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I decided to review the properties of expontents after posting a question and getting downvoted. The answer was easy; however, got a bit confused by $ n^{x^m} $ with $n,m \in \mathbb{Z}$

So I started with the most basic thing I could think of and that is to draft a picture.

I am stuck at drawing the graph of $x^y$ at $(0,0)$. Getting confused about how to take the limit to the right and left of this. Which limit do I take first, the limit of $x$ as it goes to zero or the limit of $y$ as it goes to zero, from either the right or left.

Yes, I type $x^y$ into Wolfram Alpha, however, don't understand why the first is a 3D picture and the second picture doesn't show good the point $(0,0)$. I am just deducing a guess that the two different colors on oppositite quadrants mean there are 4 cases.

I do realize the graph will contain all the 2D plane. (sorta like a slope field).

My guess and its not strong, the point $(0,0)$ doesn't exist. I have tried searching online, but from what I could understand from the articles I read (most are above my knoweldge of math), is that there are three possible values $0,1$, and DNE.

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Typically, a limit like this would make sense if no matter which line you came in on towards $(0,0)$ the limit existed and was same for any of these directions. Note that coming in on the line $y=0$ from either direction the limit is one. Coming in on $x=0$ the limit is zero. Therefore, the limit does not exist. By the look at the graph it look like some directions blow up.(Coming in on $y=-x$ from the $y$ negative direction would go to $\infty$ for instance) – user45150 Dec 10 '12 at 0:38
It might be helpful to look at a picture where the window is different:… – Cocopuffs Dec 10 '12 at 0:46
@Cocopuffs Thanks, didn't know the syntax for getting different windows. – yiyi Dec 10 '12 at 0:49
up vote 4 down vote accepted

There is a non-removable discontinuity at $(0,0)$. For example, approaching along the $x$-axis yields a different limiting value at $(0,0)$ than approaching along the $y$-axis:

$$ \lim_{t\to 0} t^0 = \lim_{t \to 0} 1 = 1. $$

$$ \lim_{t\to 0} 0^t = \lim_{t \to 0} 0 = 0. $$

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I don't think the function is meromorphic, so I find it hard to believe that there's an essential singularity... – Potato Dec 10 '12 at 0:46
@Potato: Fair enough. I didn't know a specific term for this behavior in a real function from $\mathbb{R}^2$ to $\mathbb{R}$. – Shaun Ault Dec 10 '12 at 0:53

The domain is the positive quadrant. Draw two lines: $y=m_1x$ and $y=m_2x$, where $0<m_1<m_2<\infty$. If you approach $(0,0)$ from within that sector, the limit is $1$. Moreover, I think if $f(t)$ and $g(t)$ are analytic functions of a real variable $t$ and $f(t),g(t)\to0\text{ as }t\to a$ then then $f(t)^{g(t)}\to 1\text{ as }t\to a$, even if the curve $t\mapsto(f(t),g(t))$ has a horizontal or vertical tangent at that point. But you can make $f(t)^{g(t)}$ approach $0$ or a positive number other than $1$ by making $f$ and $g$ extreme enough near that point.

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