# Limit of $\frac{n!}{n^n}$ via the Sandwich/Squeeze Theorem.

I need to find the limit as $$\lim_{n\to\infty}\frac{n!}{n^n}$$ via the Sandwich/Squeeze Theorem.

I've been stuck on this for a while as I can't say either the numerator or denominator is bound.

Edit: I'm sorry that I wasn't more explicit when I posted this, I hadn't used this site before this question. The reason why I have to use the above theorem is because it's a question set in one of university problem sheets and I've honestly been stuck on it for several weeks. My only real attempt to solve it was: $$\lim_{n \to \infty}\frac{n!}{n^n}$$ $$0\le\frac{1}{n^n}\le1$$ $$0\le\frac{n!}{n^n}\le n!$$ $$\lim_{n\to\infty}0\le\lim_{n\to\infty}\frac{n!}{n^n}\le\lim_{n\to\infty}n!$$ Which leaves the issue that $$\lim_{n\to\infty}n!$$ is undefined, so the middle limit isn't bound, and so I can't make the final conclusive statement. All my other attempts were just as unsuccessful as this one. I already know that the limit is zero, but because the question states that I must use the Sandwich/Squeeze theorem, I'm completely stuck.

• The limit is 0, though, regardless of the proof. Also, to avoid being downvoted, show what you have tried, more than just "IDK" May 2, 2017 at 2:55
• I don't see it necessary to use that theorem. The limit is 0, why? Consider the number of factors in the numerator and the number of factors in the denominator and then the size of them. It should be obvious why the limit is zero. May 2, 2017 at 3:06
• it's 0 because $x^x$ approaches infinity faster than $x!$, you can see this because $x^x$ is $x$ $x$'s multiplies, whereas $x!$ is roughly $sqrt(x)$ multiplied $x$ times May 2, 2017 at 3:06
• Possible duplicate of Using the sandwhich theorem to evaluate $\lim_{n \to \infty} \frac{n!}{n^n}$ May 2, 2017 at 11:54

Hint: Try showing that $$0 < \frac{n!}{n^n} = \boldsymbol{\frac{1}{n}} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n-1}{n} \cdot \frac{n}{n} < \frac{1}{n}$$ for $n \ge 3$.