Searching for tricks or quick techniques to identify if $f_n$ (or $f'_n$) is continuous but $f$ (or $f'$) is not continuous. I am facing problem in uniform convergence topic. 
Normally I have the basic idea of continuity, convergence and uniform convergence. While solving problems I am getting stuck in a certain area.  
Suppose a sequence of function is given. Let $$f_n(x)=x-\frac{x^n}{n}, \space x\in [0,1]$$. I have proved $f_n(x)$ converge uniformly on $[0,1]$. But when it says $f'_n(x)$ does not converge uniformly over $[0,1]$ there I get stuck.  
$f'_n(x)= 1-x^{n-1}, \space x\in[0,1],\space n\in\text{N}$  
Now $f'_n(0)=1, f'_n(1)=0$ and $$g(x)=\lim_{n\to\infty}f'_n(x){=1, \space x\in[0,1)\\=0, \space x=1}$$.  
Now a solution hint says $g(1)=0\neq f'(1)$ - here I am stuck.  
We know $f'(x)=\lim_{n\to\infty}f'_n(x)$ and here $\lim_{n\to\infty}f'_n(x)=g(x)$. Hence $g(x)$ should be same as $f'(x)$. Hence $g(1)$ should be same as $f'(1)$.  
Specifically, when it is given that $f_n$ (or $f'_n$) is continuous but $f$ (or $f'$) is not but they look almost similar within a given interval, I get stuck there.
I have gone through several texts and problem solutions given in this site but it was not clear enough for me. I know I am wrong somewhere, but if it is pointed out or explained, it will be very much worthy. Any help is greatly appreciated. 
 A: If uniform convergence holds here we must have$$\forall\epsilon>0\qquad,\qquad\exists N\qquad,\qquad\forall x\in [0,1],n>N\qquad,\qquad |f_n'(x)-g(x)|<\epsilon$$also$$|f_n'(x)-g(x)|{=x^{n-1}, \space x\in[0,1)\\=0, \space x=1}$$and $|f_n'(x)-g(x)|<\epsilon$ leads to$${x^{n-1}<\epsilon, \space x\in[0,1)\\0<\epsilon, \space x=1}$$which is impossible for $x$ sufficiently close to $1$. Therefore the convergence is not uniform.
A: No need to look at $f'$.
Use the fact that uniform convergence implies pointwise convergence, and that if a sequence of continuous functions converges uniformly, the limit function is necessarily continuous.
If we denote the pointwise limit of $(f_n')_{n=1}^\infty$ by $g$, we get
$$g(x) = \lim_{n\to\infty} f_n'(x) = \begin{cases}
1, & \text{ if $x \in [0,1\rangle$}\\
0, & \text{ if $x = 1$}
\end{cases}$$
so $g = \chi_{[0,1\rangle}$ which is not a continuous function. However, if the convergence of $(f_n')_{n=1}^\infty$ to $g$ were uniform, $g$ would have to be continuous.
