mathop {lim }limits_{n to infty } left[ {frac | vedclass

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(C) Given limit $L = \mathop {\lim }\limits_{n 	o \infty } \frac{{{1^2} + {2^2} + {3^2} + \dots + {n^2}}}{{{n^3} + 1}} = \mathop {\lim }\limits_{n

Vedclass · Accepted Answer

$1/3$

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(C) Given limit $L = \mathop {\lim }\limits_{n 	o \infty } \frac{{{1^2} + {2^2} + {3^2} + \dots + {n^2}}}{{{n^3} + 1}} = \mathop {\lim }\limits_{n 	o \infty } \frac{{\sum_{k=1}^{n} k^2}}{{{n^3} + 1}}$Using the formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,we get:$L = \mathop {\lim }\limits_{n 	o \infty } \frac{n(n+1)(2n+1)}{6(n^3+1)}$$L = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{6} \cdot \frac{n^3(1 + \frac{1}{n})(2 + \frac{1}{n})}{n^3(1 + \frac{1}{n^3})}$$L = \frac{1}{6} \cdot \frac{(1+0)(2+0)}{(1+0)} = \frac{2}{6} = \frac{1}{3}$

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{{{n^3} + 1}} + \frac{4}{{{n^3} + 1}} + \frac{9}{{{n^3} + 1}} + \dots + \frac{{{n^2}}}{{{n^3} + 1}}} \right] = $

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