Correct Answer - A
For ant integer k, we have
` kx-1 lt [kx]le kx`
`rArr Sigma_(k=1)^(n)(kx-1)lt Sigma_(k=1)^(n)[kx] le Sigma _(k=1)^(n)kx`
`rArr (1)/(n^2)Sigma_(k=1)^(n)(kx-1)lt (1)/(n^2)Sigma_(k=1)^(n) [kx] le (1)/(n^2) Sigma _(k=1)^(n) kx`
` rArr (x)/(n^2)Sigma_(k=1)^(n) k-(1)/(n^2)lt (1)/(n^2) Sigma _(k=1)^(n) [kx] le (x)/(n^2)Sigma _(k=1)^(n) k`
` rArr (x)/(2) (1+(1)/(n)) -(1)/(n^2) lt (1)/(n^2) Sigma _(k=1) ^(n) [kx] le (x)/(2) (1+(2)/(n))`
Now ,
` lim_(xto oo) {(1+(1)/(n))-(1)/(n^2)}=(x)/(2) and, lim_(nto oo) (x)/(2) (1+(1)/(n))=(x)/(2)`
` therefore lim_(xto oo) (1)/(n^2)Sigma _(k=1)^(n)[kx]=(x)/(2)" "["Using Sandwich Theorem "]`
i.e., `lim_(xtooo) ([x]+[2x]+[3x]+.....+[nx])/(n^2)=(x)/(2)`
