Consider a sample of $n$ points randomly placed on a circle of radius $1$. It can be shown that $X_n$ , the distance from the centre of the circle of the farthest point, has density function $$f_{X_n}(x)=2nx^{2n-1} \ \ \ \ \ \ 0<x<1$$ Show that $$X_n\rightarrow 1 \ \text{in probability}$$
My attempt:
I first calculated that $$F_{X_n}(x)=1 \ \ \ \ \ 0<x<1$$ Now, I want to show that $$\forall\epsilon>0, \ \ \ \lim_{n\to\infty} \mathbb{P}(\left|X_n-1\right|>\epsilon)=0$$ My question is, how do I deal with the absolute value $\left|X_n-1\right|$? For instance, if I assumed it was positive, then $$\lim_{n\to\infty} \mathbb{P}(\left|X_n-1\right|>\epsilon)=\lim_{n\to\infty} \mathbb{P}(X_n>\epsilon+1)=0$$ As $$\mathbb{P}\left(X_n>\frac{1}{2}\right)=1-F_{X_n}\left(\frac{1}{2}\right)=1-1=0$$
But why is it okay to make this assumption?
Note that $X_n \leq 1$ almost surely. (Because the density is $0$ for $x>1$). Hence $|X_n-1|>\epsilon $ iff $X_n <1-\epsilon $. Thus $P\{|X_n-1|>\epsilon\} =\int_0^{1-\epsilon } 2nx^{2n-1} \, dx =(1-\epsilon )^{2n} \to 0$ as $n \to \infty $.