$\Phi (x)$ is the distribution function of standard normal distribution. $\varepsilon$ is some positive tiny number that is less than 1.
How to prove this beautiful and important limitation: $$\mathop {\lim }\limits_{n \to \infty } {\{\Phi [(1 - \varepsilon )\sqrt {2\log n} ]\}^n}=0$$
I've already proved $\mathop {\lim }\limits_{n \to \infty } {\{\Phi [(1 + \varepsilon )\sqrt {2\log n} ]\}^n}=1$. However, I met some difficulty when proving this.