# Epsilondelta Proof

• Oct 16th 2012, 08:59 PM
brucewayne
Epsilondelta Proof
Based off of the pdf document "those who never learned to do epsilondelta." How do you solve for $\displaystyle |x-0|$ in the following $\displaystyle \lim_{x\to 0}\frac{1}{x^2-1}=1$. This is what I currently have: $\displaystyle \frac{|x^2|}{|x^2-1|}\leq \epsilon$
• Oct 16th 2012, 11:19 PM
johnsomeone
Re: Epsilondelta Proof
Quote:

Originally Posted by brucewayne
$\displaystyle \lim_{x\to 0}\frac{1}{x^2-1}=1$.

That's not true, so no correct proof can every demonstrate it.

$\displaystyle \lim_{x\to 0}\frac{1}{x^2-1}=-1.$
• Oct 17th 2012, 02:12 AM
FernandoRevilla
Re: Epsilondelta Proof
Quote:

Originally Posted by brucewayne
$\displaystyle \lim_{x\to 0}\frac{1}{x^2-1}=1$.[/TEX]

As johnsomeone pointed out, the limit is $\displaystyle -1$. Use that $\displaystyle \left |\dfrac{1}{x^2-1}+1\right |=x^2\dfrac{1}{|x^2-1|}$ and $\displaystyle \dfrac{1}{|x^2-1|}\leq \dfrac{4}{3}$ in $\displaystyle [-1/2,1/2]$.
• Oct 17th 2012, 03:46 AM
brucewayne
Re: Epsilondelta Proof
Yes, you are correct, it should equal -1.
• Oct 17th 2012, 04:11 AM
Deveno
Re: Epsilondelta Proof
also, just to point out: there is nothing special about 1/2, we can pick ANY a with 0 < a < 1, in which case:

$\displaystyle |x| < a \implies \frac{1}{|x^2 - 1|} < \frac{1}{1-a^2}$

if we call this number L, what can we say about x for which:

$\displaystyle |x| < \min\left\{a,\sqrt{\frac{\epsilon}{L}}\right\}$ ?

(choosing a close to 1 will make L very large, which will make our choice of delta fairly small. that's ok. choosing a close to 0 will make L a bit larger than 1, and whether or not a is smaller than √(ε/L) depends on how small ε is. choosing a = 1/2 represents a "middle ground" where the numbers are easy to calculate with, in this case, and we can focus on ε, rather than the expressions involving a, which can be distracting. but a = 1/3, or a = 1/4, for example, would work just as well).

slight quibble:

since we are finding a LIMIT, rather than checking for continuity, we want a DELETED neighborhood of 0, that is: 0 < |x| < δ. in this example, it doesn't matter, the function being checked:

f(x) = 1/(x2 - 1) is continuous (but unbounded) on (-1,1), so on any sub-interval.

but, in general, f(0) doesn't have to exist for the limit of f(x) as x approaches 0 to exist, and even if f(0) does exist, it may not equal the limit. that is: for CONTINUOUS functions, you can find the limit:

$\displaystyle \lim_{x \to 0} f(x)$

by just finding f(0), but if one does not know whether a function f is continuous at 0 or not, to find the limit you don't let x "go to 0", it only gets "near it".

the classic example of this is:

$\displaystyle \lim_{x \to 0} \frac{x}{x} = 1$.

note we cannot consider values in any interval containing 0, since x/x is undefined at 0. if we define:

g(x) = x/x , x ≠ 0
g(0) = 5

g is a perfectly good function, and the limit is still 1 at 0, but it isn't true that for ALL x with |x| < δ, |g(x) - 1| < ε, there is one exception: x = 0, in which case
|g(x) - 1| = 4 (and so if ε < 4, we're toast).