# Power series and the value of the expression $0^0$ [duplicate]

This question already has an answer here:

I have a doubt regarding the value of the expression $0^0$. I know this value is taken as indeterminate as far as limits are concerned. All was fine upto now. But when I encountered power series, I found out when $x=a$ in the expression summation $[b (x-a)^n]$ where $n=0$ to infinity, of the power series, then the series always converges which is understood. But what bothers me is its value converges to $b$ and not $0$. That is the first term of the power series is written as $b \cdot 0^0$ and $0^0$ is taken as $1$ and not as indeterminate.

Can anyone tell me why this is so? How is it possible at one time we define $0^0$ as indeterminate and at other time its value is taken as $1$? Could anyone help me on this one? Thanks.

-
"ow is it possible at one time we define $0^0$ as indeterminate and at other time its value is taken as $1$?" It simply reflects discontinuity of $(x, y) \mapsto x^y$ at $(0, 0)$. Nothing special or mind-blowing, you just have to remember what justifies the usual tricks in limit computation. –  Alexei Averchenko May 10 '13 at 2:20

## marked as duplicate by Nate Eldredge, Amzoti, Henry T. Horton, amWhy, Shuhao CaoJun 8 '13 at 20:38

Generally $0^0$ is taken to be $1$, so that the term of degree $0$ in a polynomial or power series can be written as $c_0x^0$, rather than having some special exception for $x=0$. It makes the notation cleaner.

As an aside, when one talks about an expression being an indeterminant form, say $E(a,b)$, one usually means that if we have functions $f(x),g(x)$ such that $\lim\limits_{x\to x_0}f(x)=a$ and $\lim\limits_{x\to x_0} g(x)=b$, we cannot conclude that $\lim\limits_{x\to x_0}E(f(x),g(x))=E(a,b)$. In the case of $0^0$, this means that even if $\lim\limits_{x\to x_0}f(x)=0$ and $\lim\limits_{x\to x_0} g(x)=0$ we cannot conclude that $\lim\limits_{x\to x_0}f(x)^{g(x)}=0^0$, no matter how $0^0$ is defined. Note that this is not the same as saying that $0^0$ is undefined. It just says the expression does not "play well" with taking limits.

-
Alex, if we take $0^0=1$ then we have $\ln(0^0)=\ln(1)$ which implies that $0 \ln(0)=0$. However the right hand side of this expression is undefined well the left is zero. What gives here? –  mtiano May 10 '13 at 1:01
@mtiano Careful, $\log a^x=x\log a$ is true when $a> 0$. –  Pedro Tamaroff May 10 '13 at 1:04
@mtiano No choice of definition for $0^0$ will make every expression work out nicely. No matter what, some definitions involving exponents will involve an exception in this case (i.e. $\ln(x^y)=y\ln x$ as long as $x>0$; in fact, any $y$ will cause problems if $x=0$, regardless of how $0^0$ is defined). The definition $0^0=1$ seems to require the fewest exceptions, making it the most convenient. –  Alex Becker May 10 '13 at 1:05
got it now... thanks for your feedback... –  under-root May 10 '13 at 1:45