How to evaluate $\lim_{n \rightarrow \infty}\left(\frac{(2n)!}{n!n^n} \right)^{1/n}$? Find$$\lim_{n \rightarrow \infty}\left(\frac{(2n)!}{n!n^n} \right)^{1/n}$$ is there some trick in this questions. seems it must simplify to something but I am unable to solve it.

Just use the following result:

If $a_{n} $ is a sequence of positive terms and $a_{n+1}/a_{n}\to L$ then $a_{n} ^{1/n}\to L$. 

The above is easily proved by taking logs and applying Cesaro-Stolz. For the current question we take $a_{n} = (2n)!/(n!n^{n})$ and then we can see that $$\frac{a_{n+1}}{a_{n}}=\frac{(2n+1)(2n+2)}{(n+1)^{2}}\cdot\dfrac{1}{\left(1+\dfrac{1}{n}\right)^{n}}\to\frac{4}{e}$$ so the desired limit is $4/e$.

On request of user "S.H.W" (via comment) I provide proof of the result used in this answer. Let $b_{n} =\log a_{n} $ then $b_{n} $ is well defined as $a_{n} $ is positive. Now we can apply Cesaro-Stolz on $b_{n} /n$ to get the relation $$\lim_{n\to\infty} \frac{b_{n}} {n} =\lim_{n\to\infty} \frac{b_{n+1}-b_{n}}{n+1-n}$$ provided the limit on right hand side exists finitely or infinitely. This means that $$\lim_{n\to\infty} \log(a_{n} ^{1/n})=\lim_{n\to\infty}\log(a_{n+1}/a_{n})$$ and on exponentiating (which is a continuous operation) we get $$\lim_{n\to\infty} a_{n} ^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$$ provided the limit on right hand exists finitely or infinitely. And thus the proof is completed via Cesaro-Stolz. 
Most textbooks provide proofs without the use of Cesaro-Stolz and logarithms and you should have a look at that proof also because it is typical of the proofs used in the theory of limits of sequences. Roughly the idea is that if the ratio $a_{n+1}/a_{n}$ lies between $L-\epsilon $ and $L+\epsilon$ for all $n\geq m$ then $a_{n} /a_{m} $ lies between $(L-\epsilon) ^{n-m} $ and $(L+\epsilon) ^{n-m} $. 
