Suppose we have a univariate random variable $X\sim\mathcal{P}$ with probability density function $f(x):\mathbb{R}\to\mathbb{R}$, $\int_{-\infty}^{\infty}f(x) dx = 1$, we then draw $n$ samples $x_1,x_2,\dots,x_n$ iid from $\mathcal{P}$. What's the probability that the next sample $x_{n+1}$ is at least $\varepsilon$ away from any of the previous samples $x_1,x_2,\dots,x_n$? $\varepsilon > 0$ is a fixed constant.

I tried to formulate this problem as: Given $x_1,x_2,\dots,x_n$ i.i.d from $\mathcal{P}$, the probability of the new sample fall outside the $\varepsilon$ ball of any previous samples is (given you know $x_{n+1}=x$, the previous n samples are all far away from it): $\prod_{i=1}^{n} Pr(\|x-x_i\|>\varepsilon)=\left[1-\int\nolimits_{x-\varepsilon}^{x+\varepsilon}f(y) dy\right]^n$, then take the expecatation of $x$, we have $\int\nolimits_{-\infty}^{\infty}\left[1-\int\nolimits_{x-\varepsilon}^{x+\varepsilon}f(y) dy\right]^nf(x)dx$ is the probability of $x_{n+1}$ fall outside of the $\varepsilon$ ball of any of $x_1,x_2,\dots,x_n$.

My question is, what conditions is required for $f(x)$ to let this probability go to $0$ as $n\to\infty$.

i.e. when does $\lim\limits_{n\to\infty}\int\nolimits_{-\infty}^{\infty}\left[1-\int\nolimits_{x-\varepsilon}^{x+\varepsilon}f(y) dy\right]^nf(x)dx = 0$ hold for $f(x)$?