limn→∞{x}+{2x}+.....+{nx}n2

0≤{x}<1 ; 0≤{2x}<1;0n2≤{x}+{2x}+.....+{nx}n2<nn2limn→∞0n2≤limn→∞{x}+{2x}+.....+{nx}n2<limn→∞1nlimn→∞f1(x)=limn→∞0n=0 ;limn→∞f2(x)=limn→∞1n=0 limn→∞{x}+{2x}+.....+{nx}n2=0

limn→∞{x}+{2x}+.....+{nx}n2

0≤{x}<1 ; 0≤{2x}<1;0n2≤{x}+{2x}+.....+{nx}n2<nn2limn→∞0n2≤limn→∞{x}+{2x}+.....+{nx}n2<limn→∞1nlimn→∞f1(x)=limn→∞0n=0 ;limn→∞f2(x)=limn→∞1n=0 limn→∞{x}+{2x}+.....+{nx}n2=0

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$\lim_{n \to \infty} \frac{{x} + {2 x} + ..... + {n x}}{n^{2}}$

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$\frac{1}{2}$

answer is B.

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Detailed Solution

$0 \leq {x} < 1 ; 0 \leq {2 x} < 1;$ $\frac{0}{n^{2}} \leq \frac{{x} + {2 x} + ..... + {n x}}{n^{2}} < \frac{n}{n^{2}}$ $\lim_{n \to \infty} \frac{0}{n^{2}} \leq \lim_{n \to \infty} \frac{{x} + {2 x} + ..... + {n x}}{n^{2}} < \lim_{n \to \infty} \frac{1}{n}$ $\lim_{n \to \infty} f_{1} (x) = \lim_{n \to \infty} \frac{0}{n} = 0;$