# Does $e^0=1$ imply that $0^0=1$?

One of the ways to define $e^{x}$ is by its power series $$\left(\ast\right)\quad e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\ldots$$ The radius of convergence of this power series is infinite so this implies to every $x\in\mathbb{R}$.

Now as $0!=1$ (by definition i guess) we get that $$1=e^0=\sum_{n=0}^\infty\frac{0^n}{n!}=\frac{0^0}{0!}+\frac{0^1}{1!}+\frac{0^2}{2!}+\ldots=\frac{0^0}{1}=0^0$$ and something definitely doesn't feel quite right when plugging that $0$ into this series.

In many places $e^x$ is written in a more specific form as $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\ldots$$ and by that it seems like it overcomes the problem.

But if we always start by first pulling the $1$ out and only then plugging the $x$ in why wouldn't $e^{x}$ be defined as $$e^x=1+\sum_{k=1}^\infty\frac{x^n}{n!}$$ instead of as $\left(\ast\right)$? Is the term $\frac{0^0}{0!}$ in the sum just a notation for $1$?

• Well, I think the general convention is to take $0^0=1$ indeed. Note however that it is not well defined for limits, ie if you are given two sequences $(a_n)$ and $(b_n)$, each converging to $0$, such that $a_n^{b_n}$ is well defined for every $n$, then you can't say $a_n^{b_n} \rightarrow 0^0=1$. You a priori have no clue whether it even converges. So, consider $0^0=1$ only as a "ponctual" convention. It has no deep meaning whatsoever, and it is not a property. Just a convenient convention. – Suzet Jul 3 '18 at 23:06
• Power series like the one you present are a good argument why we should define $0^0=1$ rather than leaving $0^0$ undefined. More generally, it could be argued that any "empty product", i.e. a product whose factors are "indexed" over the empty set in some sense, should be equal to $1$, the neutral element with respect to multiplication. There must be many other threads on Math SE about the topics I mention here! – Jeppe Stig Nielsen Jul 3 '18 at 23:11
• @JeppeStigNielsen Well i did read some about it here and as Arturo concludes it "We basically define it (or fail to define it) in whichever way it is most useful and natural to do so for the context in question." but i guess i hoped it will be something less arbitrary than that. – Jon Jul 3 '18 at 23:21
• In Apostol's Introduction to Analytic Number Theory he states that $0^0=1$. Also for $0!=1$, you should check the Gamma function, which extends the factorial function and has $\Gamma(1)=0!=1$. – Daniel Buck Jul 3 '18 at 23:27
• $0^0$ is undefined. It usually appears as $\lim {y\to 0} {x\to 0}\ x^y$, so the actual limit will depend on having a relationship between $x$ and $y$ if any. – herb steinberg Jul 4 '18 at 0:30

If you want to stay out of trouble, then the last way you wrote it may be the best. This is because the function $x^y$ for real $x$ and $y$ is generally defined as $\exp(y\log x)$, and $\exp$ and $\log$ are generally defined by power series, which require $x^k$ where $k$ is a whole number. This is easy to define as "repeated multiplication", as long as you don't let $k=0$. Then, the power series definition will imply that $x^0 = 1$ for all nonzero real $x$. However, it gives no value for when $x=0$. By limits, it's clear to see that there is no value for $0^0$ which makes $x^y$ continuous at $(0,0)$, so there's no "natural" choice in that sense. In this case, the power series for $e^x$ does not imply any value for $0^0$.

You can, of course, deal with $k=0$ at the beginning, by letting $x^0 = 1$ for all real $x$ (including $0$). This basically chooses $0^0$ to be $1$ as a convention, and one upshot of this is that it simplifies the power series definition of exponential functions. But in this case, the choice of $0^0=1$ implies the power series for $e^x$ is $\sum_{i=0}^\infty \frac{x^i}{i!}$, not the other way around.

So, no matter how you slice it, the power series for $e^x$ does not imply that $0^0 = 1$, but rather we can choose to let $0^0 = 1$ to imply the simplified power series for $e^x$.

$0^0$ is equal to $1$ because multiplying by something $0$ times amounts to not multiplying by anything, which is the same as multiplying by $1.$

However, at the same time, $\text{“ } 0^0 \text{ ''}$ is an indeterminate form because if $f\to0$ and $g\to0$ as $x\to\text{something,}$ then $f^g$ can approach $0$ or $+\infty$ or $6$ or any other member of $[0,+\infty],$ depending on what functions $f$ and $g$ are. However, in a sense, in most cases the limit of $f^g$ will be $1.$ One case is when $(f,g)\to(0,0)$ within some "sector", i.e. it stays between two positively sloped lines in the $(f,g)$-plane.

Other answers have used taylor series, mine will use more basic calc 1 math

$$\lim_{x\to0^+} 0^x=0$$

and

$$\lim_{x\to0^+}\;\; x^0=1$$

As a General rule, $0^0$ is indeterminate. However lets examine the function when both base and exponent go to zero, the function $x^x$, and take the limit as x->0

$$x^x = e^{x\ln(x)}$$

We can examine the behaviour of this function by examining the behaviour of the exponent and then use it to determine the behaviour of the function

$$x\ln(x)=\frac{\ln(x)}{x^{-1}}$$

Using L'Hopital

$$\lim_{x\to0}\;\;\frac{\frac{1}{x}}{\frac{-1}{x^2}}=0$$

Since both the left and right sided limit equal 0, the limit of this exponent goes to 0 and $0^0=1$ when examining it like this, but as a general rule, $0^0$ is indeterminate.

EDIT: Correctness of response

• Notice how the following are coded: $$\lim_{x\to0}$$ $$x\ln x$$ That is correct MathJax usage. Just right-click on those expressions and click on "show math as" and then on "TeX commands" and you'll see it. – Michael Hardy Jul 4 '18 at 0:32
• Thank you, this was my first post ever! Still learning – Colin Hicks Jul 4 '18 at 0:40
• You write $$\lim_{x\to0^-} 0^x=0$$ but for what negative exponents, $-1<x<0$ is the exponentiation $0^x$ even defined?! I do not think it makes sense. I would even say $$\lim_{x\to0^-} 0^x=\infty$$ could be better, if you insist on considering $\lim_{x\to0^-} 0^x$. – Jeppe Stig Nielsen Jul 4 '18 at 5:54
• I checked wolfram alpha, and you are correct. Only the right sided limit goes to zero, the left sided limit goes to infinity. The proof is trivial, it would be $$\lim{x\to0^-}\; 0^x=\lim{x\to0^+} \frac{1}{0^x}$$ Which, using the result from the right sided limit, goes to infinity. Thank you! – Colin Hicks Jul 4 '18 at 16:08