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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...

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$x^\infty$ 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.

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