Why is $0^0$ undefined when $x^x=1$ as $x$ approaches $0$? This question was born in another post available here.  I believe $0^0=1$, because $x^x$ is continuous as $x$ approaches $0$. 
Consider $\lim_{x \to 0}x^x$.  Let $$f(x_n)=\bigg(\frac{1}{x}\bigg)^{\frac{1}{x}}$$such that
$$
y=\bigg(\frac{1}{x}\bigg)^{\frac{1}{x}}
$$Then
$$
ln(y)=\frac{1}{x}ln\bigg(\frac{1}{x}\bigg)=\frac{1}{x}(ln1-lnx)=\frac{1}{x}(0-lnx)=-\frac{lnx}{x}
$$Now, using L'Hopital's rule,
$$
\lim_{x\to\infty}\bigg(-\frac{lnx}{x}\bigg)=\lim_{x\to\infty}\bigg(-\frac{\frac{1}{x}}{1}\bigg)=\lim_{x\to\infty}\bigg(-\frac{1}{x}\bigg)=0
$$
Now,
$$
\lim_{x\to\infty}f(x_n)=\lim_{x\to\infty}y=\lim_{x\to\infty}e^{lny}=\lim_{x\to\infty}e^{-\frac{1}{x}}=e^0=1
$$Therefore,
$$
0^0=\lim_{x\to 0}x^x=\lim_{x\to 0}f(x)=\lim_{x\to\infty}f(x_n)=1
$$For something to be well-defined, it needs to act similarly in different configurations.  $0^0=0^{c-c}=\frac{0^c}{0^c}=\frac{0}{0}$, and I understand the convention which says that $\frac{0}{0}$ is undefined, whereas $\frac{anything else}{anything else}=1$ (even $\frac{\infty}{\infty}=1$ when we are taking limits of functions), but is that purely a matter of convention?  If a majority of mathematicians decide that $\frac{0}{0}=1$, have we changed the convention?  Where is the math in all of this?  
The core of my confusion is a simple question: What IS zero?  Is it a number, or is it an altogether different concept?  
If you have previously given this issue some thought and care to share your insights, I would be most appreciative!
 A: Well you also have that $x^{\log 3 / \log x}$ is continuous as $x$ (and $\log 3 / \log x)$ approaches $0$ and it's even the constant function that is $3$ for any $x$, so by your reasoning $0^0$ should be $3$.
To put it bluntly, the function $(x,y) \mapsto x^y$ cannot be continuously extended at $(0,0)$ so you are not going to convince anyone by an argument of continuity of some function.
Also, note that $(x,y) \mapsto \frac xy$ at $(0,0)$ is in the same situation. 
Or also $(x,y) \mapsto x^\frac 1y$ at $(1,0)$.
A: View 1:
Showing that $x^{x}$ approaches 1 as the positive value x gets arbitrarily close to zero does not prove that $0^{0} = 1$. The variable x having a value close to zero is different than it having a value of exactly zero. It turns out that $0^{0}$ is undefined. $0^{0}$ does not have a value.

View 2:
For all $x>0$, we have
$$0^{x} = 0.$$
Hence,
$$\lim_{x \to 0^{+}} 0^{x} = 0$$
That is, as x gets arbitrarily close to $0$(but remains positive), $0^{x}$ stays at $0$.
On the other hand, for real numbers y such that $y \ne 0$, we have that
$$y^{0} = 1$$
Hence,
$$\lim_{y \to 0} y^{0} = 1$$
That is, as y gets arbitrarily close to $0$, $y^{0}$ stays at 1.
Therefore, we see that the function $f(x,y) = y^{x}$ has a discontinuity at the point $(x,y) = (0,0)$. In particular, when we approach $(0,0)$ along the line with $x=0$ we get
$$\lim_{y \to 0} f(0,y) = 1$$
but when we approach $(0,0)$ along the line segment with $y=0$ and $x>0$ we get:
$$\lim_{x \to 0^{+}} f(x,0) = 0$$
Therefore, the value of $$\lim_{(x,y) \to (0,0)} y^{x}$$  is going to depend on the direction that we take the limit. This means that there is no way to define $0^{0}$ that will make the function $y^{x}$ continuous at the point $(x,y) = (0,0)$.

