Is this enough to justify the change of order of limits? If I want to show that $$\lim_{x\to c}f(x)=\lim_{x\to c}\sum_{n=3}^\infty c_n x^n=\sum_{n=3}^\infty\lim_{x\to c} c_n x^n\qquad 0\leq x\leq 1$$ is it enough to say that the function is uniformly convergent on a compact interval hence it is ok  to change the order of limits? also which theorem is this due to?
Note also that $\sum_{n=3}^\infty c_n$ is convergent and $f$ is defined as the series $\sum_{n=3}^\infty c_nx^n$.
 A: In fact, it can be proved that,

Suppose $f_n \rightarrow f$ uniformly on a set $E$ in a metric space &
  $x$ is a limit point of $E$. If  $lim_{t \to x}f_n(t) = A_n$ for all
  $n$, then {$A_n$} converges & $lim_{t \to x}f(t)= \lim_{n \to
\infty}A_n$. i.e: $$lim_{t \to x}lim_{n \to \infty}f_n(t)=lim_{n \to \infty}lim_{t \to x }f_n(t)$$

In your case , let $f_N(x) = \sum_{n=3}^{N}c_nx^n$. Since the set of polynomials on $[0,1]$ is dense in $\mathbb{C}[0,1]$, there exists a continuous function $f$ on $[0,1]$ s.t $f$ is the uniform limit of the sequence of polynomials (By the Stone-Weierstrass theorem). 
So, $f_N \rightarrow f$ uniformly. Since $[0,1]$ is closed it contains all of its limit points. Hence, the above theorem can be applied to see why the limits can be interchanged.
A: Interchanging limits and summations is essentially a consequence of the Lebesgue Dominated Convergence Theorem. Here you would consider the series as an integral with respect to the counting measure. You let the functions $f_n(x)= \sum_{k=3}^n c_k x^k$ and let $f_n\to f$.
However you could also show this without using any advanced machinery. You would however need to use the Weierstrass criterion on the uniform convergence of series using the fact that you have that $\sum_{n=3}^{\infty} c_n$ converges and the $c_n$ are of course bounding $c_n x_n$ on the interval $[0,1]$.
Let $f_n=\sum_{k=3}^n c_k x^k$. You want to show that $$\lim_{x\to c} f(x)= \lim_{n\to \infty} \lim_{x\to c} f_n(x)$$ if $f_n$ is uniformly convergent to $f$.
Let $A = \lim_{x\to a} f(x)\ \  \text{ and }\ \ a_n = \lim_{x\to a} f_n(x)$
We will establish that $\lim_{n\to \infty} a_n = A$.  we have 
$$\forall x\in [0,1], |a_n -A | \le |a_n - f_n(x)| + | f_n(x) -f(x)| + |f(x) - A|.$$
1)
 As $f_n \to f$ uniformly by the Weierstrass criterion for uniform convergence of a series, there is $N\in \mathbb N$ such that $|f_n(x) - f(x)| < \epsilon/3$ for all $n\ge N$ and for all $x\in [0,1]$. 
2)
 There is $\delta>0$ so that $|f(x) - A|<\epsilon/3$ for all $x\in [0,1]$ such that $|x- a| < \delta$. 
3)
In addition, there is $\delta_n >0$ so that $|f_n(x) - a_n| < \epsilon/3$ for all $x\in [0,1]$ so that $0<|x-a|<\delta_n$. 
Now if $n\ge N$, consider $x_n$ such that $0<|x_n-a| < \min\{\delta, \delta_n\}$. Then we have $|a_n -  A| < \epsilon.$
This is true for all $n \ge N$. Hence, $\lim_{n\to \infty} a_n = A.$
Since $f_n$ converges to $f$ uniformly by the Weiestrass criterion, we are done.
