Let $f$ be a real-valued function defined on interval $(a,b)$ and let $l$ be a real number.
$1.$ $\forall t>l,\exists\epsilon >0,\forall x\in(a,a+\epsilon),f(x)<t$ and
$\ \ \ \ \forall t<l,\forall\epsilon >0,\exists x\in(a,a+\epsilon),f(x)>t$
$2. $ For all the sequences $s_n$ in $(a,b)$ such that $s_{n} \to a$ and the sequence $f(s_n)$ converges, $l$ is the largest of the limits $\lim f(s_n)$ obtainable in this way.
My thought:
The original definition of limit of right end is $\forall \epsilon>0,\exists \delta>0, \forall x\in(a,a+\delta), |f(x)-l|<\epsilon$
$\rightarrow:$ Assume $1$, suppose $2$ is false, then there is a larger limit, say $m$. But then I don't see how making $m$ a larger limit would violate the statement $1$
$\leftarrow:$ Assume $2$. Let $t$ be a limit larger than $l$ then there is no such sequence on $(a,b)$ such that $f(s_n)$ converges to $t$. Negating the original definition we get $\exists \epsilon>0,\forall \delta>0, \exists s_n\in(a,a+\delta), |f(s_n)-t|\ge\epsilon$. But I don't see a connection to statement $1$.
Could someone help?