Solved Epsilon-Delta Proof: Is it Coherent?

[razored]

[SOLVED] Epsilon Delta Proof

Does this limit proof make total sense? Given : "Show that \lim_{x \rightarrow 2} x^{2} = 4."

My attempt at it :0<|x^{2}-4|<\epsilon which can also be written as 0<|(x-2)(x+2)|<\epsilon.
0<|x-2|<\delta where \delta > 0. It appears that \delta = \frac {\epsilon}{x+2} which is the conversion factor. Which then by substitution, 0<|x-2|<\frac {\epsilon}{x+2}.

Here is the actual proof:
Choose \delta = \frac {\epsilon}{x+2}, given epsilon > 0 then if 0<|x-2|<\frac {\epsilon}{x+2} then 0<|(x-2)(x+2)|<\delta.

Is this explanation coherent? I actually have a slight idea of what I wrote. Hopefully I am on the right path.

 

[dynamicsolo]

One of the more rigorous mathematicians here may have more to say about this then I will, but I think you have the gist of it. The one safeguard that is imposed in epsilon-delta proofs when you have a non-constant upper limit on the inequality is to make that limit

min ( 1 , \frac {\epsilon}{x+2} ) ,

since the ratio can blow up at x = -2 and we've placed no restriction on the permitted values of x...

 

[razored]

One of the more rigorous mathematicians here may have more to say about this then I will, but I think you have the gist of it. The one safeguard that is imposed in epsilon-delta proofs when you have a non-constant upper limit on the inequality is to make that limit

min ( 1 , \frac {\epsilon}{x+2} ) ,

since the ratio can blow up at x = -2 and we've placed no restriction on the permitted values of x...

I can't believe I forgot to mention that!

 

[konthelion]

When you are doing \epsilon-\delta proofs, you have to say define what epsilon and delta are i.e. \forall \epsilon >0, \exists \delta such that etc.

 

[eok20]

Your \delta should not depend on x. As the post above mentions, the logic goes: for all \epsilon > 0 there exists \delta > 0 such that if |x-2| < \delta then |x^2 - 4| < \epsilon. Delta can, however, depend on epsilon. If |x-2| is bounded by \delta < 1, then what is a bound on |x+2|?

 

[dynamicsolo]

Your \delta should not depend on x.

Sorry -- quite so! It's been a while since I've looked at one of these proofs and I'm writing from someplace I don't have my books handy...

 

[razored]

My second attempt : Given any \epsilon > 0, choose \delta = \frac {\epsilon} {|x+2|} (x\neq -2) where \delta > 0, then |x-2||x+2|<\epsilon whenever |x-2|<\delta = \frac {\epsilon} {|x+2|}.

Is this right?

 

[razored]

In my revision, I've been told that I did not prove anything. Is this a complete proof?

 

[konthelion]

Let f(x)=x^2, x_{0}=2. \forall \epsilon>0,\exists \delta>0 such that |x^2-4|<\epsilon, |x-2|<\delta
Let \delta = \frac{\epsilon}{x+2}
... continue from here. "..." <\epsilon

OR let \delta=min\{ 1,\frac{\epsilon}{5} \} and use the triangle inequality.
QED

 

[razored]

Substituting \delta = \frac {\epsilon}{x+2} into |x-2|<\delta, we see that.. it works out? I do not know what to do from here.

Nor do I know the min way.

 

[konthelion]

Continue from the proof,
Let \epsilon>0, let \delta = min\{\frac{\epsilon}{5}\}

Assume that |x-2|<\delta

By Triangle inequality,
|x^2-4|=|(x+2)(x-2)|\leq |x+2||x-2|

Since |x-2|<\delta, then |x-2|<\epsilon/5

Then for 0<x<3 such that |x+2|=5 **note: this is why you get the 5 as the denominator under epsilon.

Thus, |x^2-4|\leq |x+2||x-2|&lt;5*\frac{\epsilon}{5}=\epsilon QED