# Limit question factoring [duplicate]

Possible Duplicate:
How do I find the delta analytically for $f(x)$ with a degree other than $1$

There is a question, prove that:

$$\displaystyle \lim_{x\to3} x^{2} = 9$$

in linear problems of this type I simplify $0<|f(x)-L| \lt \epsilon$ to look like $0 \lt |x-a| \lt \delta$.

Example: $$\displaystyle \lim_{x\to3} 4x - 5 = 7$$

ends up $0 \lt |x-3| \lt \delta$ then $|(4x-5)-7| \lt e$ ends up $0 \lt |x-3| \lt \delta$ then $|x-3| \lt \epsilon/4$ then $\delta = \epsilon/4$

But with quadratics I have a problem with the factoring/simplification...

step 1. $|x^{2}-9| \lt \epsilon$

step 2. $|(x+3)(x-3)| \lt \epsilon$

now I know that I have to make $|(x+3)(x-3)| \lt \epsilon$ look like $0 \lt |x-3| \lt \delta$ but I don't know how to go about it. Any help would be appreciated!

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## marked as duplicate by Ross Millikan, Henry T. Horton, 5PM, Erick Wong, ncmathsadistFeb 5 '13 at 1:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

First we give an informal description, and then condense that into a one line formal argument at the end of the post.

We want control over $|x^2-9|$, so we want control over both $x+3$ and $x-3$. First we concentrate on controlling $|x+3|$, making sure it can't get too big.

Ultimately we will insist $x$ be reasonably close to $3$, in particular within $1$ of $3$. Suppose that $\delta \le 1$. That forces $x$ to be in the interval $[2,4]$, so it forces $5\le x+3\le 7$. Now $[x+3|$ cannot be very big.

Note that $x^2-9=(x+3)(x-3)$. So if $|x-3|\lt \delta$ and $\delta\le 1$, we will have $|x^2-3|\lt 7\delta$. Thus we will be OK if we furthermore make $7\delta$ less than $\epsilon$.

It follows that if $\delta\lt \epsilon/7$ and $\delta\le 1$, we will be OK. More briefly, we will be OK if $\delta=\min(1,\epsilon/7)$.

We described the thinking that goes into the following formal argument.

Let $\delta=\min(1,\epsilon/7)$. If $|x-3|\lt \delta$, then $|x+3|\lt 7$, and therefore $$|x^2-9|=|x+3||x-3|\lt 7\delta\lt \epsilon.$$

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Very nice exposition! –  robjohn Feb 5 '13 at 0:58
Thank You for your help Andre'! –  codenamejupiterx Feb 5 '13 at 4:37