Help with proof of the uniqueness of limits.

[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..

<br /> \begin{align*}<br /> &amp;|L-M| = |L -M + f(x) - f(x)| = |(L-f(x)) + (f(x) - M)| \\<br /> &amp;|L-M| \leq |-1||f(x) - L| + |f(x) - M| = |f(x) - L| + |f(x) - M|<br /> \end{align*}<br />

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

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..)

 

[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..

 

[Kelvie]

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

Thanks for the timely reply, by the way.

 

[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.

 

[Kelvie]

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

We ignore the assumption L \neq M[/itex] and \epsilon = \frac{|L-M|}{3}[/itex]&lt;br /&gt; &amp;lt;br /&amp;gt; \begin{align*}&amp;lt;br /&amp;gt; &amp;amp;amp;|L-M| = |L -M + f(x) - f(x)| = |(L-f(x)) + (f(x) - M)| \\&amp;lt;br /&amp;gt; &amp;amp;amp;|L-M| \leq |-1||f(x) - L| + |f(x) - M| = |f(x) - L| + |f(x) - M| \\&amp;lt;br /&amp;gt; &amp;amp;amp;|L-M| \leq |f(x) - L| + |f(x) - M|&amp;lt;br /&amp;gt; \end{align*}&amp;lt;br /&amp;gt;&lt;br /&gt; &amp;lt;br /&amp;gt; \therefore |L-M| \leq 2\epsilon&amp;lt;br /&amp;gt;&lt;br /&gt; &lt;br /&gt; And since \epsilon is can be as arbitrarily small as we want it to be, L-M must equal 0. Is this also correct? &lt;br /&gt; &lt;br /&gt; Side note: When using \delta - \epsilon proofs, do we use \leq or &amp;amp;lt;, or does it not matter? My prof uses the former, and the textbook uses the latter.&lt;br /&gt; &lt;br /&gt; Can I still assume |L-M| = 0 if I use \leq ? Or do I have to have it strictly less than \epsilon?

 

[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?

 

[Kelvie]

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

Thank you for your help.

 

[mathwonk]

right you are