Find $\lim\limits_{n\to+\infty}\frac{\sqrt[n]{n!}}{n}$ 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?

This is an alternate solution for those who don't know stirling approximation yet like me
$$
\lim _{ n\longrightarrow +\infty  }{ \frac { \sqrt [ n ]{ n! }  }{ n }  } \quad =\quad \lim _{ n\longrightarrow +\infty  }{ \frac { \sqrt [ n ]{ \prod _{ k=0 }^{ n-1 }{ (n-k) }  }  }{ n }  } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty  }{ \frac { \sqrt [ n ]{ \prod _{ k=0 }^{ n-1 }{ n(1-\frac { k }{ n } ) }  }  }{ n }  } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty  }{ \frac { \sqrt [ n ]{ { n }^{ n }\prod _{ k=0 }^{ n-1 }{ (1-\frac { k }{ n } ) }  }  }{ n }  } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty  }{ \frac { n\sqrt [ n ]{ \prod _{ k=0 }^{ n-1 }{ (1-\frac { k }{ n } ) }  }  }{ n }  } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty  }{ { \left( \prod _{ k=0 }^{ n-1 }{ \left( 1-\frac { k }{ n }  \right)  }  \right)  }^{ \frac { 1 }{ n }  } } \\ \qquad \qquad \qquad \qquad =\quad \lim _{ n\longrightarrow +\infty  }{ { e }^{ \frac { 1 }{ n } \sum _{ k=0 }^{ n-1 }{ \ln { \left( 1-\frac { k }{ n }  \right)  }  }  } } 
$$
Now we will evaluate that infinite sum:
$$
\underset { n\longrightarrow +\infty  }{ lim } \frac { 1 }{ n } \sum _{ k=0 }^{ n-1 }{ \ln { \left( 1-\frac { k }{ n }  \right)  }  } \quad =\underset { n\longrightarrow +\infty  }{ lim } \quad \frac { 1-0 }{ n } \sum _{ k=0 }^{ n-1 }{ \ln { \left( 1-\frac { k }{ n }  \right)  }  } \\ \qquad \qquad \qquad \qquad \qquad =\quad \int _{ 0 }^{ 1 }{ ln(1-x)\quad dx } \\ \qquad \qquad \qquad \qquad \qquad =\quad { \left[ x\ln { (1-x) }  \right]  }_{ 0 }^{ 1 }+\int _{ 0 }^{ 1 }{ \frac { x }{ 1-x } dx } \\ \qquad \qquad \qquad \qquad \qquad =\quad { \left[ x\ln { (1-x) }  \right]  }_{ 0 }^{ 1 }+{ \left[ -\ln { (1-x)-x }  \right]  }_{ 0 }^{ 1 }\\ \qquad \qquad \qquad \qquad \qquad =\quad { \left[ (x-1)\ln { (1-x)-x }  \right]  }_{ 0 }^{ 1 }\quad =\quad -1
$$
So your limit is: ${e}^{-1}$
