
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
March 22nd, 2019, 04:52 PM  #1 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 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: 2,387 Thanks: 746 
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: 7,683 Thanks: 2664 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: 6,822 Thanks: 723 
$x^x=e^{xln(x)}$. $\lim_{x\to 0}xln(x)=0$, 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).  

Tags 
limit, limxx, power, relation, zeroinfinity 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Recurrence Relation and Closed Form Relation  uniquegel  Algebra  4  September 8th, 2014 04:18 PM 
Relation  Rohit Kakkar  Calculus  1  July 23rd, 2014 08:43 AM 
Relation?  jaredbeach  Algebra  3  August 21st, 2011 12:18 PM 
Relation?  jaredbeach  Calculus  0  December 31st, 1969 04:00 PM 
Relation  gaussrelatz  Algebra  0  December 31st, 1969 04:00 PM 