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March 22nd, 2019, 04:52 PM   #1
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What is the relation between 0^0 and limx^x?

Since we can not divide with zero, the equation below is true for every x except zero.
$\displaystyle
1 = \frac{x^1}{x^1} = x^{1-1} = x^0
$

Using limits we can see that $\displaystyle 0^0$ is actually getting close to 1.
$\displaystyle
lim_{x \to 0} x^x = 1
$

Does this say anything?
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March 22nd, 2019, 05:02 PM   #2
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Well it's interesting. The problem is that the limit depends on how you look at it.

* What is $\lim_{x \to 0} k^x$ where $k \neq 0$ is constant? It's $1$.

* But what is $\lim_{x \to 0} x^ k$ where $k \neq 0$ is constant? It's 0.

* And what is $\lim_{x \to 0} x^ x$? It's $1$ as you note.

So we can't just define $0^0$ as $1$, although if we must pick some convention, that's the one to pick.

Another moral reason for defining $0^0 = 1$ is that in set theory, $0$ is represented by the empty set $\emptyset$. And $\emptyset^\emptyset$ is the set of functions from $\emptyset$ to $\emptyset$. There does happen to be exactly one such function, namely the empty function! When you study enough math after a while you get used to this kind of reasoning. Therefore $\emptyset^\emptyset = \{\emptyset\}$; and another name for $\{\emptyset\}$ is $1$.
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March 22nd, 2019, 06:01 PM   #3
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\begin{align}\lim_{(x,y) \to (0,0)} x^y &= \lim_{(x,y) \to (0,0)} e^{y\log x} & \text{so with $y=\frac{k}{\log x}$} \\ &= \lim_{(x,y) \to (0,0)} e^{k} \end{align}
So we can get any positive value that we desire for the limit, you just have to pick the appropriate path to the origin.
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Last edited by v8archie; March 22nd, 2019 at 06:07 PM.
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March 23rd, 2019, 01:47 PM   #4
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$x^x=e^{xln(x)}$. $\lim_{x\to 0}xln(x)=0$, so $\lim_{x\to 0}x^x=1$.
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April 8th, 2019, 11:30 AM   #5
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Quote:
Originally Posted by Maschke View Post
Well it's interesting. The problem is that the limit depends on how you look at it.

* What is $\lim_{x \to 0} k^x$ where $k \neq 0$ is constant? It's $1$.

* But what is $\lim_{x \to 0} x^ k$ where $k \neq 0$ is constant? It's 0.

* And what is $\lim_{x \to 0} x^ x$? It's $1$ as you note.

So we can't just define $0^0$ as $1$, although if we must pick some convention, that's the one to pick.

Another moral reason for defining $0^0 = 1$ is that in set theory, $0$ is represented by the empty set $\emptyset$. And $\emptyset^\emptyset$ is the set of functions from $\emptyset$ to $\emptyset$. There does happen to be exactly one such function, namely the empty function! When you study enough math after a while you get used to this kind of reasoning. Therefore $\emptyset^\emptyset = \{\emptyset\}$; and another name for $\{\emptyset\}$ is $1$.
Actually, we can define $0^0=1$, not to confuse doing that with any limit. Just like we define $x^0 =1$ to make $\frac{x^m}{x^n} = x^{m-n}$ work when $m=n$. Consider the binomial expansion$$(1+x)^n = \sum_{k=0}^n \binom n k 1^k x^{n-k} = \binom n 0 1^0 x^n +\dots
\binom n n 1^nx^0$$If you want that formula to work for $x=0$ you need to define $0^0$ = 1. (Look at the last term).
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