# Proving Limits of a

• Jan 10th 2013, 02:55 PM
mathisfun26
Proving Limits of a
$\displaystyle \lim_{x \rightarrow a} x = a$
this is my scratch work *

$\displaystyle 0< \left | x-a \right | < \delta$

$\displaystyle \left | x-a \right | < \epsilon$

I let
$\displaystyle \delta = \epsilon$
Im not sure if this is right it seems to simple. I am confused by this because i have only worked with numbers.
• Jan 10th 2013, 03:13 PM
Plato
Re: Proving Limits of a
Quote:

Originally Posted by mathisfun26
$\displaystyle \lim_{x \rightarrow a} x = a$
this is my scratch work *

$\displaystyle 0< \left | x-a \right | < \delta$

$\displaystyle \left | x-a \right | < \epsilon$

I let
$\displaystyle \delta = \epsilon$

It may be too simple, but it is correct.
• Jan 10th 2013, 03:16 PM
mathisfun26
Re: Proving Limits of a
Thank you!
• Jan 11th 2013, 06:24 PM
hollywood
Re: Proving Limits of a
It is simple, and I'll bet your thinking is right, but I have a few corrections.

There's a good paper on these epsilon-delta proofs here at MHF - it's one of the four "sticky" threads in this forum (the Calculus forum).

To prove $\displaystyle \lim_{x\to{a}}f(x)=L$:

For the scratch work, you start with $\displaystyle | f(x) - L | < \epsilon$ and work backwards to $\displaystyle | x-a | <$ some function of $\displaystyle \epsilon$. That tells you what you need to use for $\displaystyle \delta$. In your case, this involves no steps (!), as you start with $\displaystyle | x-a | < \epsilon$ and it is already in the right format. So you set $\displaystyle \delta=\epsilon$, which is what you said. I'm not sure why you had $\displaystyle \delta$ earlier.

Now you need to write the proof. It looks like you might have skipped this step - which would be very important on an exam.....

I'm also a little confused as to what you mean by you've only worked with numbers.

- Hollywood
• Jan 11th 2013, 06:59 PM
Deveno
Re: Proving Limits of a
what hollywood is saying is: to prove a limit exists, GIVEN epsilon, you have to somehow produce delta. you can't "start with delta".

a proper proof (which hides the "scratch work", but that's ok, if sometimes mysterious) would go something like:

let ε > 0. then for δ = ε,

we have: 0 < |x - a| < δ implies |x - a| < ε.

that is, for f(x) = x, we have:

$\displaystyle \lim_{x \to a} f(x) = \lim_{x \to a} x = a$. since this also equals f(a), this shows that the identity function f(x) = x, is continuous (which really shouldn't be that surprising...if x is near a, then ...x is near a!).