Consider $$\lim_{n\to{\infty}}{\Phi_{0,1}^n{(x\ln{n}})}$$where x>0.
i.e. $$\lim_{n\to{\infty}}[\int_{-\infty}^{x\ln{n}}\frac{1}{\displaystyle\sqrt{2\pi}}e^{\frac{-y^2}{2}}dy]^n$$ I think the answer would be 1, but I don’t know how to get it. Help me please.
I'll call $\Phi(x)=\Phi_{0,1}(x)=\int_{-\infty}^x(2\pi)^{-1/2}e^{-s^2/2}\,ds$. You can take the logarithm of that quantity of yours and consider $$\lim_{y\to\infty}y\ln\Phi(x\ln y)=\lim_{t\to0^+}\frac{\ln\Phi(-x\ln t)}{t}$$
Notice that for $x>0$ the limit on the RHS is an indeterminate form $\left[\frac 00\right]$ and since \begin{align}\lim_{t\to0^+}\frac d{dt}\ln\Phi(-x\ln t)&=\lim_{t\to 0^+}-\frac1{\Phi(-x\ln t)}\frac xt\frac1{\sqrt{2\pi}}e^{-\frac12x^2\ln^2t}\\&=\lim_{t\to0^+}-\frac x{\sqrt{2\pi}\Phi(-x\ln t)}e^{-\ln t-\frac12 x^2\ln^2t}\\&=\left[-\frac x{\sqrt{2\pi}}e^{-\infty}\right]=0^-\end{align} you can use l'Hopital, which yields $$\lim_{t\to 0^+}\frac{\ln\Phi(-x\ln t)}{t}=0$$