Clarification of Delta-Epsilon Proof

[Dethrone]

I read this somewhere:

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.

Somewhere in the middle, you have $\displaystyle -6<x-3<-4$, which, in my textbook says we can rewrite as $|x-3|<6$. Why does that hold? Doesn't that interval consist of $-6<x-3<6$, so you are actually enlarging the domain? Or is that why we need the notation, $\displaystyle \delta=\min\left(1,\frac{\epsilon}{6}\right)$ to say that even if it does takes on values of $-4 < x < 6$ which were part of the added domain, it must always be less than $1$ too?

For the notation $\delta=\min\left(1,\frac{\epsilon}{6}\right)$ , what does that actually mean? Does it mean that the value of which we are actually restricting $\delta$ should actually be the smaller of the two conditions, whatever it may be?

 

[Euge]

I read this somewhere:
Somewhere in the middle, you have $\displaystyle -6<x-3<-4$, which, in my textbook says we can rewrite as $|x-3|<6$. Why does that hold? Doesn't that interval consist of $-6<x-3<6$, so you are actually enlarging the domain? Or is that why we need the notation, $\displaystyle \delta=\min\left(1,\frac{\epsilon}{6}\right)$ to say that even if it does takes on values of $-4 < x < 6$ which were part of the added domain, it must always be less than $1$ too?

For the notation $\delta=\min\left(1,\frac{\epsilon}{6}\right)$ , what does that actually mean? Does it mean that the value of which we are actually restricting $\delta$ should actually be the smaller of the two conditions, whatever it may be?

Hi Rido12,

The inequality $-6 < x - 3 < -4$, although not equivalent to $|x - 3| < 6$, does imply $-6 < x - 3 < 6$ and therefore $|x - 3| < 6$.

Having $\delta = \min\left(1,\frac{\epsilon}{6}\right)$ means that $\delta$ is the smaller of the numbers $1$ and $\frac{\epsilon}{6}$. In particular, $\delta \le 1$ and $\delta \le \frac{\epsilon}{6}$. More precisely, if $\epsilon < 6$, then $\epsilon/6 < 1$ and thus $\delta = \frac{\epsilon}{6}$. On the other hand, if $\epsilon \ge 6$, then $\frac{\epsilon}{6} \ge 1$ and so $\delta = 1$.

 

[Dethrone]

Ah, that makes sense...while $-6 < x - 3 < -4$ implies $|x - 3| < 6$, the opposite isn't true, which is because we're saying that $x-3$ is within -6 and 6, which is true. Makes a lot of sense now!

My professor does/teaches these proofs in like 1 minute each, he doesn't even take time to breathe in the 50 minutes he lectures nor does he stop to clarify anything. So basically my class just scribbles everything down not knowing what is happening, which is also hard because he uses chalk faster than we can scribble on paper...(Crying)

 

[Euge]

Ah, that makes sense...while $-6 < x - 3 < -4$ implies $|x - 3| < 6$, the opposite isn't true, which is because we're saying that $x-3$ is within -6 and 6, which is true. Makes a lot of sense now!

My professor does/teaches these proofs in like 1 minute each, he doesn't even take time to breathe, nor does he stop to clarify anything. So basically my class just scribbles every down not knowing what is happening, which is also hard because he uses chalk faster than we can scribble on paper...(Crying)

Sounds like a fast paced class, but don't worry because there are people here who can break these concepts down for you so you can understand! :)