Good evening,(adsbygoogle = window.adsbygoogle || []).push({});

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..)

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Help with proof of the uniqueness of limits.

**Physics Forums | Science Articles, Homework Help, Discussion**