My Math Forum What is the relation between 0^0 and limx^x?

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 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. $\displaystyle 1 = \frac{x^1}{x^1} = x^{1-1} = x^0$ 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?
 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 $1$. Thanks from babaliaris
 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. Thanks from topsquark and babaliaris 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
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 Originally Posted by Maschke 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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