# How can I show using the $\epsilon-\delta$ definition of limit that $u(x,y_0)\to l$ as $x\to x_0?$

Let $u(x,y)\to l$ as $(x,y)\to z_0=(x_0,y_0).$ How can I show using the $\epsilon-\delta$ definition of limit that $u(x,y_0)\to l$ as $x\to x_0?$

Choose $\epsilon>0.$ Then $\exists~\delta>0$ such that $|u(x,y)-l|<\epsilon~\forall~(x,y)\in (B(z_0,\delta)-\{(x_0,y_0\})\cap D$ ($D$ being the domain of $u$).

Of course the domain of $u(x,y_0)$ is $D_1=\{x:(x,y_0)\in D\}$

Thus for all $x\in((x_0-\delta,x_0+\delta)-\{x_0\})\cap D_1, |u(x,y_0)-l|<\epsilon.$ This far is easy. But how can I say that $x_0$ is a limit point of $D_1?$

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You're right, this is not automatic. You could have $D_1$ empty. For instance, $u(x,y):=3$ on $\{x=0\}\setminus \{(0,0)\}$ has limit $3$ when $(x,y)\longrightarrow (0,0)$ along the domain. But the limit of $u_1(x)=u(x,0)$ does not exist at $(0,0)$. As the domain is empty, in the first place. –  1015 Apr 29 '13 at 10:55
I even removed $(0,0)$, if you notice. So the domain of $u_1$ is empty. If you are trying to prove CR equations, I guess you are working on a open domain of $\mathbb{C}$, no? Then it can't happen. –  1015 Apr 29 '13 at 11:16
In general, when one defines differentiability, one requires the function to be defined in an open neighborhood of the point. And then one takes the limit. So it is ok. –  1015 Apr 29 '13 at 11:56
One good reason to define differentiability with this requirement is that then, at a local extremum, the derivative is zero. And this is a fundamental principle. Of course, there are some situations/people for which one makes sense of the derivative at, say, $0$, for a function defined on, say, $[0,+\infty)$. It suffices to consider the right limit. But then you loose the fundamental principle above at $0$. E.g. $f(x)=x$ on $[0,+\infty)$. So you want to be careful with such boundary points. If nothing is said, I believe most people mean the open condition. –  1015 Apr 29 '13 at 12:17
Let us take a concrete example. If $f:[0,+\infty)\longrightarrow \mathbb{R}$ is "right-differentiable" at $0$, then $f(x)-f(0)\sim f'(0^+)x$, so $f$ is continuous at $0^+$. –  1015 Apr 30 '13 at 3:05