MCQ
$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^3} + 1}} + \frac{4}{{{n^3} + 1}} + \frac{9}{{{n^3} + 1}} + ........ + \frac{{{n^2}}}{{{n^3} + 1}}} \right] = $
  • A
    $1$
  • B
    $2/3$
  • $\frac{1}{3}\,$
  • D
    $0$

Answer

Correct option: C.
$\frac{1}{3}\,$
c
(c) Given limit $ = \mathop {\lim }\limits_{n \to \infty } \,\,\frac{{{1^2} + {2^2} + {3^2} + ..... + {n^2}}}{{1 + {n^3}}} = \mathop {\lim }\limits_{n \to \infty } \,\,\frac{{\Sigma {n^2}}}{{1 + {n^3}}}$

$ = \mathop {\lim }\limits_{n \to \infty } \,\,\frac{1}{6}\frac{{n\,(n + 1)\,(2n + 1)}}{{1 + {n^3}}}$

$ = \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{6}\frac{{\left( {1 + \frac{1}{n}} \right)\,\left( {2 + \frac{1}{n}} \right)}}{{\left( {\frac{1}{{{n^3}}} + 1} \right)}}$

$ = \frac{1}{6}\,.\,1\,.\,\frac{2}{{(1)}} = \left( {\frac{1}{3}} \right).$

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