limits - Find $limlimits_{nto+infty}frac{sqrt[n]{n!}}{n}$

I tried using Stirling's approximation and d'Alambert's ratio test but can't get the limit. Could someone show how to evaluate this limit?

Answer

Use equivalents: $$\frac{\sqrt[n]{n!}}n\sim_{\infty}\frac{\bigl(\sqrt{2\pi n}\bigr)^{\tfrac 1n}}{n}\cdot\frac n{\mathrm{e}}=\frac 1{\mathrm{e}}\bigl({2\pi n}\bigr)^{\tfrac 1{2n}}$$ Now $\;\ln\bigl({2\pi n}\bigr)^{\tfrac 1{2n}}=\dfrac{\ln\pi+\ln 2n}{2n}\xrightarrow[n\to\infty]{}0$, hence $$\frac{\sqrt[n]{n!}}n\sim_{\infty}\frac 1{\mathrm{e}}. $$