Proving $\lim \limits_{n\to +\infty } \left(1+\frac{x}{n}\right)^n=\text{e}^x$

Question

Here: Link between $\lim \limits_{n \to \infty} (1+{1/n})^n$ and $\lim \limits_{n \to \infty} (1+{x/n})^n$

is a proof of the formula for $e^x$. The thing causing my doubt is that we cannot assume that $u$ is an integer. The most common definition of limit of sequence concerns integers only. How to fix the original proof?

Answer 1

I am not sure if this is what you want but one can prove from the sequence definition that $$\lim\limits_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$$ as follows, given $x$ there is $n$ such that $$n\leq x<n+1$$ and thus $$\frac{1}{n+1}<\frac{1}{x}\leq\frac{1}{n}$$ and it follows $$\left(1+\frac{1}{n+1}\right)^n<\left(1+\frac{1}{x}\right)^x\leq \left(1+\frac{1}{n}\right)^{n+1}$$ And since the terms on the left and right limit to $e$ we see that the continuous limit and sequence limits are the same.

Answer 2

The definition of a limit at infinity for reals is that for any $n > N$, the difference between the function evaluated at $n$ and the limit is less than $\epsilon$, with every $\epsilon$ having a suitable $N$. If it's true for all reals greater than $N$, it must be true for all integers greater than $N$, since the integers are a subset of the reals.

Answer 3

If we use the usual limit

$$\lim_{X\to 0}\frac{\ln(1+X)}{X}=1$$

which comes the derivative,

we get that $\forall x\in \Bbb R$

$$\lim_{\alpha\to +\infty}\alpha\ln(1+\frac{x}{\alpha})=x$$

thus by taking the exponential,

$$\lim_{\alpha\to+\infty}(1+\frac{x}{\alpha})^{\alpha}=e^x.$$

Answer 4

I suppose I was looking for a theorem like this one: https://proofwiki.org/wiki/Limit_of_Sequence_is_Limit_of_Real_Function