$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^3}}}\left[ {{1^2}\sin \frac{1}{n} + {2^2}\sin \frac{2}{n} + {3^2}\sin \frac{3}{n} + ....+{n^2}\sin \frac{n}{n}} \right]$ equals

  • A
    $\cos1 + 2\sin1$
  • B
    $2\sin1 - 2$
  • C
    $\cos1 - 2\sin1 - 2$
  • D
    $\cos1 + 2\sin1 - 2$

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Similar Questions

$\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}}} = $

Evaluate the following definite integral as a limit of sums:
$\int_{2}^{3} x^{2} d x$

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{\left( {n + 2} \right)}^{1/3}}}}{{{n^{4/3}}}} + \dots + \frac{{{{\left( {2n} \right)}^{1/3}}}}{{{n^{4/3}}}}} \right)$ is equal to

By the definition of the definite integral,the value of $\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2-1^2}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)$ is equal to

The value of $\lim_{n \to \infty} \left[ \frac{1}{n}\sin \left( \frac{1}{n} \right)\left( \cos \left( \frac{1}{n} \right) \right)^2 + \frac{1}{n}\sin \left( \frac{2}{n} \right)\left( \cos \left( \frac{2}{n} \right) \right)^2 + \dots + \frac{1}{n}(\sin 1)(\cos 1)^2 \right]$ is

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