1. ## Epsilon delta proofs

I'm having trouble understanding this proof. I understand what the e-d definition is trying to say, and I understand the geometric argument, but... well, consider an example:

Prove that $\displaystyle \lim_{x \to 4} 4x-5 = 7$

So, we take the definition and plug in:

For $\displaystyle e, d > 0$

$\displaystyle 0 < |x-3| < d \to |f(x)-7| < e$

$\displaystyle 0 < |x-3| < d \to 4|x-3| < e$

$\displaystyle 0 < |x-3| < d \to |x-3| < \frac{e}{4}$

So, at this point, we have $\displaystyle d = \frac{e}{4}$.

So what we're saying is:

If the distance between x and 3 is less than $\displaystyle \frac{e}{4}$, then the distance between f(x) and 7 is less than e.

So? If e is 100, and if |x-3| < 25, then |f(x)-7| < 100.

If e is 10, and if |x-3| < 2.5, then |f(x)-7| < 10.

I can see intuitively that as x gets closer to 3, so does f(x) close in on 7, but it's exactly the same kind of reasoning we used in our informal idea of limits.

I dont' see how this is any more rigorous?

Can I just be mindless about this and find an equivalent statement for the definition (with values plugged in) that is easy to prove? Is that basically the method?

2. Originally Posted by rowe
I'm having trouble understanding this proof. I understand what the e-d definition is trying to say, and I understand the geometric argument, but... well, consider an example:

Prove that $\displaystyle \lim_{x \to 4} 4x-5 = 7$

So, we take the definition and plug in:

For $\displaystyle e, d > 0$

$\displaystyle 0 < |x-3| < d \to |f(x)-7| < e$

$\displaystyle 0 < |x-3| < d \to 4|x-3| < e$

$\displaystyle 0 < |x-3| < d \to |x-3| < \frac{e}{4}$

So, at this point, we have $\displaystyle d = \frac{e}{4}$.

So what we're saying is:

If the distance between x and 3 is less than $\displaystyle \frac{e}{4}$, then the distance between f(x) and 7 is less than e.

So? If e is 100, and if |x-3| < 25, then |f(x)-7| < 100.

If e is 10, and if |x-3| < 2.5, then |f(x)-7| < 10.

I can see intuitively that as x gets closer to 3, so does f(x) close in on 7, but it's exactly the same kind of reasoning we used in our informal idea of limits.

I dont' see how this is any more rigorous?

Can I just be mindless about this and find an equivalent statement for the definition (with values plugged in) that is easy to prove? Is that basically the method?
It is more "rigorous" because it is saying exactly what is meant, mathematically, by "x gets closer to 3" and "f(x) closes in on 7".

3. Forget about it. I can't explain why I don't understand this, my brain is too fried. Is the following proof correct:

Prove $\displaystyle \lim_{x \to 3} x^2 = 9$

$\displaystyle 0 < |x-3| < d \to |x^2 - 9| < e$

Thus:

$\displaystyle |(x+3)||(x-3)| < e$

$\displaystyle |x-3| < \frac{e}{|x+3|}$

Therefore:

$\displaystyle 0 < |x-3| < \frac{e}{|x+3|} \to |x^2 - 9| < e$

And then work backwards to prove? (Another thing I don't understand, isn't working backwards redundant since you just went the way you came?)

4. Originally Posted by rowe
Forget about it. I can't explain why I don't understand this, my brain is too fried. Is the following proof correct:

Prove $\displaystyle \lim_{x \to 3} x^2 = 9$

$\displaystyle 0 < |x-3| < d \to |x^2 - 9| < e$

Thus:

$\displaystyle |(x+3)||(x-3)| < e$

$\displaystyle |x-3| < \frac{e}{|x+3|}$

Therefore:

$\displaystyle 0 < |x-3| < \frac{e}{|x+3|} \to |x^2 - 9| < e$

And then work backwards to prove? (Another thing I don't understand, isn't working backwards redundant since you just went the way you came?)
As long as it is clear that every step is "reversible"- that is that you can work backwords, you don't have to actually do that. This is sometimes refered to as "synthetic proof".