lim _{n rightarrow infty} frac{sqrt{1}+sqrt{2 | vedclass

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(C) We are given the limit $L = \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1} \sqrt{r}}{n \sqrt{n}}$.To evaluate this,we can write the sum as $\

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(C) We are given the limit $L = \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1} \sqrt{r}}{n \sqrt{n}}$.To evaluate this,we can write the sum as $\sum_{r=1}^{n} \sqrt{r} - \sqrt{n}$.Thus,$L = \lim _{n \rightarrow \infty} \left( \frac{\sum_{r=1}^{n} \sqrt{r}}{n \sqrt{n}} - \frac{\sqrt{n}}{n \sqrt{n}} \right)$.$L = \lim _{n \rightarrow \infty} \left( \frac{1}{n} \sum_{r=1}^{n} \sqrt{\frac{r}{n}} - \frac{1}{n} \right)$.Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(\frac{r}{n}) = \int_{0}^{1} f(x) dx$.Here,$f(x) = \sqrt{x}$.So,$L = \int_{0}^{1} \sqrt{x} dx - \lim _{n \rightarrow \infty} \frac{1}{n}$.$L = \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{1} - 0 = \frac{2}{3} (1)^{3/2} = \frac{2}{3}$.

$\lim _{n \rightarrow \infty} \frac{\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n-1}}{n \sqrt{n}}$ is equal to

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