mathop {lim }limits_{n to infty } frac{{[{1^2 | vedclass

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(C) The given limit is $L = \mathop {\lim }\limits_{n 	o \infty } \frac{{\sum_{k=1}^{n} [k^2 x + k^2]}}{n^3}$.Using the property $[a+b] = [a] + b$ if

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$\frac{x}{3} + \frac{1}{3}$

Vedclass · Answer

(C) The given limit is $L = \mathop {\lim }\limits_{n 	o \infty } \frac{{\sum_{k=1}^{n} [k^2 x + k^2]}}{n^3}$.Using the property $[a+b] = [a] + b$ if $b$ is an integer,we have $[k^2 x + k^2] = [k^2 x] + k^2$.Thus,$L = \mathop {\lim }\limits_{n 	o \infty } \frac{{\sum_{k=1}^{n} [k^2 x]}}{n^3} + \mathop {\lim }\limits_{n 	o \infty } \frac{{\sum_{k=1}^{n} k^2}}{n^3}$.For the first part,using the squeeze theorem,$\sum [k^2 x] \approx \sum k^2 x = x \frac{n(n+1)(2n+1)}{6} \approx x \frac{2n^3}{6} = \frac{x n^3}{3}$.So,$\mathop {\lim }\limits_{n 	o \infty } \frac{\sum [k^2 x]}{n^3} = \frac{x}{3}$.For the second part,$\mathop {\lim }\limits_{n 	o \infty } \frac{\sum_{k=1}^{n} k^2}{n^3} = \mathop {\lim }\limits_{n 	o \infty } \frac{n(n+1)(2n+1)}{6n^3} = \frac{2}{6} = \frac{1}{3}$.Therefore,$L = \frac{x}{3} + \frac{1}{3}$.

$\mathop {\lim }\limits_{n \to \infty } \frac{{[{1^2}x + {1^2}] + [{2^2}x + {2^2}] + [{3^2}x + {3^2}] + \dots + [{n^2}x + {n^2}]}}{{{n^3}}}$ is equal to :- (where $[.]$ denotes the greatest integer function)

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