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

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(B) The given limit is $L = \lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}} \sum_{r=1}^{n} \frac{1}{\sqrt{r}}$.We can rewrite the expression as $L = \

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(B) The given limit is $L = \lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}} \sum_{r=1}^{n} \frac{1}{\sqrt{r}}$.We can rewrite the expression as $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{\sqrt{r/n}}$.Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$.Here,$f(x) = \frac{1}{\sqrt{x}}$.Thus,$L = \int_0^1 x^{-1/2} dx$.Evaluating the integral: $L = \left[ \frac{x^{1/2}}{1/2} \right]_0^1 = [2\sqrt{x}]_0^1 = 2(1) - 2(0) = 2$.

$\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}}\right]=$

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