I'm working on a larger problem and it would be very helpful if the following were true:

For $0\leq s < \frac{1}{2}$ and $s \leq t \leq 1-s$, for every $\varepsilon > 0$ there exists a $\delta > 0$ such that if $\sqrt{(t-\frac{1}{2})^{2}+(s-\frac{1}{2})^{2}} < \delta$ then either $$\left|\frac{t-s}{1-2s} \right| < \varepsilon \text{$\quad$ or $\quad$} \left|\frac{t-s}{1-2s} - 1 \right| < \varepsilon$$

That is, as $s$ and $t$ both approach $\frac{1}{2}$, the function is either very close to $1$ or very close to $0$. It's not obviously true to me, but it would wrap the problem up nicely and I haven't been able to think of a counter example path.


1 Answer 1


Short Answer: No, the conjecture is not true unless you restrict your domain to the two line segments at the boundary.

Draft: The problem is actually quite standard in multivariable calculus after some reformulation. To get some idea, it is better to draw it in the plane. The region you are considering looks like a triangle. However, let's do the following change of variable to see what happens:

$$x = s-\frac{1}{2}, \quad y = t-\frac{1}{2}$$ With this pure translation, it becomes the triangle $-1/2\leq x<0$ and $ x \leq y \leq -x$.

In polar coordinates, it would be $-1/2\leq x<0$, $\ 3\pi/4\leq \theta \leq 5\pi/4$.

What is more of interest is the function. It becomes

$$\frac{x - y}{2x}=\frac{\cos \theta - \sin \theta}{2\cos\theta}$$

And we are taking limit tends to origin, i.e. $r \rightarrow 0$. But the above function clearly depends on $\theta$ and in the domain we are considering, $\theta$ is not fixed. Then the picture is quite clear now:

If we take $\theta = \pi$, then the "limit" is $-1-0/2(-1) = 1/2$. We will use this to rigorously show that the conjecture is wrong.

However, if you "evaluate the limit" at $\theta = 3\pi/4$, that translates to $s \in [0,1/2), \ t= 1-s$, then the limit is $1$ and if you do it at $\theta = 5\pi/4$, then the limit is $0$.

Long Answer: We are going to show that the conjecture is wrong by taking $\epsilon = 1/2$.

Pick any $\delta >0$. Take $t_0 = 1/2$ and $s_0 = \max\{0,1/2-\delta/2\}$. $s_0 \in [0,1/2)$ and $t_0 \in [s_0,1-s_0]$. $\sqrt{(t_0-1/2)^2-(s_0-1/2)^2} \leq \delta/2<\delta$. But

$$\frac{t-s}{1-2s} = \frac{1/2-s}{1-2s} = \frac{1}{2} $$ , for $s \neq 1/2$. But $|1/2 - 0| = |1/2 - 1| =1/2 \geq \epsilon$. Contradiction arises. And we are done.

  • $\begingroup$ Thanks. I hadn't thought of switching to polar coordinates. Unfortunately this means I'll likely be posting the full problem. $\endgroup$
    – user367387
    May 2, 2017 at 15:19
  • $\begingroup$ @JPelter Polar coordinates is just one approach and it is not magic solution to all limit problem:-) for example $xy^2/(x^2+y^4)$ has no limit at (0,0) since if you do it on the path $y=x^2$… $\endgroup$ May 2, 2017 at 15:32
  • $\begingroup$ But using polar coordinates tell you that the limit is 0…this is not sufficient since only the straight line paths that pass through origin are considered with polar coordinates. This example is taken from Rudin and I am doing analysis this semester…from this, I learn that the key to do such problem is to be skeptical :-) $\endgroup$ May 2, 2017 at 15:40
  • $\begingroup$ I meant $x=y^2$ $\endgroup$ May 2, 2017 at 15:41

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