If $[x]$ denotes the greatest integer $\le x$,then $\mathop {\text{Limit}}\limits_{n \to \infty } \frac{1}{n^4} \left( [1^3 x] + [2^3 x] + \dots + [n^3 x] \right)$ equals
- A$x/2$
- B$x/3$
- C$x/6$
- D$x/4$
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Similar Questions
$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{2 n} \frac{r}{\sqrt{n^2+r^2}}=$
$\mathop {\lim }\limits_{n \to \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}} = $
DifficultAIEEE 2002
View SolutionFor 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:
$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{r = 1}^{2n} {\frac{r}{{\sqrt {{n^2} + {r^2}} }}} $ equals
DifficultIIT 1997
View Solution$\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}}\right]=$
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