# How does $e^0 = 1$ if you define $e^x = \sum_{n = 0}^\infty x^n/n!$

How does $e^0 = 1$ if you define $e^x = \sum_{n = 0}^\infty x^n/n!$ since $e^0 = 0^0$, and we know the right hand side is undefined?

• $$0^0=1{}{}{}$$ – Asaf Karagila May 20 '13 at 1:01
• Also, the right side is not undefined, because by definition $0!=1$. There are probably as many arguments for this convention as for $0^0=1$. – Gyu Eun Lee May 20 '13 at 1:30
• But you defined the exponential function wrong. – mick Dec 13 '13 at 20:40

## 2 Answers

Actually, there are many arguments to define $0^0 := 1$. This is one of the many compelling reasons.

Arguably the most persuasive is the fact that if $f$ and $g$ are analytic functions with $f(a)=g(a)=0$, not identically zero in a neighborhood of $a$, then $$\lim_{x\to a} f(x)^{g(x)} = 1.$$

In this situation, you should view that $0^0$ as

$$\lim_{x \to 0} x^0,$$

which is pretty easily seen to be $1$.

In situations of continuity, it is often better do define $0^0$ as $1$. One reason I've heard for this is that continuity of $x^0$ is more useful than continuity of the related function $0^x$.

• Or $\lim_{x\to 0}x^x$. – Berci May 20 '13 at 1:20
• @Berci In this context our exponents remain fixed, while our base varies with the center of expansion. Thus, it would be strange to view both varying simultaneous. – awwalker May 20 '13 at 2:42