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$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{2 n} \frac{r}{\sqrt{n^2+r^2}}$ equals
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The correct answer is:
$-1+\sqrt{5}$
$L=\lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{1}{n} \cdot \frac{r / n}{\sqrt{1+(r / n)^2}}=\int_0^2 \frac{x}{\sqrt{1+x^2}} d x=\sqrt{5}-1$
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