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?