continuity of a two variables function in a particular point

Question

I have a function with two variables $f(x,y)= \frac{x}{y}$ and a point, $(0,1)$. It is simple to see that this function is continuous in $(0,1)$ but I have to prove directly this continuity applying the definition and finding a value for $\delta$ as a conseguence of a value for $\epsilon$.

I've seen the solution in my book and I'm trying to understand at least what it says.

In the neighbourhood of $(0,1)$ it occurs that $|y-1|\le \frac{1}{2} \rightarrow y \ge \frac{1}{2}\Longrightarrow |y| \ge \frac{1}{2} \rightarrow |\frac{x}{y}|\le 2|x|$.

As a conseguence of this $\delta =\min\left\{\frac{\epsilon}{2}, \frac{1}{2}\right\}$ I can't understand this last passage. Could someone explain more clearly why $\delta =\min\left\{\frac{\epsilon}{2}, \frac{1}{2}\right\}$ and what is the line of reasoning of that?

Answer 1

Suppose that $\delta=\min\left\{\frac\varepsilon2,\frac12\right\}$. Then, asserting that $\bigl\|(x,y)-(0,1)\bigr\|<\delta$ means two things:

• $\bigl\|(x,y)-(0,1)\bigr\|<\frac\varepsilon2$;
• $\bigl\|(x,y)-(0,1)\bigr\|<\frac12$.

From the second fact, you deduce that $|y-1|<\frac12$ and that therefore $y>\frac12$. So, $\left|\frac xy\right|<2|x|$.

Therefore,\begin{align}\left|f(x,y)-f(0,1)\right|&=\left|\frac xy-0\right|\\&=\left|\frac xy\right|\\&<2|x|\\&<2\frac\varepsilon2\text{ (since $|x|<\frac\varepsilon2$)}\\&=\varepsilon.\end{align}

Answer 2

Well, the correct answer probably depends on what norm on $\mathbb{R}^2$ you are using. The good news is that on $\mathbb{R}^2$ you can use whatever norm you need (so that the proof goes smoothly), since $\mathbb{R}^2$ is a finite-dimensional space (I can elaborate a bit later).

So suppose, the norm is $\| (x,y)\| = |x| + |y|.$ You want to prove that given an arbitrary $\epsilon>0,$ the difference $|f(x,y) - (0,1)|$ (here you use just ordinary absolute value since your function is a scalar function) is less than $\epsilon$ whenever $\|(x,y) - (0,1) \| < \delta.$ The point in tkaing $\delta = \min{(1/2, \epsilon/2)}$ is that you can use the inequalities $|x| < \epsilon/2$ and $|y-1|<1/2$ simultaneously.

Answer 3

Let $\epsilon >0$ be given .

Have to show:

There is a $\delta >0$ such that

$\sqrt{x^2+(y-1)^2}\lt \delta$ implies

$|x/y| \lt \epsilon.$

$ |x/y| =|√x^2/y| \lt |\dfrac{\sqrt{x^2 +(y-1)^2}}{y}|.$

Now consider :

1) $y$ such that $|y-1|\lt 1/2$.

Then $-1/2+1 \lt y$ , or $1/2 \lt y.$

We then have:

$|x/y| \lt \dfrac{\sqrt{x^2+(y-1)^2}}{y} \lt $

$\dfrac{\sqrt{x^2+(y-1)^2}}{1/2}.$

Choose $\delta =\min(1/2, \epsilon/2)$.

Then

2) $|x/y| \lt 2\delta \epsilon$.

Reasoning:

If a) $1/2 \le \epsilon/2$ we can use

1) and in the last step of 2) replacing $\delta$ by

$\epsilon/2$ we enlarge.

Similar argument for :

b): If $\epsilon/2 \le 1/2$.