So I need to prove/disprove that $f_n(x)$ converge.
I have proved it but I don't really like my solution and I hope to see $\epsilon$ proof ($\exists \epsilon>0$ such that $|f_n(x)-f(x)|<\epsilon$)
My solution without using $\epsilon$ proof
$$\lim_{n \to \infty}f_n(x)=\lim_{n \to \infty}\sin(x)^{\frac {1}{n}}=1 \forall x\in I$$
Let $\epsilon >0$
$$\sup_{x\in I}|f_n(x)-f(x)|=\sup_{x\in I}|\sin(x)^{\frac {1}{n}}-1|$$
$$\forall x\in\left[\frac{\pi}{4},\frac{3\pi}{4}\right] : \frac {\sqrt2}{2}\le \sin(x)\le1$$
$$\left(\frac{\sqrt2}{2}\right)^{\frac{1}{n}}-1
\le\sin(x)^{\frac{1}{n}}-1\le0$$
$$0\le\left|\sin(x)^{\frac{1}{n}}-1\right|
\le\left|\left(\frac{\sqrt2}{2}\right)^{\frac {1}{n}}-1\right|$$
So we get that the supremum is $|1-1|=0$ as $n\to\infty$. Which means our sequence function is uniformly convergent.
Is there any other proof with $\epsilon$ and $N>n$ usage?
Thank you and a have a nice day!
Over the interval $I=[\pi/4,3\pi/4]$ the function since is bounded below by $\sin(\pi/4)=1/\sqrt{2}$, that is $1/\sqrt{2}\leq \sin(x)\leq 1$. Then $$2^{-1/(2n)}\leq \sqrt[n]{\sin (x)}\leq 1,\qquad x\in I$$
Since $2^{-1/(2n)}\xrightarrow{n\rightarrow\infty}1$ (no dependence on $x$) uniform convergence follows. No need for $\varepsilon-\delta$ arguments here, unless you want to prove formally that $2^{-1/(2n)}$ indeed converges to $1$.