# 0 to the power of 0, what does the essential discontinuity actually look like?

So having watch this clip by Numberphile which explains why $0^0$ is undefined https://www.youtube.com/watch?v=BRRolKTlF6Q

And also this

http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

And also this

Zero to the zero power - is $0^0=1$?

I understand how when given a function of the form $x^y$ then you have the following results

$$\lim_{x\rightarrow 0} x^0=1,$$ $$\lim_{y\rightarrow 0^+} 0^y=0,$$

and both the video and the mathforum link (and numerous others) mentioned about how when approaching form different directions you get different answers

However, both google and mathematica failed to actually show what the discontinuity look like

While a similar problematic case of "$\frac{0}{0}$" indeterminate form gives rather consistently the shape of the essential singularity attempt to do so for the case $x^y$ in google and mathematica does not really illuminate the shape of the essential singularity

By approaching zero from the x axis, and also analysing the x and y partial derivatives of the function $x^y$ I can clearly see a jump  However I don't quite get what the result for the x^y cases as you approach from directions that are not x or y axis geometrically look like

For example, for the "$\frac{0}{0}$" case it is easy to see why Numberphile said the value of $\frac{0}{0}$ depends on the angle you approach the origin But for $0^0$... While algebraically the limits clearly evaluated to different values depending on how we approach it, geometrically it does not seeemed to agree with what the limits said (the curve look smooth and continuous...probably due to the limitation of the graphing programs)

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So to sum up my question,what does the discontinuity of $0^0$ actually look like, is it like an oscillation, a jump, a dot or something more complicated? "==============================================="

UPDATE to ask for clarification, which should be in the comment section had it has a "posting image function")

Using the answers of Aes and Meelo, and examining the plots of the curves and parametric curves used by them to investigate the limiting behavior, I got the following  So the singularity has a shape of the vertical line interval (0,1) as mentioned by Aes and its neighborhood is as mentioned by Meelo has a very steep but non vertical "fault like structure"near the x axis which using the general form of a curve that can give limits that are between 0 and 1, as shown by Aes, explains why we need to travel along these curves (so that part of our journey is on that steep fault) in order not not get dragged into the value of 1

Or in short, based on the answers, is my understanding as shown below the correct way to understand it? • It doesn't make sense to plot $x^y$ when $x$ is negative. Try setting the range of $x$ to $[0,2]$. – Rahul Nov 23 '14 at 5:53
• Your understanding looks good. I've added a few more comments in my post. Note that singularity over $(0,0)$ is the interval $[0,1]$ since you can limit to $0$ and $1$ as well. – aes Nov 23 '14 at 16:48

A contour plot is the easiest to look at for this. If you approach along the contours, you'll get the value for that contour.

Note that $x^y$ can be undefined for $x < 0$ (any rational power with an even denominator won't work, e.g.), which is part of the strange behavior you're getting. I'll plot only for the positive values, though there's a symmetry to the negative $y$ values as Meelo mentions.

Here's a simple contour plot (the light color is near 1 and the dark color is near 0). It looks very clean and simple actually. As we'll see below, the 3-D discontinuity is a straight line over $x = y = 0$ going from $z = 0$ to $z = 1$. You can sort of make that out in the contour plot. Let's understand this in terms of approaching from different directions. Take $f(t) = (x(t),y(t))$ as a curve to approach on, with $x(t)$ and $y(t)$ approaching zero as $t \rightarrow 0$. We could jump right now to finding the contours and approaching along those, but let's explore the limits first. If you like, jump to the end for the contours.

To find $\lim x(t)^{y(t)}$, let's take logs. I'm also going to leave off $t$ so the notation is easier, but derivatives will be with respect to $t$. Then we get $\lim (y \log x) = \lim \frac{\log x}{1/y}$.

Apply l'Hôpital to get $\lim \frac{\dot x/x}{-\dot y/y^2} = \lim \frac{-\dot x}{\dot y} \frac{y^2}{x}$.

If we let $x(t)$ and $y(t)$ be nonzero and linear in $t$, we'll get zero for this limit, which means $1$ for $x^y$ (because we took a log). Looking at the picture, you can see $1$ is the predominant behavior.

