Problem in finding $\delta$ to prove that $\lim_{x\to a}[\lim_{y\to b}f(x,y)]=\lim_{y\to b}[\lim_{x\to a}f(x,y)]=L$ Theorem: Let $f$ be a vector field. If $\lim_{(x,y)\to(a,b)}f(x,y)=L$ and $1$D limits $\lim_{x\to a}f(x,y)$ and $\lim_{y\to b}f(x,y)$ exist then, $\lim_{x\to a}[\lim_{y\to b}f(x,y)]=\lim_{y\to b}[\lim_{x\to a}f(x,y)]=L$.
I tried to prove it as follows:
Geometrically, let $\lim_{x\to a}f(x,y)=g(y)$, whence it follows that we fix some $y$ arbitrarily and then approach $a$ along $x$-axis to get $\lim_{x\to a}f(x,y)$. Similarly for the other $1$D limit.  Then since the limit $\lim_{(x,y)\to(a,b)}f(x,y)=L$ exists, it doesn't matter how we approach $(a,b)$ and hence the result is verified.Is my interpretation correct? 
Now coming to the proof: Let $\lim_{y\to b}f(x,y)=h(x)$. Since $\lim_{(x,y)\to(a,b)}f(x,y)=L$, it follows that $\forall \epsilon_1 \gt 0, \exists \delta_1 \gt 0$ such that if $0\lt \sqrt{(x-a)^2+(y-b)^2}\lt \delta_1$, then $||f(x,y)-L||\lt \epsilon_1 \tag{1}$ 
Since $\lim_{x\to a}f(x,y)=g(y)$, it follows that $\forall \epsilon _2\gt 0, \exists \delta_2\gt 0$ such that if $0\lt |x-a|\lt \delta_2$, then $||f(x,y)-g(y)||\lt \epsilon_2 \tag{2}$ 
Similarly, $\forall \epsilon _3\gt 0, \exists \delta_3\gt 0$ such that if $0\lt |y-b|\lt \delta_3$, then $||f(x,y)-h(x)||\lt \epsilon_3\tag {3}$ 
From (2)& (3), it follows that $\require{enclose}
     \enclose{horizontalstrike}{||g(y)-h(x)||\le ||f(x,y)-g(y)||+||f(x,y)-h(x)||\lt\epsilon_2+\epsilon_3}$, whenever $\require{enclose}
     \enclose{horizontalstrike}{0\lt |y-b|\lt \delta_3}$ and $\require{enclose}
     \enclose{horizontalstrike}{0\lt |x-a|\lt \delta_2}$ 
That is by taking $\require{enclose}
     \enclose{horizontalstrike}{\epsilon_2=\epsilon_3=\epsilon/2}$, we have $\require{enclose}
     \enclose{horizontalstrike}{||g(y)-h(x)||\lt\epsilon}$, whenever $\require{enclose}
     \enclose{horizontalstrike}{0\lt \sqrt{(x-a)^2+(y-b)^2} \lt  \sqrt{\delta_2^2+\delta_3^2}}$  
Since  $\lim_{x\to a}[\lim_{y\to b}f(x,y)]=\lim_{x\to a}h(x)$, consider 
$||h(x)-L||\lt||h(x)-f(x,y)||+||f(x,y)-L|\lt \epsilon_3+\epsilon_1$, whenever $0\lt |y-b|\lt \delta_3$ and $0\lt \sqrt{(x-a)^2+(y-b)^2}\lt \delta_1$ whence by taking $\epsilon_3=\epsilon_1=\epsilon/2$, it would follow that 
$||h(x)-L||\lt \epsilon$ whenever $0\lt \sqrt{(x-a)^2+(y-b)^2}\lt \delta$ and similarly for $g(y)$. 
But the problem is how to get this $\delta$ from $(1)$ and $(3)$? 
I tried the following here: since $0\lt |y-b|\lt \sqrt{(x-a)^2+(y-b)^2}$, taking $\delta=\min\{\delta_1,\delta_3\}$ will cure the problem. Is this correct?
Please help. Thanks for your time.
 A: We have $|f(x,y) - L| < \epsilon_1$ when  $\sqrt{(x-a)^2 + (y-b)^2} < \delta_1$.  If $|x-a| < \delta_1/\sqrt{2}$ and $|y-b| < \delta_1/\sqrt{2}$, then $\sqrt{(x-a)^2 + (y-b)^2} < \delta_1$  holds and it is again true that $|f(x,y) - L| < \epsilon_1$
It also is enough that $|x-a| < \delta_1/\sqrt{2}$ for $|h(x) - L| \leqslant \epsilon_1$ to hold as well.
To prove this claim, note that by the reverse triangle inequality
$$||f(x,y) -L| - |h(x) - L|| \leqslant |f(x,y) - h(x)|$$
Since $h(x) = \lim_{y \to b} f(x,y)$, for any $\epsilon > 0$ and each fixed $x$ there exists $\delta(\epsilon,x) > 0$ such that if $|y-b| < \delta(\epsilon,x)$, then
$$||f(x,y) -L| - |h(x) - L|| \leqslant|f(x,y) - h(x)| < \epsilon$$
Assume that for some fixed $x$ where $|x-a| < \delta_1/\sqrt(2)$ we have  $|h(x) - L| > \epsilon_1$.
Take $\epsilon = \frac{1}{2}(|h(x) - L| -\epsilon_1)$ and $|y-b| < \min(\delta(\epsilon,x), \delta_1/\sqrt{2})$.  It follows that
$$|f(x,y) -L| - |h(x) -L| > - \epsilon = -\frac{|h(x) - L|}{2}+ \frac{\epsilon_1}{2},$$
implying
$$|f(x,y) -L| > \frac{|h(x) - L|}{2}+ \frac{\epsilon_1}{2} > \epsilon_1$$
This contradicts $|f(x,y) - L| < \epsilon_1$ and we can conclude that $|h(x) - L| \leqslant \epsilon_1$, Therefore,
$$\lim_{x \to a} \lim_{y\to b} f(x,y) = \lim_{x \to a}h(x) = L$$

The proof that $\lim_{y \to b} \lim_{x\to a} f(x,y) = L$ is handled in the same way.
A shorter proof is to simply say that $|h(x) - L| = \lim_{y \to b}|f(x,y) - L|$ and $|f(x,y) - L| < \epsilon_1$ implies that $|h(x) - L| \leqslant \epsilon_1$.  These steps are justified by well-known limit theorems which are embedded in the proof above.
