lim _{n rightarrow infty}left[frac{sqrt{n^2-1 | vedclass

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Question

(A) The given expression is $S = \lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{\sqrt{n^2-r^2}}{n^2}$.We can rewrite this as $S = \lim _{n \rightar

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) The given expression is $S = \lim _{n ightarrow \infty} \sum_{r=1}^{n} \frac{\sqrt{n^2-r^2}}{n^2}$.We can rewrite this as $S = \lim _{n ightarrow \infty} \sum_{r=1}^{n} \frac{n \sqrt{1-(r/n)^2}}{n^2} = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sqrt{1-(\frac{r}{n})^2}$.Using the definition of a definite integral as the limit of a sum,$\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f(\frac{r}{n}) = \int_{0}^{1} f(x) dx$.Here,$f(x) = \sqrt{1-x^2}$.Thus,$S = \int_{0}^{1} \sqrt{1-x^2} dx$.Using the standard integral formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a})$,we get:$S = [\frac{x}{2} \sqrt{1-x^2} + \frac{1}{2} \sin^{-1}(x)]_{0}^{1}$.Evaluating at the limits: $S = (0 + \frac{1}{2} \sin^{-1}(1)) - (0 + \frac{1}{2} \sin^{-1}(0)) = \frac{1}{2} 	imes \frac{\pi}{2} = \frac{\pi}{4}$.

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

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