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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..)

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# Help with proof of the uniqueness of limits.

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