How Does Epsilon-Delta Definition Prove a Limit?

[misogynisticfeminist]

This is a worked example in the book, but i can't figure out why it is this way.

Using 0 < (x-a) < \delta and (f(x)-L) < \epsilon for lim_(x\rightarrow a) f(x) = L.

Prove that lim_{x\rightarrow2} (4x-3) = 5

I can understand the steps until

{4} (x-2) < \epsilon and 0 < (x-2) < \delta

then, they suddenly get, \epsilon = \delta/4...

Cant really understand how they arrive at there, and they don't even show that the the limit is 5.

I'm self taught too, so my understanding might be a little wrong.

 

[lurflurf]

\lim_{x\rightarrow a}f(x)=L
means in words
f(x) can be made as close to L as desired by making x sufficiently close to a.
(I will use d and h instead of greak leters)
so we want f(x) close to L
that is we chose some h>0 and we desire that
|f(x)-L|<h
when x is close to a
we make this concrete by saying
|f(x)-L|<h
whenever
0<|x-a|<d(h)
that is we look for a condition on the closeness to a that will assure f(x) is close to L. In general this condition depends upon h.
so in the problem at hand
f(x)=4x-3
a=2
L=5
so
|f(x)-L|=|4x-3-5|=|4x-8|=4|x-2|<h
this is assured if
|x-2|<h/4
by definition (if the limit exist) the condition is expressed
0<|x-2|<d(h)
d(h)=h/4
thus we have proven the given limit
any smaller d(h) would also do
so d(h)=h/9999999999999999
for example would do
the idea is just we want to bound f(x) near a
we know f(1.999)=4.996 say
or f(about 2)=about 5
so we want to know for what x is 4x-3 within h of 5
we see that it will be true when |x-2|<h/2
or we could say
5-|h|<f(2+h/4)<5+|h|
so we have a relation between closeness of x to a and closeness of f(x) to L
for example
5-.004<f(1.999)<5+.004

 

[HallsofIvy]

This is a worked example in the book, but i can't figure out why it is this way.

Using 0 &lt; (x-a) &lt; \delta and (f(x)-L) &lt; \epsilon for lim_(x\rightarrow a) f(x) = L.

Prove that lim_{x\rightarrow2} (4x-3) = 5

I can understand the steps until

{4} (x-2) &lt; \epsilon and 0 &lt; (x-2) &lt; \delta

then, they suddenly get, \epsilon = \delta/4...

Cant really understand how they arrive at there, and they don't even show that the the limit is 5.

First- BE CAREFUL! It is important to state definitions precisely. There are two errors I see in:

Using 0 &lt; (x-a) &lt; \delta and (f(x)-L) &lt; \epsilon for lim_(x\rightarrow a) f(x) = L.

The first is that you don't say how those inequalities are to be used! The second is that you are missing the absolute values signs.
The precise definition is: lim_(x\rightarrow a) f(x) = L if and only if, for any \epsilon&gt;0, there exist \delta&gt; 0 such that if 0&lt;|x-a|&lt; \delta then |f(x)-L|&lt; \epsilon.

Here, a= 2, f(x)= 4x- 3, and L= 5. In order to show that that definition holds (that the limit really is 5) we must show that, for any \epsilon&gt; 0 such a \delta[/b] really does exist. The best way to do that is to show how to find it!<br /> <br /> We <b>want[\b] |f(x)-L|= \epsilon. <br /> But |4x-3-5|= |4x- 8|= 4|x-2| so that is the same as 4|x-2|&amp;lt; \epsilon or |x-2|&amp;lt; \frac{\epsilon}{4}. <br /> Now compare that with |x-2|&amp;lt; \delta! Looks like a good choice for \delta would be \frac{\epsilon}{4} doesn&#039;t it?<br /> <br /> In fact, we can do exactly that. A &quot;strict&quot; proof would be: given \epsilon&amp;gt;0, let \delta= \frac{\epsilon}{4}. Then if |x-2|&amp;lt; \delta, |x- 2|&amp;lt; \frac{\epsilon}{4} so that 4|x-2|&amp;lt; \epsilon.<br /> But then |4x- 8|= |4x-3-5|= |f(x)- L|&amp;lt;\epsilon which what we needed to show.<br /> <br /> You are <b>right</b> to be a little puzzled because your book did not do it that way. Instead of starting from &quot;let \delta= \frac{\epsilon}{4}&quot;, as I did here, your text, like most texts, did the &quot;preliminary part&quot; &quot;If I want |f(x)- L|&amp;lt; \epsilon, how can I find \delta&quot; and worked the <b>opposite</b> way. That is sometimes called &quot;synthetic proof&quot;: start from what you want to show and arrive at something you know (or can make) true. Strictly speaking, a proof should go the <b>other</b> way- from what you know is true to what you want to prove! <br /> <br /> However, as long as everything you do is <b>reversible</b> (If you can go from 4|x-2|&amp;lt; \epsilon to |x-2|&amp;lt; \frac{\epsilon}{4}, you can certainly go from |x-2|&amp;lt; \frac{\epsilon}{4} to 4|x-2|&amp;lt; \epsilon.) then just doing it one way implies the other way.</b>