Is $f_{n}(x)=x^{n}$ uniformly convergent on the interval $[0,1]$? Let $f_{n}(x)=x^{n}$.
I know that $f_{n}(x)$ converges to $0$ on the interval $[0,1)$ and converges to $1$ on $x=1$.
But is $f_{n}(x)=x^{n}$ uniformly convergent on the interval $[0,1]$?
 A: Hint: 
When $x = 1$ we have $$\lim_{n\rightarrow \infty}{f_n} = 1$$
When $x\in [0,1)$
$$\lim_{n\rightarrow \infty}{f_n} = 0$$
A: Let $\varphi$ be the pointwise limit of $(f_n)$ : 
$\varphi(x) = 0$ if $x\neq 1$, $\varphi(1) = 1$.
You want to show that the convergence isn't uniform, ie.
$$\exists \varepsilon > 0,~\forall N\in\mathbb N,~\exists n>N,~\exists x\in[0,1],~|f_n(x) - \varphi(x)| > \varepsilon.$$
Pick $\varepsilon = \frac14$. Let $N\in\mathbb N$, $n>N$ and $x=1-\frac1n$ (we'll choose $n$ later).
Then $|f_n(x) - \varphi(x)| = f_n(x) = (1-\frac1n)^n$. For $n$ large enough, $(1-\frac1n)^n$ is close to $e^{-1} > 1/4$. So, taking $n$ large enough, you have a $x$ such that $|f_n(x) - \varphi(x)| > \varepsilon$.
Therefore, the convergence is not uniform.
A: Hint: Generally, $(f_{n}) \to f$ uniformly on a set $A$ if and only if 
$$
\sup_{x\in A} |f_{n}(x) - f(x)| \to 0.
$$
(This is little more than a restatement of the definition of uniform convergence.)
If $f_{n}(x) = x^{n}$ and $A = [0, 1)$ ([sic], right endpoint omitted), then $f(x) = 0$ for all $x$ in $A$. What is $\sup\limits_{x \in A} |f_{n}(x)|$?
