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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?

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

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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 $0^x$.

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  • $\begingroup$ Or $\lim_{x\to 0}x^x$. $\endgroup$ – Berci May 20 '13 at 1:20
  • $\begingroup$ @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. $\endgroup$ – awwalker May 20 '13 at 2:42

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