$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{2 n} \frac{r}{\sqrt{n^2+r^2}}=$

For a sufficiently large value of $n$,the sum of the square roots of the first $n$ positive integers,i.e.,$\sqrt{1} + \sqrt{2} + \sqrt{3} + \dots + \sqrt{n}$,is approximately equal to:

For a positive integer $n$,define $f(n) = n + \sum_{r=1}^n \frac{16r + (9-4r)n - 3n^2}{4rn + 3n^2}$. Then,the value of $\lim_{n \rightarrow \infty} f(n)$ is equal to

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

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:

$\lim _{n}$ ${\rightarrow \infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\ldots+\frac{n}{n^2+(2n)^2}\right)=$