We also have the simple behavior you noticed approaching with $x = 0$ (limit is $0$) or with $y = 0$ (limit is $1$).

How do we get other values for the limit? We play with the above limit. Let's let $y(t) = t$, so it becomes $\lim \frac{- t^2 \dot x}{x}$.

Notice that if we approach with $x = t^r$ ($r > 0$) we'll get $0$ (i.e. $x^y$ goes to $1$). We have to do something more drastic.

If we approach with $x = e^{-a/t}$ then $\dot x = \frac{a}{t^2} e^{-a/t}$ so everything cancels and we get $-a$ for the limit.

The values you can get for $-a$ here are any negative value, which corresponds to the interval $(0,1)$ after exponentiating: this gives the "line over $x = y = 0$" I referred to in the beginning. We also had $0$ and $1$ above, so we're getting the closed interval $[0,1]$.

Here's a plot of $x = e^{-1/y}$, which, when approached along, has a limit of $-1$ for the log, or $1/e$ for $x^y$. This is in fact one of the contours. Indeed if you solve for the contours, $c = x^y$ so $\log c = y \log x$ so $\log c / y = \log x$ so $x = e^{\log c / y}$. • It's worth noting that, even not having a plot of $x^y$ for negative $y$, that it would look similar, since $x^{-y}=\frac{1}{x^y}$. – Milo Brandt Nov 23 '14 at 6:15

An annoying thing about this discontinuity, is that if we approach it by any line, we get an answer. In particular, the line with slope $\beta$ through the origin always has the limit $$\lim_{t\rightarrow 0}t^{\beta t}$$ and this always equals $1$. This can be proven, since $t^{\beta t}=(t^t)^{\beta}$, and since $1^{\beta}=1$, if we can show $$\lim_{t\rightarrow 0}t^t=1$$ then we show for all $t$. In particular, it's rather obvious that $t^t<1$ for all $t\in (0,1)$, so the limit is at most $1$. It actually approaches $1$ as can be established by a simple argument (which isn't quite formal) relating $x_1=t^t$ and $x_2=\frac{t}{2}^{\frac{t}2}$. Note that $x_2$ is a value of $t^t$ closer to $t=0$ than $x_1$. In particular, notice that $x_2=\sqrt{t}\cdot \left(\frac{1}2\right)^t$. Since the $\left(\frac{1}2\right)^t$ term goes to $1$ as $t$ does, this basically means that, if we halve $t$, the value of $t^t$ will be affected by a square root - and repeatedly applying the square root iterates any value in $(0,1)$ towards $1$, eventually. So $\lim\limits_{t\rightarrow 0}t^t=\lim\limits_{t\rightarrow 0}t^{\beta t}=1$.

This is interesting, since it means that the lines approaching the origin have limit $1$, except for the limit along the axis of the base, which has limit $0$. However, we could go on to compute other limits - like when we approach the origin by a parabola or otherwise. Unfortunately, it seems, to my estimation, that no particularly natural example exists, since the above prove easily extends to show that if we approach $(0,0)$ in $x^y$ by setting $x$ and $y$ to some polynomial of a dummy variable $t$, the limit is still $1$ (or $0$ in that one case). This would mean that $x^y$ is, in some sense, almost seems to approach $1$ near $(0,0)$, but, near the $x$ axis takes a very sharp twist towards $0$ (for $y>0$) or $\infty$ (for $y<0$). However, suppose we move, in $x^y$, across the curve $x=t$ and $y=\log_t(\frac{1}2)$. It's rather obvious that $x^y=\frac{1}2$ everywhere on this curve and, moreover, both $x$ and $y$ approach $0$. So, along this curve, we get a limit of $\frac{1}2$. It's a fairly contrived example, but it's certainly valid.

• I do not think it is a contrived example, and it might seem less contrived (for any positive constant $k$) with $x = t, y=\dfrac{\log_e(k)}{{\log_e(t)}}$ so you get $x^y=k$ for all positive $t\not=1$ – Henry Jun 21 '18 at 7:14