- #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..)
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: