What is $\lim_{x=0\rightarrow1}x^\infty$? I know it's all about definition...
But still I want to know whether the answer is $0$, $1$, impossible to say or something else, like that the mathematical statement is wrong.
However, to clarify, this normally applies:
\begin{equation*}
x^\infty=
\begin{cases}
\vdots& \vdots \\
0& \text{if }\space 0\leq x<1\\
1& \text{if }\space x=1\\
\vdots& \vdots
\end{cases}
\end{equation*}
But I don't know what happens when you use limitations like this:
\begin{equation*}
\lim_{x=0\rightarrow1}x^\infty
\end{equation*}
Maby the answer can be everything from $0$ to $1$, just speculating...
 A: $x^\infty$ doesn't have any generally-accepted meaning, but $\lim_{y \rightarrow \infty} x^y$ does, so let's use that instead.
Lemma: If $0 < x < 1$, then
$$\lim_{y \rightarrow \infty} x^y = 0$$.
Proof: We must show that $\forall \epsilon > 0.\ \exists c>0.\ \forall y>c.\ \left| x^y - 0 \right|<\epsilon$. Given $\epsilon$, let $c = \log_x \epsilon$. Then choosing $k>0$ such that $y=k+c$, $\left | x^y-0 \right | = \left | x^y\right | = \left | x^{k + \log_x \epsilon} \right | = \left | x^k \cdot \epsilon \right |$. By hypothesis $0 < x < 1$, $k > 0$, and $\epsilon > 0$, so $0 < x^k < 1$, so $\left| x^k \cdot \epsilon \right| < \epsilon$, which completes the proof.
Theorem:
$$\lim_{x \rightarrow 1^{-}} \lim_{y \rightarrow \infty}x^y = 0$$
Proof: We must show that $\forall \epsilon > 0.\ \exists \delta<1.\ \delta < x < 1 \rightarrow \left|\left(\lim_{y \rightarrow \infty} x^y\right)-0\right| < \epsilon $. For any choice of $\epsilon$, let $\delta = 0$. Then $0 < x < 1$ so our result follows immediately from the lemma.
