I had bounded $|x+1|$ below by $\frac{5}{2}$ and bounded $|x-3|$ above by $\frac{3}{2}$ – fdart17 Nov 26 '10 at 3:04
• By picking $\delta = min(\frac{1}{2}, \frac{3}{5}\epsilon)$ it worked out. A little note, don't we want to bound $|x+1|$ below since it is in the denominator? – fdart17 Nov 26 '10 at 3:11
• @Moron: Yes, that's what I meant. Thanks. – Arturo Magidin Nov 26 '10 at 4:57
• @Fdart17: You want to bound $|x+1|$ below, or equivalently, $\frac{1}{|x+1|}$ above; I talked about the latter, not the former. – Arturo Magidin Nov 26 '10 at 4:57

I'm going to go out on a limb and guess that you're trying to show the limit is 3 and that $f(x) = {x^2 - 2x + 9 \over x + 1} - 3$. I suggest trying to translate what you've done into the fact that $|{x^2 - 2x + 9 \over x + 1} - 3| < {3 \over 5}|x - 2|$ whenever $|x - 2| < {1 \over 2}$.

This means that if you choose any $\epsilon < {1 \over 2}$, then you have that $|{x^2 - 2x + 9 \over x + 1} - 3| < {3 \over 5}\epsilon$ whenever $|x - 2| < \epsilon$. So, given $\epsilon$, the natural choice for $\delta$ is ${3 \over 5}\epsilon$. (you got the $\delta$ and $\epsilon$ reversed.)

Now verify that the "for every $\epsilon$ there is a $\delta$" definition is satisfied in this way.