Let $f_n(x) =x^n$ for $n \in \mathbb{ N.}$ Which of the following statements are true? Let $f_n(x) =x^n$  for $n \in \mathbb{ N.}$ Which of the following statements are true?
$1)$ The sequence $\{f_n\}$ converges  uniformly on $[ \frac{1}{6},\frac{1}{3}]$
$2)$ The sequence $\{f_n\}$ converges  uniformly on $[ 0,1]$
$3)$ The sequence $\{f_n\}$ converges  uniformly on $( 0,1)$
My answer  is   option 1)  and option 3) because if  $-1 <x <1$ ,then $|x|^n$   is converges to $0$  when  n tend to infinity
Is  it correct ??
 A: The convergence is uniform only for $(1)$.
We have $$\sup_{x \in \left[\frac16, \frac13\right]}|x^n| = \frac1{3^n} \xrightarrow{n\to\infty} 0$$
so $f_n \to 0$ uniformly on $\left[\frac16, \frac13\right]$.
For $(2)$, the pointwise limit is $$f(x) = \begin{cases} 0, &\text{ if $x \in [0,1)$} \\
1, &\text{ if $x = 0$}\end{cases}$$
which is not a continuous function. However, uniform limit of a sequence of continuous functions is always continuous.
For $(3)$ the pointwise limit is $0$ but for $x_n = 1- \frac1n$ we have
$$\sup_{x \in (0,1)} |x^n| \ge \left(1-\frac1n\right)^n \xrightarrow{n\to\infty} \frac1e$$
so $\sup_{x \in (0,1)} |x^n| \not\to 0$. Hence the convergence is not uniform.
A: The sequence $(f_n)_{n\in\mathbb N}$ converges pointwise to$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\mathbb{R}\\&x&\mapsto&\begin{cases}0&\text{ if }x<1\\1&\text{ otherwise.}\end{cases}\end{array}$$So, if it converges uniformly on some $A\subset[0,1]$, it must converge uniformly to $f|_A$. So, this excludes (2), since $f$ is discontinuous, but each $f_n$ is continuous. And (3) is false to, since $\lim_{n\to\infty}f_n\left(\sqrt[n]{1-\frac1n}\right)=1$, whereas it should be $=$, if the convergence was uniform.
But (1) holds, since $\left(\forall x\in\left[\frac16,\frac13\right]\right):\bigl\lvert f_n(x)\bigr\rvert\leqslant\frac1{3^n}\to0$.
