$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+{{3}^{99}}+......{{n}^{99}}}{{{n}^{100}}}=$

$\mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 0}^n {\frac{n}{{{{\left( {2r + n} \right)}^2}}}} $ is equal to

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

The value of $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$ is:

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: