Showing $(f'_n)$ converges I am trying to show that the sequence of derivatives $(f_n')$ when,
$$f_n(x)=\frac{x^n}{n}$$
for $x\in [0,1]$
So I calculated the derivative with respect to x and got the following,
$$f_n'(x) = x^{n-1}$$
So $f_n'(x)$ converges pointwise to 0 if x is in $[0,1)$ and $1$ if $x=1$
I believe I have done this part correct and now I am trying to prove whether or not it has uniform convergence. 
To do so I want to use the fact that 
$$sup|f'n(x)-f'(x)| <= b_n$$
where $limb_n = 0$ then it would be uniformly convergent.
so for $x=1$
$$sup|f'n(x)-f'(x)|$$
$$=sup|x^{n-1}-1|$$
and for $x\in [0,1)$
$$sup|f'n(x)-f'(x)|$$
$$=sup|x^{n-1}|$$
How do I proceed from here and determine if it is uniformly convergent or not?
 A: A uniformly convergent sequence of continuous functions has a continuous limit, which is not the case here, so we know the sequence does not converge uniformly. 
To show this, it would suffice to show that, for all $n\in \Bbb{N}$, there exists $x\in (0,1)$ such that $f_n(x)=x^{n-1} > 1/2$ (since this would mean that, for $\varepsilon=1/2$, and for all $n$, there exists $x$ such that $\vert f_n(x) - f(x)\vert = x^{n-1} > \varepsilon$). Indeed, since $f_n(0) = 0, f_n(1) = 1$, there exists $x\in (0,1)$ such that $f_n(x) = 3/4$ by the intermediate value theorem. 
A: I'd like to give a theorem which may give you a hint, I will also prove the theorem.
Theorem: if $f_n$ uniformly converge to $f$ on $[a,b]$ and each is continuous, then $f$ is continuous.
Proof. Let $I=[a,b]$ Let $x_0 \in I$. Let $\epsilon > 0$. Let N be so that for every $n > N$ and $x \in I$ we have $|f_n(x)-f(x)|<\epsilon$. Fix $n >N$.
The function $f_n$ is continuous. So, there is $\delta > 0$ so that if $|x-y|<\delta$ then $|f(x)-f(y)| < \delta$, Thus,
$|f(x)-f(y)|\le|f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)| < 3\epsilon$
QED.
We conclude that $x^{n-1}$ does not uniformly converge on $[0,1]$ 
