Show that $F=\{f_n\;|\; n\in N, \; f_n:[0,1]\to R \; \ni f_n(x)=x^n\}$ is not equicontinuous at $x=0$

Question

Show that $F=\{f_n\;|\; n\in N, \; f_n:[0,1]\to R \; \ni f_n(x)=x^n\}$ is not equicontinuous at $x=0$

I am asked to prove it is not equicontinous at $x=0$

But I can't see how, if I take any $x$ such that $|x|<\delta$, then clearly $|\delta^n - 0| \to 0$

I think the problem would come at $x=1$

because $|1^n - (1-\delta)^n|\to1\neq 0$ so $\;\epsilon = \frac12$ work here

Can somebody check if this is correct?

Answer 1

Given $e\in (0,1).$ For each $n\in \Bbb N$ there exists $d_n\in (0,1)$ such that $\forall x\in (1-d_n,1]\,(|f_n(x)-f_n(1)|<e).$

But there does not exist $d\in (0,1)$ such that $\forall n\in \Bbb N\,\forall x\in (1-d,1]\,(|f_n(x)-f_n(1)|<e).$

Because for any $d\in (0,1)$ we have

$1-d/2\in (1-d,1]$

and $\lim_{n\to \infty}f_n(1-d/2)=0$

while $f_n(1)=1$ for all $n\in \Bbb N.$

So there are only finitely many $n\in \Bbb N$ for which $f_n(1-d/2)\ge 1-e.$ So there are infinitely many $n\in \Bbb N$ for which $|f_n(1-d/2)-f_n(1)|>e.$