What Is the Delta Epsilon Proof in Terms of Pizza Cooking?

[ineedhelpnow]

can someone explain it?

 

[MarkFL]

I can give you an example I posted as a solution at another forum a couple of years ago:

We are given to prove:

\(\displaystyle \lim_{x\to-2}\left(x^2-x-3\right)=3\)

For any given \(\displaystyle \epsilon>0\) we wish to find a \(\displaystyle \delta\) so that:

\(\displaystyle \left|x^2-x-3-3\right|=\left|x^2-x-6\right|<\epsilon\) whenever \(\displaystyle 0<|x+2|<\delta\)

To do this, consider:

\(\displaystyle \left|x^2-x-6\right|=|x+2||x-3|\)

Thus, to make:

\(\displaystyle |x+2||x-3|<\epsilon\)

we need only make:

\(\displaystyle 0<|x+2|<\frac{\epsilon}{|x-3|}\)

Now, we want to define \(\displaystyle \delta=\frac{\epsilon}{|x-3|}\), but we cannot because \(\displaystyle \delta\) is supposed to be dependent on \(\displaystyle \epsilon\) only. To get around this, we will replace \(\displaystyle |x-3|\) with a number \(\displaystyle M\) which satisfies:

\(\displaystyle |x-3|\le M\)

So now we may write:

\(\displaystyle |x+2|<\frac{\epsilon}{M}\)

and proceed as before, taking \(\displaystyle \delta=\frac{\epsilon}{M}\). But, we have a problem, as there is no number \(\displaystyle M\) that satisfies:

\(\displaystyle |x-3|\le M\) for all real numbers $x$. However, we are only interested in those close to \(\displaystyle x=1\). It doesn't matter how close, we just want to bound \(\displaystyle |x-3|\) by restricting $x$ near 1, and any restriction will do. For example, if we require:

\(\displaystyle 0<|x+2|<1\), i.e., $x$ should be less than 1 unit away from 1, or equivalently \(\displaystyle \delta=1\), then we have:

\(\displaystyle -1<x+2<1\)

\(\displaystyle -6<x-3<-4\)

We can now take \(\displaystyle M=6\), so we should let \(\displaystyle \delta=\frac{\epsilon}{6}\). Remember that we also need:

\(\displaystyle |x+2|<1\) so that we define \(\displaystyle \delta=\min\left(1,\frac{\epsilon}{6}\right)\), then:

\(\displaystyle 0<|x+2|<\delta\) implies \(\displaystyle |x+2|<1\) and \(\displaystyle |x+2|<\frac{\epsilon}{6}\). We can now write the proof.

Let \(\displaystyle \epsilon>0\) and define \(\displaystyle \delta=\min\left(1,\frac{\epsilon}{6}\right)\). Then if \(\displaystyle 0<|x+2|<\delta\), we have:

\(\displaystyle \left|x^2-x-6\right|=|x+2||x-3|\)

\(\displaystyle \left|x^2-x-6\right|<6|x+2|\) since \(\displaystyle |x+2|<\delta\) and \(\displaystyle \delta\le1\)

\(\displaystyle \left|x^2-x-6\right|<6\left(\frac{\epsilon}{6}\right)=\epsilon\) since \(\displaystyle |x+2|<\delta\) and \(\displaystyle \delta\le\frac{\epsilon}{6}\)

Therefore, we have shown that \(\displaystyle 0<|x+2|<\delta\) implies \(\displaystyle \left|x^2-x-6\right|<\epsilon\) which shows

\(\displaystyle \lim_{x\to1}\left(x^2-x-3\right)=3\) by definition.

 

[Prove It]

Think of yourself as working in a pizza shop. When the pizzas go in the oven, everybody has a picture of the perfectly cooked pizza. We could call this level of cooking "the limit". This level of cooking has a certain time associated to it.

Obviously it is going to be impossible to have the pizzas be in the oven for this exact ideal amount of time 100% of the time (it might not even be possible to do at all). But we each allow a certain TOLERANCE before we consider the pizza overcooked or undercooked.

The idea is, that as long as you are SUFFICIENTLY CLOSE to the time required for the ideal level of cooking, then you will be in this tolerance.

Also, as you gain experience, you are likely to get better at picking the right amount of time, and thus are going to decrease your tolerance. So your tolerance should decrease as you close in on the correct amount of time. This means that the amount of time is related to the tolerance.

That is all the precise definition of a limit is. If we set $\displaystyle \begin{align*} x \end{align*}$ to represent the amount of time in the oven, then $\displaystyle \begin{align*} f(x) \end{align*}$ represents the level of cooking.

We set $\displaystyle \begin{align*} L \end{align*}$ to represent the ideal level of cooking, and $\displaystyle \begin{align*} \epsilon \end{align*}$ to represent our tolerance (notice that this value would be freely chosen).

We can use $\displaystyle \begin{align*} a \end{align*}$ to represent the time associated with the ideal level of cooking, and $\displaystyle \begin{align*} \delta \end{align*}$ to represent the amount of leeway in time.

So going back to our original statement, we choose our cooking tolerance, some $\displaystyle \begin{align*} \epsilon > 0 \end{align*}$, and as long as you have the pizzas in the oven for an amount of time that is sufficiently close to the ideal time - in other words, if the absolute difference between the actual time and our ideal time is within a sufficiently close leeway, so $\displaystyle \begin{align*} 0 < |x - a| < \delta \end{align*}$ (we have to put 0 there because there's no guarantee it actually can be in there for the ideal amount of time), then we are certain that our pizza will have a level of cooking within our tolerance, in other words, $\displaystyle \begin{align*} \left| f(x) - L \right| < \epsilon \end{align*}$. And since the level of cooking is related to the time in the oven, then $\displaystyle \begin{align*} \delta \end{align*}$ must also be related to $\displaystyle \begin{align*} \epsilon \end{align*}$.

So putting it all together, when we think of a limit as an "ideal value" that gets squeezed in on, then $\displaystyle \begin{align*} \lim_{x \to a} f(x) = L \end{align*}$ if we can show that for all $\displaystyle \begin{align*} \epsilon > 0 \end{align*}$ that there exists a value $\displaystyle \begin{align*} \delta >0 \end{align*}$ which is a function of $\displaystyle \begin{align*} \epsilon \end{align*}$ such that $\displaystyle \begin{align*} 0 < |x - a| < \delta \implies \left| f(x) - L \right| < \epsilon \end{align*}$.