$Problem:$ 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.