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

**jakobsandberg** · July 29th 2010, 01:08 PM

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?

**Math Major** · July 29th 2010, 08:12 PM

So, the definition of $\mathop{\lim}\limits_{x \to c} f(x) = L$ is that $\forall \epsilon > 0$ $\exists \delta > 0 : \forall x, 0 < | x - c| < \delta \rightarrow |f(x) - L| < \epsilon$ .

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

Think of the condition like a game. Someone gives you a number $\epsilon$ . All you know about it is that it's positive. You get no other information. What you're trying to do is pick a $\delta$ (you pick $\delta$ ) so that the condition of the limit is satisfied.

**Math Major** · July 29th 2010, 08:20 PM

Just to clarify, let me work out an easy example. Say you want to compute $\mathop{\lim}\limits_{x \to 1} x = 1$ . Let $\epsilon > 0$ be given (I don't know what it is, but it's a fixed number). Now, I need to pick a $\delta > 0$ so that my limit is satisfied. But, in this case, $f(x) = x$ , so let $\delta = \epsilon$ . That is, for whatever number $\epsilon$ you give me, I'll pick my $\delta$ to be the same thing!

So, let's test that this works. Suppose that $|x - 1| < \delta = \epsilon$ . Then, quite necessarily, $|f(x) - 1| = |x - 1| < \epsilon$ precisely because I'm assuming that $|x - c| < \delta$ . I hope that clears up how this process works.

**HallsofIvy** · July 30th 2010, 02:13 AM

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".

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