Sorry, this is probably a really stupid question. But I'd appreciate an answer :)

What is the difference between DELTA Y and EPSILON when solving limit proofs?

- Jul 29th 2010, 01:08 PMjakobsandbergWhat is the difference between DELTA Y and EPSILON when solving proofs?
Sorry, this is probably a really stupid question. But I'd appreciate an answer :)

What is the difference between DELTA Y and EPSILON when solving limit proofs? - Jul 29th 2010, 08:12 PMMath Major
So, the definition of $\displaystyle \mathop{\lim}\limits_{x \to c} f(x) = L $ is that $\displaystyle \forall \epsilon > 0$ $\displaystyle \exists \delta > 0 : \forall x, 0 < | x - c| < \delta \rightarrow |f(x) - L| < \epsilon$.

Whew, that's a mouthfull, so let's break that down some. Firstly, the condition is say that for any positive number $\displaystyle \epsilon $, (no matter how super-small!), I can come up with another number $\displaystyle \delta > 0 $ so that when my x is a $\displaystyle \delta$-distance away from c, $\displaystyle f(x) $ will be within $\displaystyle \epsilon$-distance from L.

Think of the condition like a game. Someone gives you a number $\displaystyle \epsilon$. All you know about it is that it's positive. You get no other information. What you're trying to do is pick a $\displaystyle \delta$ (you pick $\displaystyle \delta$) so that the condition of the limit is satisfied. - Jul 29th 2010, 08:20 PMMath Major
Just to clarify, let me work out an easy example. Say you want to compute $\displaystyle \mathop{\lim}\limits_{x \to 1} x = 1$. Let $\displaystyle \epsilon > 0 $ be given (I don't know what it is, but it's a fixed number). Now, I need to pick a $\displaystyle \delta > 0 $ so that my limit is satisfied. But, in this case, $\displaystyle f(x) = x $, so let $\displaystyle \delta = \epsilon $. That is, for whatever number $\displaystyle \epsilon $ you give me, I'll pick my $\displaystyle \delta $ to be the same thing!

So, let's test that this works. Suppose that $\displaystyle |x - 1| < \delta = \epsilon$. Then, quite necessarily, $\displaystyle |f(x) - 1| = |x - 1| < \epsilon $ precisely because I'm assuming that $\displaystyle |x - c| < \delta $. I hope that clears up how this process works. - Jul 30th 2010, 02:13 AMHallsofIvy
Can you give an example of a problem "solving limits" that involves a "delta y"? I have seen many proofs that a limit is as claimed that involved "epsilon" and "delta" but I have never seen one that involved "delta y".