Let $b \in \mathbb{R}$, then $ \forall n \in \mathbb{N}(\frac{\lfloor b \cdot 10^n\rfloor}{10^n} \in \mathbb{R})$, but does $\frac{\lfloor b \cdot 10^n\rfloor}{10^n}$ have a particular name?
And is the following correct?
Let $a,b \in \mathbb{R}$, then $a^b=\lim\limits_{n \to +\infty}a^\frac{\lfloor b \cdot 10^n\rfloor}{10^n}$?
Thanks in advance!
By definition, $\lfloor b\cdot 10^n\rfloor$ is the floor of $b\cdot 10^n$, that is, the unique integer $n \in \Bbb Z$ such that $n \le b\cdot 10^n < n+1$.
In particular, then, $\dfrac{\lfloor b\cdot10^n\rfloor}{10^n}$ is a rational number.
Moreover, we have that:
$$b-\frac{\lfloor b\cdot10^n\rfloor}{10^n} = \frac{b \cdot 10^n - \lfloor b\cdot10^n\rfloor}{10^n}$$
and it is not hard to see that for any $x \in \Bbb R$, $0 \le x - \lfloor x \rfloor<1$. That is, $\dfrac{\lfloor b\cdot10^n\rfloor}{10^n}$ is a rational number that differs from $b$ by less than $\dfrac{1}{10^n}$.
In other words:
$$0 \le \lim_{n\to\infty} b- \dfrac{\lfloor b\cdot10^n\rfloor}{10^n} < \lim_{n\to\infty} \frac1{10^n} = 0$$
so by the squeeze theorem:
$$\lim_{n\to\infty} b-\dfrac{\lfloor b\cdot10^n\rfloor}{10^n} = 0$$
meaning that $\dfrac{\lfloor b\cdot10^n\rfloor}{10^n}$ converges to $b$ as $n \to\infty$.
The continuity of $a^x$ in $x$ then assures that:
$$a^b = \lim_{n\to\infty} a^{\frac{\lfloor b\cdot10^n\rfloor}{10^n}}$$
as desired.
On the other hand, this convergence can be taken as the definition of $a^b$ for real $b$ and $a > 0$: for rational $b = \frac pq$, we can use roots and powers to calculate $a^b = \sqrt[q]{a^p}$.
One then defines $a^b = \lim_{n\to\infty} a^{b_n}$ for any sequence $(b_n)_n$ that converges to $b$. (Of course, it then has to be shown that this does not depend on the particular choice of $(b_n)_n$, which requires some work.)
I hope that clarifies things for you.