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I've been searching for a rigorious proof of the existence of $\lim\limits_{n\to\infty} (1+\frac{1}{n})^n$ for $n \in \mathbb{R}$, but the proofs I came across only allow $n$ to take integer values (Some examples are What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists? and https://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function#Equivalence_of_the_characterizations)

However, the fact that $\lim\limits_{n\to\infty} f(x)$ exists, where $f:\mathbb{Z} \to \mathbb{R}$, doesn't necessarily imply that $\lim\limits_{n\to\infty} g(x)$ also exists, where $g:\mathbb{R} \to \mathbb{R}$ and $\forall x \in \mathbb{Z}\,(g(x) = f(x))$

An example would be $\sin(\pi x)$, which $\lim\limits_{x\to\infty} \sin{\pi x} = 0$ if we add the restriction $x \in \mathbb{Z}$, but not for $x \in \mathbb{R}$

Therefore is there any way to truly prove $\lim\limits_{n\to\infty} (1+\frac{1}{n})^n$, letting $n$ take real values?

gldanoob
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  • How have you defined raising a real number to an irrational exponent? The proof will depend on this definition. – Alann Rosas Oct 14 '21 at 04:20
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    FYI, there is a reason why this is usually considered as a sequence, not as a real function. In order to talk about this limit for real $n$, you need to first define what it means to raise one real number to the power of another. We typically define $x^y = \exp(y \ln(x))$, which means we also need a definition of the natural exponential $\exp$. One of the ways we do this is to define it as a (sequential) limit: $\exp(x) = \lim_{n \to \infty} (1 + x/n)^n$ (note: we only raise a number to the power of positive integers, so repeated multiplication is fine). – Theo Bendit Oct 14 '21 at 04:29
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    This is where the interest in these sequential limits arose: to construct $e$ and the exponential, from which we construct the logarithm, and finally exponentiation of real numbers (or even complex numbers). As such, we really only can be interested in the real limit after we have settled the sequential limit (or used another way of defining the exponential/logarithm). – Theo Bendit Oct 14 '21 at 04:32
  • You can actually define real exponents without $exp$, by raising to fractional powers that approximate the real number then taking the limit – gldanoob Oct 14 '21 at 04:36
  • @gldanoob There are a few ways to navigate this, to be sure. The approximation via fractional powers is considered one of the less elegant ways of doing this. You would need to establish that this construction is well-defined, as there are many ways to approximate an irrational number by rational numbers (e.g. show that $x^y$ as a function of rationals is locally uniformly continuous). This construction doesn't generalise easily to the complex numbers too, which is another reason why defining $\exp$ first is generally preferred. – Theo Bendit Oct 14 '21 at 18:53

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