# Thread: Using epsilon-delta definition of a limit

1. ## Using epsilon-delta definition of a limit

Hi there
I'm doing a worked example in a Larson book. I don't understand some of it. I have scanned this excerpt from Larson.

So what I don't understand is "Thus, letting $\delta$ be the minimum of $\varepsilon/5$ and 1, it follows that ....."
I don't understand why he is saying letting delta be the minimum of epsilon/5 and 1, and where he is getting 1 from? I'm thinking 1 is related to the solution of |x - 2| in the interval (1,3). In that interval the largest number that |x-2| can evaluate to is 1, or almost 1 and when that happens |x+2| = 5 or almost 5 because it's an open interval.
Am I thinking correctly? And does anybody else have anything to add to enhance my understanding becauuse even if I'm thinking right there, I still don't feel fully understand what he is doing in this example.
Dave.

2. ## Re: Using epsilon-delta definition of a limit

Do you under why any of this is done. We want to make the distance between $4$ and $x^2$, $|x^2-4|$, as small as necessary when $x$ is close to $2$, that is the distance $|x-2|$ small enough to make that happen. Because $x\to 2$ you can control the distance $x$ is from $2$.

Now Larson used 1, let use use 3. So we start with $|x-2|<3$, $x$ is within three units of $2$.
Basic algebra: $\left\{ \begin{array}{l}|x-2|<3\\-3<x-2<3\\-1<x<5\\-7<1<x+2<7\\|x+2|<7\end{array} \right.$
The last step may take some thought.

Most students can follow algebra, even the fact $-7<1$ is true, but WHY go after $|x+2|~?$
The answer is so mundane: $|x^2-4|=|x+2|\cdot |x-2|$ that is why. We want to make that as small as possible.

If we insure that $|x-2|<3$ we know that $|x+2|<7$.
Thus take any positive $c>0$ if we say $d=\min\{c/7,3\}$ then $|x-2|<d\le 3$ insures that $|x+2|<7$.

Put that together to get $|x^2-4|=(|x-2|)(|x+2|)<(c/9)(9)<c$ DONE.

3. ## Re: Using epsilon-delta definition of a limit

I am still trying to digest this to be honest. I tink I need to go and do some more reading and come back to this, and I am totally confused as to where the 9s came from.

4. ## Re: Using epsilon-delta definition of a limit

I can understand everything up to "Thus take an positive c > 0......"

5. ## Re: Using epsilon-delta definition of a limit

Originally Posted by Bucephalus
I am still trying to digest this to be honest. I tink I need to go and do some more reading and come back to this, and I am totally confused as to where the 9s came from.
It could be any number greater or equal to then seven. Once again, we need to bound the term $|x+2|$ because we have control over the term $|x-2|$.

Originally Posted by Bucephalus
I can understand everything up to "Thus take an positive c > 0......"
It is part of the definition: $\displaystyle{\lim _{x \to a}}f(x) = L$ means
if $c>0$ there is a positive number $d$ such that if $|x-a|<d$ then $|f(x)-L|<c.$

7. ## Re: Using epsilon-delta definition of a limit

Originally Posted by Plato

It is part of the definition: $\displaystyle{\lim _{x \to a}}f(x) = L$ means
if $c>0$ there is a positive number $d$ such that if $|x-a| then $|f(x)-L|
Yes i'm familiar with this definition. I'm just not getting the d=min{...} stuff, but I think it's only because I'm not conceptualising the |x-2| < d relationship with c yet. I have to go and concentrate, but before I do that I will go and read that document that Jeff posted.

8. ## Re: Using epsilon-delta definition of a limit

Originally Posted by JeffM
Thanks JeffM for the reference. It filled in all the holes that were troubling me.
Thanks Plato for helping me out too.
Dave.