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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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    I would go accept some answers to your other question if you want some answers2012-10-27
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    6 questions, zero accepted answers. Don't you like the answers you get here?2012-10-27
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    Also, where are you working? In general topological spaces this result isn't necessarily true.2012-10-27
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    i just had this theorem in front of me and I thought of breaking it in parts and tried finding some alternatives.Couldn't find solutions but found questions !2012-10-27
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    I would suggest that you revisit the definition of limit of a sequence. The definition of the limit of a sequence goes as "A sequence $\{x_n \}_{n=1}^{\infty}$ converges to a limit $x$, if given ***any*** $\epsilon>0$, ***there exists*** $N(\epsilon) \in \mathbb{N}$, such that ***for all*** $n > N(\epsilon)$, we have that $$\vert x_n - x \vert < \epsilon.$$" Your problem is to prove that $x$ is unique.2012-10-27
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    I know that I have to prove that $x$ is unique, I have written this proof directly from a book, I can assure you that I have not added or removed anything from it.2012-10-27
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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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