Help with finding Limit What is the limit  of  $$\lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{\frac{3n^3+4}{4n^4-1}}$$
Does any one can help, I am not sure how to solve this.
 A: hint: Use the famous Sterling inequality:  
$$ \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n \le n! \le \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n\cdot e^{\frac{1}{12n}}$$
 and use the Squeeze lemma to find the limit. Can you manage to take it from here?
A: $$
\begin{aligned}
\lim _{n\to \infty }\left(\frac{n!}{n^n}\right)^{\frac{3n^3+4}{4n^4-1}} &= \lim _{n\to \infty}\exp\left(\frac{3n^3+4}{4n^4-1}\ln\left(\frac{n!}{n^n}\right)\right)\\ &\approx
\lim _{n\to \infty }\exp \left(\frac{3n^3+4}{4n^4-1}\ln \left(\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n}{n^n}\right)\right)\\ &=
\lim _{n\to \infty }\exp \left(\frac{3n^3+4}{4n^4-1}\ln \left(\frac{\sqrt{2\pi n}}{e^n}\right)\right)\\ &=
\lim _{n\to \infty }\exp \left(\frac{3n^3+4}{4n^4-1}\left(\frac{1}{2}\ln \left(2\pi n\right)-n\right)\right)\\ &=
\color{red}{e^{-\frac{3}{4}}}
\end{aligned}$$
Solved with Stirling approximation 
$$x! \approx \sqrt{2\pi x}\left(\frac{x}{e}\right)^x, \text{ for } x \to \infty$$
A: Hint (without Stirling). Note that, as $n\to +\infty$,
$$\left(\frac{n!}{n^n}\right)^{\frac{3n^3+4}{4n^4-1}}=\exp\left(\frac{3n^3+4}{4n^4-1}\,\ln\left(\frac{n!}{n^n}\right)\right)=
\exp\Big(\underbrace{\frac{3+\frac{4}{n^3}}{4-\frac{1}{n^4}}}_{\to 3/4}\cdot \underbrace{\frac{1}{n}\sum_{k=1}^n\ln\left(\frac{k}{n}\right)}_{\text{Riemann sum}}\Big).$$
Can you take it from here?
A: Let $a_n=n! /n^n$ so that $a_{n+1}/a_n=(n/(n+1)) ^n\to 1/e$ and therefore $b_n=a_n^{1/n}\to 1/e$. Let $c_n=(3n^4+4n)/(4n^4-1)$ so that $c_n\to 3/4$. The given sequence is $x_n=b_n^{c_n} $ and therefore it tends to $(1/e)^{3/4}=e^{-3/4}$.
