questions about how to show sequence of functions are uniform convergent There is a theorem :
Assume that
(1) $\{f_n\}$ sequence of functions in $A$ point-wise converge to $f$
(2) for any $n$, $\{f_n\}$ is of class $C^1$ function
(3) $\{f_n'\}$ converge uniformly to $g$. 
Then :
(1) $f$ is differentiable.
(2) $f'(x)=g(x)$.
I don't quit understand this theorem, does it mean to take derivative when checking if the sequence function is uniformly convergent? or we take a limit as $n$ goes to infinity, and check if the function go to a constant?
 A: The theorem is just saying this: Suppose that $f_n \rightarrow f$ and $f'_n \rightarrow g$ uniformly.  Then $f$ must be nice enough that we can take its derivative, and we also have $$\frac{d}{dx} f = \frac{d}{dx} \lim f_n = \lim \frac{d}{dx} f_n = g.$$
In other words, we can swap the order of taking limits and taking derivatives whenever the derivatives converge uniformly to something ($g$ in this case).
A: We know that if a series of functions $\{f_n\}$ converges uniformly to some function $f$, then it also converges $\textit{pointwise}$ to that function. A good place to start when trying to calculate uniform limits of functions is to first compute the pointwise limit, then try to prove uniform convergence to that limit.
This theorem does not necessarily mean you should take the derivative to figure out if a function is uniformly convergent. This theorem says that if a series of functions $\{f_n\}$ as well as the series of its derivatives $\{f'_n\}$ converges uniformly, then $f$ is differentiable. 
Checking if a function "goes to a constant" as you asked, is not very relevant here, because the uniform limit of a series of functions (or a series of derivatives of functions) is $\textit{not}$ necessarily always a constant. For example, the uniform limit of $f_n(x) = \sqrt{x^2 + \frac 1n}$ on $[0,10]$ is $f(x) = |x|$, which is clearly not constant on this domain.
