Interchangeable limits The following exercise 2.2.9 is borrowed from Terence Tao's Analysis II, page 33. While I understand the problem completely, I lack the technique to attack it. So I will really appreciate a small hint on how to start the proof.
Let $f : \mathbb{R}^2 \to \mathbb{R}$ be a function. Let $(x_0 , y_0 ) \in \mathbb{R}^2$ be a point. 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
$$\lim_{x\to x_0 } \liminf_{y \to y_0} f (x, y) = \lim_{y\to y_0 } \liminf_{x \to x_0} f (x, y) = f (x_0 , y_0 )$$
In particular, we have
$$\lim_{x\to x_0 } \lim_{y \to y_0} f (x, y) = \lim_{y\to y_0 } \lim_{x \to x_0} f (x, y) = f (x_0 , y_0 )$$
whenever the limits on both sides exist.
We have defined $\limsup$ as
$$\limsup_{x\to x_0} f(x,y) = \inf \left\{ \sup \{f(x,y): |x-x_0| <r \}: r>0\right\}$$
 A: We only prove the first claim, others follow in a similar manner.
Our goal is to show that 
$$\lim_{x\to x_0} \limsup_{y\to y_0} f(x,y) = f(x_0,y_0)$$
i.e., that
$$\forall \epsilon>0 \quad \exists \delta_{\epsilon}>0 \quad \forall x \in \mathbb{R}: \quad |x-x_0|<\delta_{\epsilon} \implies |\limsup_{y\to y_0} f(x,y) - f(x_0,y_0)|<\epsilon$$
Pick an arbitrary $\epsilon>0$. Since $f(x,y)$ is continuous at $(x_0,y_0)$, we have
$$\exists \gamma_{\epsilon}>0 \quad \forall (x,y) \in \mathbb{R}^2: \quad d((x,y),(x_0,y_0))<\gamma_{\epsilon} \implies d(f(x,y),f(x_0,y_0))<\epsilon$$
or, written alternatively,
$$\exists \gamma_{\epsilon}>0 \quad \forall (x,y) \in \mathbb{R}^2: \quad \sqrt{(x-x_0)^2 + (y-y_0)^2}<\gamma_{\epsilon} \implies |f(x,y) - f(x_0,y_0)|<\epsilon$$
We set $\delta_{\epsilon} := \frac{\gamma_{\epsilon}}{2}$ and consider only $x,y \in \mathbb{R}$ such that $|x-x_0|<\delta_{\epsilon}$ and $|y-y_0|<\delta_{\epsilon}$. Due to the continuiuty assumption
$$\forall (x,y) \in \mathbb{R}^2: \quad |x-x_0|<\delta_{\epsilon} \text{ and } |y-y_0|<\delta_{\epsilon}\implies f(x_0,y_0) - \epsilon < f(x,y) < f(x_0,y_0)+\epsilon$$
for all such $x,y$. Then $f(x_0,y_0)+\epsilon$ is an upper bound for $f(x,y)$ on this interval; hence,
$$\forall (x,y) \in \mathbb{R}^2: \quad |x-x_0|<\delta_{\epsilon} \text{ and } |y-y_0|<\delta_{\epsilon}\implies f(x_0,y_0) - \epsilon < \sup\{f(x,y)\} \leq f(x_0,y_0)+\epsilon$$
or, written alternatively,
$$\forall x \in \mathbb{R}: \quad |x-x_0|<\delta_{\epsilon}\implies f(x_0,y_0) - \epsilon < \sup\{f(x,y):|y-y_0|<\delta_{\epsilon}\} \leq f(x_0,y_0)+\epsilon$$
Taking the infimum with respect to $r>0$ yields
$$\forall x \in \mathbb{R}: \quad |x-x_0|<\delta_{\epsilon}\implies f(x_0,y_0) - \epsilon \leq \inf\left\{ \sup\{f(x,y):|y-y_0|<\delta_{\epsilon}\}:r>0 \right\} \leq f(x_0,y_0)+\epsilon$$
Therefore, for our arbitrarily chosen $\epsilon>0$ we have found a $\delta_{\epsilon}>0$ such that whenever $|x-x_0|<\delta_{\epsilon}$ we have $f(x_0,y_0) - \epsilon \leq \limsup_{y\to y_0} f(x,y) \leq f(x_0,y_0)+\epsilon$, as desired.
