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March 22nd, 2019, 04:52 PM  #1 
Senior Member Joined: Oct 2015 From: Greece Posts: 115 Thanks: 8  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^{11} = 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? 
March 22nd, 2019, 05:02 PM  #2 
Senior Member Joined: Aug 2012 Posts: 2,265 Thanks: 690 
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 $1$. 
March 22nd, 2019, 06:01 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,640 Thanks: 2624 Math Focus: Mainly analysis and algebra 
\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. Last edited by v8archie; March 22nd, 2019 at 06:07 PM. 
March 23rd, 2019, 01:47 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,732 Thanks: 689 
$x^x=e^{xln(x)}$. $\lim_{x\to 0}xln(x)=0$, so $\lim_{x\to 0}x^x=1$.

April 8th, 2019, 11:30 AM  #5  
Newbie Joined: Oct 2018 From: Arizona Posts: 4 Thanks: 0  Quote:
\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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limit, limxx, power, relation, zeroinfinity 
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