જો $[x]$ એ $x$ થી નાનો અથવા તેના જેટલો મહત્તમ પૂર્ણાંક દર્શાવે,તો $\mathop {\text{Limit}}\limits_{n \to \infty } \frac{1}{n^4} \left( [1^3 x] + [2^3 x] + \dots + [n^3 x] \right)$ ની કિંમત શોધો.

$\lim _{n \rightarrow \infty} n\left[\frac{1}{3 n^2+8 n+4}+\frac{1}{3 n^2+16 n+16}+\ldots+\frac{1}{15 n^2}\right]=$

$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^3}\right)^{\frac{1}{n^3}}\left(1+\frac{8}{n^3}\right)^{\frac{4}{n^3}}\left(1+\frac{27}{n^3}\right)^{\frac{9}{n^3}} \ldots \left(1+\frac{n^3}{n^3}\right)^{\frac{n^2}{n^3}}\right]=$

$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots+\frac{1}{n} \sec ^2 1\right]=$

$\mathop {\lim }\limits_{n \to \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}} = $

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