I am taking an analysis course at university and one of the recent problems we had was about defining $e^x$ by
$$e^x = \lim_{n\to \infty} \Big(1 + \frac{x}{n}\Big)^n$$
Now, I've seen elsewhere that you can define e^x by
$$e^x = \lim_{n\to 0} \Big(1 + nx\Big)^\frac{1}{n}$$
and I can understand how this is derived from the first definition with
$$\lim_{n\to \infty} \Big(1 + \frac{x}{n}\Big)^n = \lim_{n\to 0} \Big(1 + \frac{x}{1/n}\Big)^\frac{1}{n} = \lim_{n\to 0} \Big(1 + nx\Big)^\frac{1}{n}$$
but I was wondering if it is true in general that for any uniformly convergent functional sequence $(f_n)$ we can say that $(f_\frac{1}{n})$ is also uniformly convergent with
$$\lim_{n\to \infty}f_n = \lim_{n\to 0}f_\frac{1}{n}$$
After a bit of looking, I haven't been able to find any documentation on this or proof of this proposition, although intuitively it seems correct. Any guidance on where to look or start would be appreciated.