lim _{n rightarrow infty}left(frac{1}{sqrt{n^ | vedclass

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(C) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{1}{\sqrt{n^2-r^2}}$.We can rewrite the expression by taking $n$ common

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(C) The given limit is $L = \lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{1}{\sqrt{n^2-r^2}}$.We can rewrite the expression by taking $n$ common from the square root in the denominator:$L = \lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{1}{n \sqrt{1-(\frac{r}{n})^2}}$.This is a Riemann sum of the form $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(\frac{r}{n})$,which is equivalent to the definite integral $\int_0^1 f(x) dx$.Here,$f(x) = \frac{1}{\sqrt{1-x^2}}$.Thus,$L = \int_0^1 \frac{1}{\sqrt{1-x^2}} dx$.The integral of $\frac{1}{\sqrt{1-x^2}}$ is $\sin^{-1}(x)$.Evaluating the definite integral from $0$ to $1$:$L = [\sin^{-1}(x)]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$.

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

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