How do we prove $\forall t>f(x),\exists\delta>0,\forall y\in$ dom($f$), $|y-x|<\delta\implies f(y)<t$ if and only if (whenever a sequence $s_n$ of points in dom $(f)$ converges to $x$, we have $\lim\sup f(s_n)\le f(x)$)?
we know the definition of $\lim\sup s_n=l$ is that $\forall t>l, \exists N, n>N\implies s_n<t$.
The $\le$ really stops me from going. Could someone help?
Let us start with: Whenever a sequence $s_n$ of points in dom $(f)$ converges to $x$, we have $\lim\sup f(s_n)\le f(x)$
Suppose there exists $ t>f(x)$, such that $\forall\delta>0$,$\exists y\in$ dom($f$), $|y-x|<\delta$ and $ f(y)\geq t$
Then you can have a sequence of $y_n$s converging to $x$, such that $f(y_n)\geq t>f(x) \forall n$, which brings a contradiction to $\limsup f(y_n)\le f(x)$.
Now can you go the other way?
Hint for the other direction:
Suppose, $\forall t>f(x)$,$\exists\delta^*>0$ such that $\forall y\in$ dom($f$), $|y-x|<\delta$, we have $f(y)<t$
Pick any $t>f(x)$. Now take a sequence of points in domain of $f$ that converge to $x$, i.e, $s_n \to x$. There would exist $N$ such that $|s_n-x |<\delta^*\ \ \forall n\geq N$.
Using what is given, $f(s_n)<t\ \ \forall n\geq N$. Now use the definition of limsup to get your proof? Use the definition that limsup is the sup of all the subsequential limits if you want.