View 3:
Zero raised to the zero power is one. No really, it’s true. 
Let’s consider the problem of defining the function $f(x,y) = y^x$ for positive integers y and x. There are a number of definitions that all give identical results. For example, one idea is to use for our definition:
$$y^x := 1 \times y \times y \cdots \times y$$
where the y is repeated x times. In that case, when x is one, the y is repeated just one time, so we get
$$y^{x} = 1 \times y$$
However, this definition extends quite naturally from the positive integers to the non-negative integers, so that when x is zero, y is repeated zero times, giving
$$y^{0} = 1$$
which holds for any y. Hence, when y is zero, we have
$$0^0 = 1$$
We've just proved that $0^0 = 1$ But this is only for one possible definition of $y^x$. What if we used another definition? For example, suppose that we decide to define $y^x$ as
$$y^x := \lim_{z \to x^{+}} y^{z}$$
In words, that means that the value of $y^x$ is whatever $y^z$ approaches as the real number z gets smaller and smaller approaching the value x arbitrarily closely.Interestingly, using this definition, we would have
$$0^0 = \lim_{x \to 0^{+}} 0^{x} = \lim_{x \to 0^{+}} 0 = 0$$
Hence, we would find that $0^0 = 0$ rather than $0^0 = 1$. Granted, this definition we’ve just used feels rather unnatural, but it does agree with the common sense notion of what $y^x$ means for all positive real numbers x and y, and it does preserve continuity of the function as we approach $x=0$ and $y=0$ along a certain line.
So which of these two definitions (if either of them) is right? What is $0^0$ really? Well, for $x>0$ and $y>0$ we know what we mean by $y^x$. But when $x=0$ and $y=0$, the formula doesn’t have an obvious meaning. The value of $y^x$ is going to depend on our preferred choice of definition for what we mean by that statement, and our intuition about what $y^x$ means for positive values is not enough to conclude what it means for zero values.
But if this is the case, then how can mathematicians claim that $0^0=1$? Well, merely because it is useful to do so. Some very important formulas become less elegant to write down if we instead use $0^0=0$ or if we say that $0^0$ is undefined.
$$(a+b)^x = \sum_{k=0}^{\infty} \binom{x}{k} a^k b^{x-k}$$ 
Hence, if we do not use $0^0 = 1$ then the binomial theorem does not hold when $a=0$ because then $b^x$ does not equal $0^0 b^{x}$.
If mathematicians were to use $0^0 = 0$, or to say that $0^0$ is undefined, then the binomial theorem would continue to hold (in some form), though not as written above. In that case though the theorem would be more complicated because it would have to handle the special case of the term corresponding to $k=0$. We gain elegance and simplicity by using $0^0 = 1$.
There are some further reasons why using $0^0 = 1$ is preferable, but they boil down to that choice being more useful than the alternative choices, leading to simpler theorems, or feeling more "natural" to mathematicians. The choice is not "right", it is merely nice.
A: One problem, $\frac{\infty}{\infty}$ is not always $1$, example, $\lim_{x\to\infty}\frac{e^{x+a}}{e^{x}}=e^a\neq1$ unless $a=0$. 
So why should $\frac{0}{0}$ be any different?
Additionally, consider $0=5\cdot0$, always true for real numbers, but in this case $\frac{0}{0}=5$ not $1$
A: $0^0$ is well defined if it is defined as $0^0=\lim_{x\to 0}(x^x)$ . In this case $0^0=1$.
$0^0$ is undefined if it is defined as $0^0=\lim_{x\to 0 \text{ and } y\to 0}(x^y)$ , if no relationship between $x$ and $y$ is specified.
If a relationship between $x$ and $y$ is specified, the limit becomes defined. Then the corresponding value of $0^0$ becomes defined and can be any number or $\infty$ , depending on the relationship. Several examples were shown in the previous answers and many can be found in the publications. For example:
http://fr.scribd.com/doc/14709220/Zero-puissance-zero-Zero-to-the-Zeroth-Power
It is a non-sens to ask for the value of $0^0$ without refering to a particular and complete definition of $0^0$. Each particular definition of $0^0$ can be used in a limited domain of mathematics, according to the specified definition of $0^0$.
A: We know that $x^0 = 1$ for all $x \neq 0$ and $0^y = 0$ for all $y \neq 0$. 
What happens when these two cases collide? 

 We get different limits, and so $0^0$ is not well-defined

A: $0^0$ is considered an indeterminant form of a limit because $\lim_{x\to c} f(x)^{g(x)}$ is not always zero when $\lim_{x\to c} f(x)= 0$ and $\lim_{x\to c} g(x)=0$.  It depends on the behaviour of the two functions.
For example: $\lim_{x\to 0^+} 0^x = 0, \lim_{x\to 0^+} x^0 = 1$
