I'm stuck on the following exercise (from T.Tao's Analysis 2 book):
"Let $f\colon\mathbb{R}^2\to\mathbb{R}$ be a function from $\mathbb{R}^2$ to $\mathbb{R}$ and $(x_0,y_0)$ be a point in $\mathbb{R}^2$.
If $f$ is continuous at $(x_0,y_0)$, show that $\lim_{x\to x_0}\limsup_{y\to y_0}f(x,y)=\lim_{y\to y_0}\limsup_{x\to x_0}f(x,y)=f(x_0,y_0)$ and
$\lim_{x\to x_0}\liminf_{y\to y_0}f(x,y)=\lim_{y\to y_0}\liminf_{x\to x_0}f(x,y)=f(x_0,y_0)$".
(note: $\limsup_{x\to x_0}f(x):=\inf_{r>0}\sup_{|x-x_0|<r}f(x)$ and $\liminf_{x\to x_0}f(x):=\sup_{r>0}\inf_{|x-x_0|<r}f(x)$)
I've first tried to use the hypothesis that $f$ is continuous but I haven't got anywhere so I tried to argue by contradiction but this approach too didn't led me anywhere and now I'm stuck. Any hints?
I will prove the first equality. Take $\varepsilon>0$. You need to prove that there is a $\delta>0$ such that$$|x-x_0|<\delta\implies\left|\limsup_{y\to y_0}f(x,y)-f(x_0,y_0)\right|<\varepsilon.\tag{1}$$Take $\delta>0$ such that\begin{align}|x-x_0|<\delta\wedge|y-y_0|<\delta&\implies\bigl|f(x,y)-f(x_0,y_0)\bigr|<\varepsilon\\&\iff f(x_0,y_0)-\varepsilon<f(x,y)<f(x_0,y_0)+\varepsilon.\end{align}But then$$|x-x_0|<\delta\implies f(x_0,y_0)-\varepsilon\leqslant\limsup_{y\to y_0}f(x,y)\leqslant f(x_0,y_0)+\varepsilon.$$Therefore, you have $(1)$, expect that $<$ becames $\leqslant$, but that is easy to fix.
The other equalities have similar proofs.