What is $\lim_{x \rightarrow 0} x^0$?
What is $$\lim_{x \rightarrow 0} x^0$$? Would this equal to $\lim_{x \rightarrow 0}x^x = 1$?
If the limit is undefined, would $\lim_{x \rightarrow 0^+} x^0$ be defined?
$\endgroup$ 74 Answers
$\begingroup$$x^0=1$ for all $x$, in particular for $x\neq 0$. So this is the same as $\lim_{x\to 0} 1 = 1$.
$x^x$ is not well-defined on any $(-\delta,\delta)\setminus\{0\}$, so you have to be careful. $\lim_{x\to 0^+} x^x$ does exist.
$\endgroup$ 8 $\begingroup$(i) $$\lim_{x \to 0} x^0 = 1$$
because $a^0 = 1$ for any $a \neq 0$. But with the limit, we are considering the neighborhood of $0$, and not $0$ itself.
(ii)
$$\lim_{x \to 0^+} x^x = \exp \left( \lim_{x \to 0^+} x \log x \right) \stackrel{\mathcal{L}}{=} \exp(0)=1 .$$
$\endgroup$ 1 $\begingroup$$$x^0=1$$was based on the assumption that $x\neq 0$ and only then we could show that $$\frac{x^m}{x^m}=x^{m-m}=x^0=1$$ Therefore since when x goes to 0 x is never 0 then $x^0$goes to 1.
$\endgroup$ 1 $\begingroup$Answer:
$$\lim_{x \to 0} x^0 = 1 $$
because anything to the zero power will equal 1.
/edit
$\lim_{x\to 0^+} x^x$ does exist.
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