Help with proof of the uniqueness of limits.

In summary, the conversation discusses a problem related to proving the uniqueness of limits. The textbook provides a starting point by defining limits and assuming their non-equality. The conversation then dives into the proof, using delta-epsilon notation and discussing different approaches. The conversation concludes with a solution to the problem and a side note about the use of inequality symbols in proofs.
  • #1
Kelvie
11
0
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
[tex]\lim_{x\to a} f(x) = L [/tex]
[tex]\lim_{x\to a} f(x) = M[/tex]

Assume [itex]L \neq M [/itex] and let [itex]\frac{|L - M|}{3} = \epsilon[/itex]

So here goes my proof..

[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|
\end{align*}
[/tex]

So by definition..
[tex]
\begin{align*}
0 \leq |x-a| \leq \delta (\epsilon) \implies \substack{|f(x) - L| \leq \epsilon \\ and \\ |f(x) - M| \leq \epsilon} \\
\end{align*}
[/tex]
[tex]
\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*}
[/tex]

Which can not possibly be true, so I conclude that our initial assumption [itex]L \neq M[/itex] 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:
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  • #2
"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
So this is "hand-in"able? :P

Thanks for the timely reply, by the way.
 
  • #4
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:
  • #5
Alright, this also makes sense, doesn't it?

We ignore the assumption [tex]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*}
[/tex]
[tex]
\therefore |L-M| \leq 2\epsilon
[/tex]

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

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

Can I still assume [itex]|L-M| = 0[/itex] if I use [itex]\leq[/itex] ? Or do I have to have it strictly less than [itex]\epsilon[/itex]?
 
  • #6
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
If by the second statement you meant [itex]0 \leq L < a[/itex] then yeah, I guess they are equivalent statements.

Thank you for your help.
 
  • #8
right you are
 

1. What is the uniqueness of limits?

The uniqueness of limits refers to the property of a limit having only one possible value. In other words, if a function approaches a certain value as it gets closer to a specific point, then that value is the only limit that the function can have at that point.

2. Why is it important to prove the uniqueness of limits?

Proving the uniqueness of limits is important because it ensures that the limit of a function at a particular point is well-defined and consistent. This allows for more accurate and reliable calculations and interpretations in various mathematical and scientific contexts.

3. How is the uniqueness of limits proven?

The uniqueness of limits is typically proven using the epsilon-delta definition of a limit. This involves showing that for any small positive number (epsilon), there exists a corresponding small positive number (delta) such that the function's output will be within epsilon of the limit as the input approaches the specific point.

4. What are some examples of unique limits?

Some examples of unique limits include:

  • The limit of the function f(x) = 2x+3 as x approaches 5 is 13.
  • The limit of the function g(x) = 1/x as x approaches 0 is undefined.
  • The limit of the function h(x) = sin(x) as x approaches 0 is 0.

5. Can a function have more than one limit at a specific point?

No, a function can only have one limit at a specific point. If a function has more than one limit, it means that the function is not continuous at that point and violates the uniqueness of limits property.

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