If $f: R \rightarrow R$ is given by $f(x)=x+1$, then the value of $\lim _{n \rightarrow \infty} \frac{1}{n}\left[f(0)+f\left(\frac{5}{n}\right)+f\left(\frac{10}{n}\right)+\ldots .+f\left(\frac{5(n-1)}{n}\right)\right]$, is :

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Limits

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}) + \dots . + f (\frac{5 ( n - 1 )}{n})]

, is :

Answer: (C)

Since

f (x) = x + 1

, the bracket is

\sum_{k = 0}^{n - 1} (\frac{5 k}{n} + 1)

, and

\frac{1}{n} k = 0 \sum n - 1 (\frac{5 k}{n} + 1) = \frac{5}{n ^{2}} \cdot \frac{n ( n - 1 )}{2} + 1 = \frac{5 ( n - 1 )}{2 n} + 1

which is the Riemann sum for

\int_{0}^{1} (5 x + 1) d x

written out. As

n \to \infty

the limit is

\frac{5}{2} + 1 = \frac{7}{2}

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option

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