$\lim _{n \rightarrow \infty} \frac{1}{n} [(n+1)(n+2) \cdots (2n)]^{\frac{1}{n}} = $

Given that $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x$. If $f: R \rightarrow R$ is defined by $f(x)=x^2+2$, then $\lim _{n \rightarrow \infty} \frac{3}{n}\left[f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right)+\ldots+f(7)\right]=$

If $f(n) = \frac{1}{n} [(n+1)(n+2)(n+3) \ldots (2n)]^{\frac{1}{n}}$, then $\lim_{n \rightarrow \infty} f(n) =$

$\mathop {\lim }\limits_{n \to \infty } {\left\{ {\left( {1 + \frac{{{1^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{2^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{3^2}}}{{{n^2}}}} \right) \dots \left( {1 + \frac{{{{(n - 1)}^2}}}{{{n^2}}}} \right)} \right\}^{1/n}}$ equals to:

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \dots + \frac{1}{{2n}}} \right] = $

For $a \in \mathbb{R}$ (the set of all real numbers),$a \neq -1$,if $\lim_{n \to \infty} \frac{1^a + 2^a + \dots + n^a}{(n+1)^{a-1}[(na+1) + (na+2) + \dots + (na+n)]} = \frac{1}{60}$,then $a$ is equal to: