In some stuff I have this step: $$\lim_{g\to 0}\int_0^1\bigg(\frac{1-x^{\frac{g}{1+g}}}{\frac{g}{1+g}}\bigg)^ndx=\int_0^1\bigg(\lim_{g\to 0}\frac{1-x^{\frac{g}{1+g}}}{\frac{g}{1+g}}\bigg)^ndx$$ where $n$ is some not so important parameter. I'd like to show that this step IS actually valid. What I have to show is that $f_g(x)=\frac{1-x^{\frac{g}{1+g}}}{\frac{g}{1+g}}$ converges uniformly on $[0,1]$. But I'm confused about WHAT to limit, where, what to maximize. The definition says: Sequence of functions $f_n$ converges uniformly on some interval $I$ to $f$ iff $$\lim_{n\to\infty}\sup_{x\in I}|f_n(x)-f(x)|=0$$ This confuses me, do I limit the $g$ to infinity, or to zero?... Thanks for any useful tips, links and hints. My further steps:
Let $t=\frac{g}{1+g}$, thus $f_t(x)=\frac{1-x^t}{t}$ we also know that $\lim_{t\to 0}f_t(x)=-\ln{x}$. Now let $g_t(x)=\frac{1-x^t}{t}+\ln{x}$ we are to find $$\sup_{0\le x\le 1}{g_t(x)}=\max\{g_t(0),g_t(1),g_t(x_0)\}=g_t(1)$$ where $g'(x_0)=0$ is $x_0=1$. Thus we have $$\lim_{t\to\infty}\sup_{0\le x \le 1}|g_t(x)|=\lim_{t\to\infty}|0|=|0|=0$$ thus integrand converges uniformly and thus the step of interchanging limit and integral is valid. IS that correct?
You do not have uniform continuity, as @zhw. points out in the comments, but it is a straightforward application of the dominated convergence theorem and a little convex observation.
If $f$ is concave, then $f'(0) \ge {f(t)-f(0)\over t} \ge f'(t)$. (See https://en.wikipedia.org/wiki/Convex_function#Functions_of_one_variable, for example.)
For $x \in (0,1]$, the function $f(t) = - x^t = - e^{t \log x}$ is concave, hence $-\log x \ge {1-x^t \over t} \ge 0$.
Hence $({1-x^t \over t})^n \le |\log x|^n$.
It is straightforward to check that $\int_0^1 (\log x)^n dx = (-1)^n n!$.
Furthermore, for $x \in (0,1]$, $\lim_{t \downarrow 0} {1-x^t \over t} = - \log x$.
The dominated convergence theorem shows that $\lim_{t \downarrow 0} \int_0^1 ({1-x^t \over t})^n dx = \int_0^1 \lim_{t \downarrow 0} ({1-x^t \over t})^n dx = \int_0^1 (- \log x)^n dx = n!$.
Aside:
Using integration by parts, we get $\int_0^1 \log x dx = (x \log x) \mid_0^1 - \int_0^1 1dx = -1$, and $\int_0^1 (\log x)^n dx = (x (\log x)^n) \mid_0^1 - n\int_0^1 (\log x)^{n-1}dx = -n \int_0^1 (\log x)^{n-1}dx$. Induction shows that $\int_0^1 (\log x)^n dx = (-1)^n n!$.