How to prove definition of exponential function as limit of powers converges

[brian44]

I've tried and searched for a long time, and I haven't been able to prove or find a proof that the following sequence converges (without using another definition of the exponential function):

\forall x \in \mathbb{R}. Prove that:

\lim_{n \rightarrow \infty} (1+ x/n)^n exists.

I can prove that it is monotonic and using binomial theorem I can show that it is bounded for x=1. However if I try to use the same approach for general x, I get the power series for e^x and can only say it is bounded if I can prove the power series converges, which I don't know how to do. But even if I did is there any way to prove this limit exists without proving the power series converges (the other definition of e^x)?


Another related question I can't figure out is, how can I prove that


\lim_{n \rightarrow \infty} (1+ x/n + o(x/n))^n = \lim_{n \rightarrow \infty}(1 + x/n)^n

where o(x/n) is any function that goes to 0 more quickly as (x/n) \rightarrow 0 than (x/n)?

 

[deluks917]

I admit I don't know much real analysis. However if we have the proof for x =1 can't we let u= n/x. Then u goes to inf. as n does, x/n = 1/u and n = u*x. If we call the result for x=1 e, we get (lim u->inf (1 + 1/u))^x = e^x.

 

[brian44]

The problem with such an approach is that it uses the exponential function and its properties in the definition of the exponential function itself - so I would consider it circular reasoning.

Before defining e^x we have definitions for integer powers as products, but without defining the exponential function and logarithms, we can't say what something^x, where x is a real number, means, specifically I believe we define these general powers as a\in \mathbb{R^+} x\in\mathbb{R}, a^x := e^{x \log(a)}

 

[Gib Z]

Without ever knowing anything about the exponential function you can expand the limmand into a series using the binomial theorem and show that series is bounded by \sum \frac{x^n}{n!} which converges for all x by the ratio test.