Question:
Is $$\lim_{n\to\infty}\left(\left(1+\frac{1}{n}\right)^{nx}\right)=\lim_{n\to\infty}\left(\left(1+\frac{x}{n}\right)^{n}\right)$$
Background:
I am trying to show the equivalence of definitons of $e^x$ starting with $$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$
I proceeded as follows $$e^x=\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\right)^x$$ and as the limit does not depend on $x$ then$$e^x=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}$$
Next I expanded using the binomial expansion:
$$e^x=\lim_{n\to\infty}\left(1+nx\frac{1}{n}+\frac{nx\left(nx-1\right)}{2!}\left(\frac{1}{n}\right)^2+...+\right)$$
$$e^x=\lim_{n\to\infty}\left(1+x+\frac{x^2-\frac{x}{n}}{2!}+...+\right)$$ $$e^x=1+x+\frac{x^2}{2!}+...+$$ I have seen $e^x$ defined as $$\lim_{n\to\infty}\left(\left(1+\frac{x}{n}\right)^{n}\right)$$ and am trying to reconcile that fact.
Using the binomial expansion $$\lim_{n\to\infty}\left(\left(1+\frac{x}{n}\right)^{n}\right)=\lim_{n\to\infty}\left(1+n\left(\frac{x}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{x}{n}\right)^2+...+\right)$$ $$=1+x+\frac{x^2}{2!}+...+=e^x$$
So it seems the two are equivalent as $n\to\infty$ but not before. Is this correct? I feel it is and that the two functions are equivalent in the limit.
A broader question therefore is:
Are there extra algebraic manipulations such as moving the $x$ inside that are valid in limit situations? Perhaps under given constraints?
Thanks for your help in advance. I have scanned the site for the answer but didn't see anything but could have missed it.
You can make the following change of variable $nx=k$ so $\lim\limits_{n\rightarrow \infty} \left(1+\dfrac{1}{n} \right)^{n x}=\lim\limits_{k\rightarrow \infty} \left(1+\dfrac{x}{k} \right)^k$.