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$Problem:$ Every convergent sequence has a unique limit.
$Proof:$ Let's assume that a sequence $\langle a_n\rangle$ converges to two distinct limits $l$ and $l^{'}$.
Let's assume $\epsilon=\frac{1}{2} |l-l^{'}|$.Since, $l\neq l^{'}$,$|l-l^{'}|>0$ so that $\epsilon>0$.
Given $\epsilon>0$, $\exists$ a $+\text{ive}$ integer $m_1$ such that $|a_n-l|<\frac{\epsilon}{2}$ for every $n\geq m_1$.
Similarly, for second limit $l^{'}$
Given $\epsilon>0$, $\exists$ a $+\text{ive}$ integer $m_2$ such that $|a_n-l^{'}|<\frac{\epsilon}{2}$ for every $n\geq m_2$.
Let $m=\max(m_1,m_2)$
Now, $|l-l^{'}|=|(l-a_n)+(a_n-l^{'})|\leq |l-a_n|+|a_n-l^{'}|$
$|l-l^{'}|<\epsilon$ for every $n \geq m$ which contradicts the assumption above. Hence $l = l^{'}$.

My question here is: a) If the limits are different than how can we assume that $\epsilon$ is same for both the limits. Infact, if $l^{'}\gg l$ than $\epsilon^{'}\ll\epsilon$ as n gets larger and larger.
b) How this method to prove theorem is appropriate, if I tend to use $\epsilon$ and $\epsilon^{'}$ rather than common $\epsilon$ at both places.
c) Can I expect to prove it by working out from $x$-axis rather than working on it with the assumption for $\epsilon$ from $y$-axis.
d) In case of $\epsilon=\frac{1}{2} |l-l^{'}|$, what is the $y$ here for which $y+\epsilon$ and $y-\epsilon$ can be considered.

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    In the definition of limit, one must be able to show that for **any** positive $\epsilon$, there is an $m$ such that $\dots$. So in order to prove a contradiction, you are free to choose $\epsilon$. The **geometric** intuition is very simple. If $l\ne l'$, then $a_n$ cannot be simultaneously ridiculously close to $l$ and to $l'$.2012-10-27

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There are really two important ideas working in this proof: The first is that we can use $\mathbf{any}$ $\epsilon$ we want by the definition of convergence. The second is the triangle inequality. To see exactly how we take advantage of both of these I'll write the proof slightly differently (no longer by contradiction).

Fix $\mathbf{any}$ $\epsilon >0$ and by the definition of convergence we can find $N$ and $M$ large so that $\left|l - a_n\right| < \epsilon \text{ for any } n \geq N;$ $ \left| l' - a_m\right| < \epsilon \text{ for any } m\geq M.$So now fix a specific $k\geq \max(N,M)$ then, since $k$ is fixed, we can use the triangle inequality $\left|l - l'\right| \leq \left| l - a_k \right| + \left| a_k - l'\right| \leq \epsilon + \epsilon = 2 \epsilon.$

And since we chose $\epsilon > 0$ arbitrary, this means that $\left|l - l'\right| = 0$ and so $l = l'$.

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    It's probably worth saying that this is identically the proof in baby Rudin.2016-01-17