Let be $f_n:(0,1)\to\mathbb{R}$, where $f_n(x)=x^n$ for each $n\in\mathbb{N}$ and $\left(f_n\right)_{n\in\mathbb{N}}$ a sequence of functions. (We already know that it converges pointwisely)
Show that both limits $\lim\limits_{n\to\infty}\lim\limits_{x\to c}x^n$ and $\lim\limits_{x\to c}\lim\limits_{n\to\infty}x^n$ exist for an arbitrary $c\in(0,1)$ and that $\lim\limits_{n\to\infty}\lim\limits_{x\to c}x^n=\lim\limits_{x\to c}\lim\limits_{n\to\infty}x^n$.
My approach:
$1.)$ First we check $\lim\limits_{n\to\infty}\lim\limits_{x\to c}x^n$. We define $L_n:=\lim\limits_{x\to c}x^n$ and notice that this limit exists, due to continuity, for each $n\in \mathbb{N}$. We show the existence of $\lim\limits_{n\to\infty}\lim\limits_{x\to c}x^n$ by proving that $\left(L_n\right)_{n\in\mathbb{N}}$ is a Cauchy sequence and hence convergent (completeness of $\mathbb{R}$). Let be $\epsilon>0$ arbitralily chosen.
We know that for each $m,n$ there exist a $\delta_1>0$ and $\delta_2>0$ such that for all $x$ with $|x-c|<\delta_1\implies |L_n-x^n|<\frac{\epsilon}{3}$ and for all $x$ with $|x-c|<\delta_2\implies |L_m-x^m|<\frac{\epsilon}{3}$. As $c\in(0,1)$ we can find a $\delta_3>0$ such that $c+\delta_3<1$.We set $\delta:=\min\{\delta_1,\delta_2,\delta_3\}$. As $\lim\limits_{n\to\infty}(c+\delta_3)^n=0$ we know that there exists a $n_0\in\mathbb{N}$ such that for all $n>n_0\implies(c+\delta_3)^n<\frac{\epsilon}{6}$. Finally, for all $m,n>n_0$ we can draw the following conclusion: $$ |L_m-L_n|=|L_m-x^n+x^n-x^m+x^m-L_n|\leq|L_m-x^n|+|x^n-x^m|+|x^m-L_n|\\ <\frac{\epsilon}{3}+|x^n-x^m|+\frac{\epsilon}{3}<\frac{\epsilon}{3}+(c+\delta_3)^n+(c+\delta_3)^m+\frac{\epsilon}{3}<\epsilon. $$ Hence, $\lim\limits_{n\to\infty}\lim\limits_{x\to c}x^n$ exists. If I exchange $L_m$ in the above argumentation with $0$ then we immediately see that it must be $\lim\limits_{n\to\infty}\lim\limits_{x\to c}x^n=0$.
$2.)$ Now we check $\lim\limits_{x\to c}\lim\limits_{n\to\infty}x^n$. We define $L_x:=\lim\limits_{n\to\infty}x^n$ and notice that this limit exists, due to pointwise convergence, for each $x\in (0,1)$. We show the existence of $\lim\limits_{x\to c}L_x$ by applying the Cauchy criterion of limits of functions, namely for an arbitrary $\epsilon>0$ there must exist a $\delta>0$ such that for all $x,y\in(0,1)$ with $|x-c|<\delta$ and $|y-c|<\delta\implies |L_x-L_y|<\epsilon$. Let be $\epsilon>0$ arbitralily chosen.
We know that for each $x$ and $y$ there exist a $n_1>0$ and $n_2>0$ such that for all $n>n_1\implies |L_x-x^n|<\frac{\epsilon}{3}$ and for all $n>n_2\implies |L_y-x^m|<\frac{\epsilon}{3}$. As $c\in(0,1)$ we can find a $\delta>0$ such that $c+\delta<1$. As $\lim\limits_{n\to\infty}(c+\delta)^n=0$ we know that there exists a $n_3\in\mathbb{N}$ such that for all $n>n_3\implies(c+\delta_3)^n<\frac{\epsilon}{6}$. Finally, for all $x,y$ with $|x-c|<\delta$ and $|y-c|<\delta$ and $m,n> \max\{n_1,n_2,n_3\}$ we can draw the following conclusion: $$ |L_x-L_y|=|L_x-x^n+x^n-x^m+x^m-L_y|\leq|L_x-x^n|+|x^n-x^m|+|x^m-L_y|\\ <\frac{\epsilon}{3}+|x^n-x^m|+\frac{\epsilon}{3}<\frac{\epsilon}{3}+(c+\delta)^n+(c+\delta)^m+\frac{\epsilon}{3}<\epsilon. $$ Hence, $\lim\limits_{x\to c}\lim\limits_{n\to\infty}x^n$ exists. If I exchange $L_y$ in the above argumentation with $0$ then we immediately see that it must be $\lim\limits_{x\to c}\lim\limits_{n\to\infty}x^n=0$.
Is this attempt correct? Maybe there is a faster or more elegant way? Any comments are appreciated :)