Uniform convergence of $f_n(x) = x^n$ on $[0,c]$ 
Let $c \in (0,1)$ be fixed. Let $$f_n(x) = x^n,\quad x \in [0,1)$$and$$f(x) = 0,\quad x \in [0,1)$$ Show that $f_n$ converges uniformly to $f$ on $[0,c]$. 

So, we have, $f_n(0) = 0,  f_n( c) = c^n,f_n'(x) = nx^{n-1}, f_n'(0) = 0, f_n'( c) = nc^{n-1}.$ Since I don't know which one is supremum of $f_n(x)$, I've divided this into two cases:
Case 1: If $f_n'(c ) > f_n(c )$, then  $$\lim_{n \to \infty} \sup_{\{0\leq x \leq c\}}\ | f_n(x) - f(x)|= | nc^{n-1} - 0|= 0,$$ since $0<c<1$. 
Case 2: If $f_n(c ) > f_n'(c )$, $$\lim_{n \to \infty} \sup_{\{0\leq x \leq c\}}|f_n(x) - f(x)| = | c^n - 0|= 0,$$ again because $0<c<1$. 
Thus, $f_n$ converges uniformly to $f$. What do you think? Is it correct? 
 A: Since $f_n'(x) = nx^{n-1}>0, \forall x\in[0,c]$, $f_n(x)$ is increasing and $|f_n(x)|\leqslant c^n$ for $x\in[0,c]$. 
Since $c^n\to0$, by Weierstrass's M test, $f_n$ converges uniformly to $f$.
A: This is more a comment but ...
To show that
$c^n \to 0$,
since $0 < c < 1$,
$c = \frac1{1+b}$,
where
$b
=\frac1{c}-1
$.
Then
$c^n
=\frac1{(1+b)^n}
$.
But,
by Bernoulli's inequality,
$(1+b)^n
\ge 1+nb
=1+n(\frac1{c}-1)
>n(\frac1{c}-1)
$.
Therefore
$c^n
\le \frac1{n(\frac1{c}-1)}
= \frac{c}{n(1-c)}
$.
Note: 
nothing original here.
I read this many years ago
in
"What is Mathematics?"
by Courant and Robbins.
Highly recommended,
and currently available
for an extremely reasonable $14
from Amazon.
A: An explicit argument: if $x \in (0,1)$ and $\varepsilon>0$ then $x^n<\varepsilon$ is equivalent to $n \ln(x) < \ln(\varepsilon)$ or $n>\frac{\ln(\varepsilon)}{\ln(x)}$. Now check that if $n>\frac{\ln(\varepsilon)}{\ln(c)}$ then $x^n<\varepsilon$ for every $x \in [0,c]$. This will necessarily work because the limit is zero and $x^n$ is increasing (so the convergence is slowest at the rightmost point).
