# Help with proof of the uniqueness of limits.

1. Sep 20, 2004

### Kelvie

Good evening,
I am a first year engineer here and a first time poster also.

I had a problem that has been bugging me for the last few days; after much head-scratching and tree-killing, I may have solved it. I am, however, not sure at all if all my assumptions along the way are correct. So I am here to seek wisdom.

The question deals with proving the uniqueness of limits.

Prove that all limits are unique.

The textbook got me started, it said to define
$$\lim_{x\to a} f(x) = L$$
$$\lim_{x\to a} f(x) = M$$

Assume $L \neq M$ and let $\frac{|L - M|}{3} = \epsilon$

So here goes my proof..

\begin{align*} &|L-M| = |L -M + f(x) - f(x)| = |(L-f(x)) + (f(x) - M)| \\ &|L-M| \leq |-1||f(x) - L| + |f(x) - M| = |f(x) - L| + |f(x) - M| \end{align*}

So by definition..
\begin{align*} 0 \leq |x-a| \leq \delta (\epsilon) \implies \substack{|f(x) - L| \leq \epsilon \\ and \\ |f(x) - M| \leq \epsilon} \\ \end{align*}
\begin{align*} \therefore |L-M| &\leq 2\epsilon \\ |L-M| + \epsilon &\leq 3\epsilon \\ |L-M| + \epsilon &\leq 3 \left(\frac{|L-M|}{3}\right) \\ |L-M| + \epsilon &\leq |L-M| \end{align*}

Which can not possibly be true, so I conclude that our initial assumption $L \neq M$ was false, and therefore L must equal M.

Is this not the way to answering the question? If not, how should I look at this problem? What should I have done differently? What other approaches should I take?

(Side note... I REALLY hate delta-epsilon proofs..)

Last edited: Sep 20, 2004
2. Sep 20, 2004

### arildno

"So I am here to seek wisdom."
Seek and ye shall recieve! (Welcome to PF!)
Yep, you've got it.
Nope, those proofs are cuties..

3. Sep 20, 2004

### Kelvie

So this is "hand-in"able? :P

Thanks for the timely reply, by the way.

4. Sep 20, 2004

### mathwonk

It is much easier in words. i.e. if L is the limit of f(x) as x approaches a, then the inverse image of every interval centered at L contains a punctured interval centered at a. same for M. But this is a contradiction if Ldoes not equal M, since then L,M are centers of disjoint intervals whose inverse images are thus disjoint. But no two punctured intervals centered at a are disjoint.

the only place epsilon is needed is to describe the size of the disjoint intervals centered at L,M, namely |L-M|/3.

Last edited: Sep 21, 2004
5. Sep 21, 2004

### Kelvie

Alright, this also makes sense, doesn't it?

We ignore the assumption L \neq M[/itex] and [tex]\epsilon = \frac{|L-M|}{3}[/itex] [tex] \begin{align*} &|L-M| = |L -M + f(x) - f(x)| = |(L-f(x)) + (f(x) - M)| \\ &|L-M| \leq |-1||f(x) - L| + |f(x) - M| = |f(x) - L| + |f(x) - M| \\ &|L-M| \leq |f(x) - L| + |f(x) - M| \end{align*}
$$\therefore |L-M| \leq 2\epsilon$$

And since $\epsilon$ is can be as arbitrarily small as we want it to be, $L-M$ must equal 0. Is this also correct?

Side note: When using $\delta - \epsilon$ proofs, do we use $\leq$ or $<$, or does it not matter? My prof uses the former, and the textbook uses the latter.

Can I still assume $|L-M| = 0$ if I use $\leq$ ? Or do I have to have it strictly less than $\epsilon$?

6. Sep 21, 2004

### mathwonk

that looks nice. as to your question, ask your self: is it true that if L is such that
0<= L <= a, for all positive a, then L is zero?

also: is it true that if L is such that
0<= L > a, for all positive a, then L is zero?

7. Sep 21, 2004

### Kelvie

If by the second statement you meant $0 \leq L < a$ then yeah, I guess they are equivalent statements.