If f:R→R is given by fx=x+1 then the value of limn→∞1nf0+f5n+f10n+......+f5n−1n is

fx=x+1limn→∞1n1+5n+1+10n+1+........+5n−1n+1=limn→∞1nn+51+2+3+.....+n−1n=limn→∞ 1+5n−12n=1+52=72

If f:R→R is given by fx=x+1 then the value of limn→∞1nf0+f5n+f10n+......+f5n−1n is

fx=x+1limn→∞1n1+5n+1+10n+1+........+5n−1n+1=limn→∞1nn+51+2+3+.....+n−1n=limn→∞ 1+5n−12n=1+52=72

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If $f : R \to R$ is given by $f (x) = x + 1$ then the value of $\lim_{n \to \infty} \frac{1}{n} [f (0) + f (\frac{5}{n}) + f (\frac{10}{n}) + ...... + f (\frac{5 (n - 1)}{n})]$ is

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$\frac{5}{2}$

$\frac{7}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = x + 1 \\ \lim_{n \to \infty} \frac{1}{n} [1 + (\frac{5}{n} + 1) + (\frac{10}{n} + 1) + ........ + (\frac{5 (n - 1)}{n} + 1)] \\ = \lim_{n \to \infty} \frac{1}{n} [n + 5 (\frac{1 + 2 + 3 + ..... + n - 1}{n})] = \lim_{n \to \infty} 1 + \frac{5 (n - 1)}{2 n} = 1 + \frac{5}{2} = \frac{7}{2} \end{array}$