Limit 0 < ( x − a ) < δ Help

[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 $$0 < (x-a) < \delta$$ 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.

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

Using $$0 < (x-a) < \delta$$ and $$(f(x)-L) < \epsilon$$ 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 $$\epsilon>0$$, there exist $$\delta> 0$$ such that if $$0<|x-a|< \delta$$ then $$|f(x)-L|< \epsilon$$.

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 $$\epsilon> 0$$ such a [tex]\delta[/b] really does exist. The best way to do that is to show how to find it!

We want[\b] $$|f(x)-L|= \epsilon$$.
But |4x-3-5|= |4x- 8|= 4|x-2| so that is the same as $$4|x-2|< \epsilon$$ or $$|x-2|< \frac{\epsilon}{4}$$.
Now compare that with $$|x-2|< \delta$$! Looks like a good choice for $$\delta$$ would be $$\frac{\epsilon}{4}$$ doesn't it?

In fact, we can do exactly that. A "strict" proof would be: given $$\epsilon>0$$, let $$\delta= \frac{\epsilon}{4}$$. Then if $$|x-2|< \delta$$, $$|x- 2|< \frac{\epsilon}{4}$$ so that $$4|x-2|< \epsilon$$.
But then $$|4x- 8|= |4x-3-5|= |f(x)- L|<\epsilon$$ which what we needed to show.

You are right to be a little puzzled because your book did not do it that way. Instead of starting from "let $$\delta= \frac{\epsilon}{4}$$", as I did here, your text, like most texts, did the "preliminary part" "If I want $$|f(x)- L|< \epsilon$$, how can I find $$\delta$$" and worked the opposite way. That is sometimes called "synthetic proof": start from what you want to show and arrive at something you know (or can make) true. Strictly speaking, a proof should go the other way- from what you know is true to what you want to prove!

However, as long as everything you do is reversible (If you can go from $$4|x-2|< \epsilon$$ to $$|x-2|< \frac{\epsilon}{4}$$, you can certainly go from $$|x-2|< \frac{\epsilon}{4}$$ to $$4|x-2|< \epsilon$$.) then just doing it one way implies the other way.