We consider $f_n(x)=x^n$ on $[0,1]$. Each function $f_n(x)$ is continuous, but the limit function $f(x)$ is not continuous: $$ f(x)=\left\{ \begin{array}{ll} 0, 0\leq x<1\\ 1, x=1\\ \end{array} \right. $$
Question 1: How can we prove that: $$ \lim_{n\rightarrow \infty}(\lim_{x\rightarrow 1}f_n(x))=1 $$ $$ \lim_{x\rightarrow 1}(\lim_{n\rightarrow \infty}f_n(x))=0 $$
Question 2: Is uniform convergence of $f_n(x)\rightarrow f(x)$ a sufficient condition to get the same result if we interchange the limits?
Question 1:
$$\lim_{x\to 1} f_n(x) = \lim_{x\to 1} x^n = 1^n = 1$$
because, for every $n$, the function $x\mapsto x^n$ is continuous.
Therefore,
$$\lim_{n\to\infty} \lim_{x\to 1} f_n(x) = \lim_{n\to\infty} 1 = 1$$
On the other hand, if $x<1$ (we can assume this when calculating $\lim_{x\to 1}$), then
$$\lim_{n\to\infty} f_n(x) = \lim_{n\to\infty}x^n = 0$$
So $$\lim_{x\to 1}\lim_{n\to\infty} f_n(x) =\lim_{x\to 1} 0 = 0$$
Question 2: Yes.