A) 0
B) \[x\]
C) \[\frac{x}{2}\]
D) \[\frac{{{x}^{2}}}{2}\]
Correct Answer: C
Solution :
[c] \[nx-1<\left| nx \right|\le nx.\] Putting \[n=1,2,3,...,n\] and adding them, \[x\Sigma n-n<\sum [nx]\le x\Sigma n\] \[\therefore x.\frac{\Sigma n}{{{n}^{2}}}-\frac{1}{n}<\frac{\Sigma [nx]}{{{n}^{2}}}\le x.\frac{\Sigma n}{{{n}^{2}}}\] ? (i) Now, \[\underset{x\to \infty }{\mathop{\lim }}\,\left\{ x.\frac{\Sigma n}{{{n}^{2}}}-\frac{1}{n} \right\}=x.\underset{n\to \infty }{\mathop{\lim }}\,\frac{\Sigma n}{{{n}^{2}}}-\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}=\frac{x}{2}\] As the two limits are equal, by (i) \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{\Sigma [nx]}{{{n}^{2}}}=\frac{x}{2}.\]
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