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Zero to zero power

From what I understand $0^0$ is indeterminate, yet when you evaluate $\lim\limits_{x\to 0}x^0$ you get 1 (given on Something seems wrong about this.

If wolframalpha is correct, how can this be?

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marked as duplicate by Arturo Magidin, JavaMan, Austin Mohr, Zev Chonoles Jan 14 '12 at 7:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See this previous question. If $x\gt 0$, then $x^0=1$, so $\lim\limits_{x\to 0^+}x^0 = \lim\limits_{x\to 0^+}1 = 1$. – Arturo Magidin Jan 13 '12 at 17:31
$x^0$ is not the only limit of this form. Indeterminate means that other limits might lead to other answers. For example, what is $\lim_{x \to 0} 0^x$? – N. S. Jan 13 '12 at 17:35
Indeterminate, incidentally, does not mean undefined. $0^0=1$ by definition, at least when the exponent is an integer $0$. There are plenty of reasons for this, from convenience to counting arguments. On the other hand $\frac{1}{0}$ is undefined but not indeterminate. – Thomas Andrews Jan 13 '12 at 20:37
up vote 7 down vote accepted

Your limit is correct. The trouble is that is not the only way to approach $0^0$.

Consider $0^x$, $x>0$. Then the limit as $x$ approaches $0$ from above is equal to $0$. For the limit to exist, every pair of functions $f(x), g(x)$, with both functions approaching $0$ as $x \to 0$, must agree on $$\lim_{x \to 0} f(x)^{g(x)}$$

and these two examples show there are distinct pairs of functions which disagree on the limit.

(in this context, we might restrict to $f, g$ being nonnegative functions, but the same principle applies)

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