lim _{n} {rightarrow infty}left[left(1+frac{1 | vedclass

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(C) Let $A = \lim _{n \rightarrow \infty}\left[\prod_{r=1}^{n} \left(1+\frac{r^2}{n^2}\right)\right]^{\frac{1}{n}}$.Taking the natural logarithm on bo

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$2 e^{\frac{\pi-4}{2}}$

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(C) Let $A = \lim _{n ightarrow \infty}\left[\prod_{r=1}^{n} \left(1+\frac{r^2}{n^2}ight)ight]^{\frac{1}{n}}$.Taking the natural logarithm on both sides:$\log A = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \log \left(1+\frac{r^2}{n^2}ight)$.Using the definition of the definite integral as the limit of a sum,$\lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}ight) = \int_0^1 f(x) dx$,we get:$\log A = \int_0^1 \log(1+x^2) dx$.Integrating by parts,let $u = \log(1+x^2)$ and $dv = dx$. Then $du = \frac{2x}{1+x^2} dx$ and $v = x$.$\int_0^1 \log(1+x^2) dx = [x \log(1+x^2)]_0^1 - \int_0^1 \frac{2x^2}{1+x^2} dx$.$= [1 \cdot \log(2) - 0] - 2 \int_0^1 \left(\frac{1+x^2-1}{1+x^2}ight) dx$.$= \log 2 - 2 \int_0^1 \left(1 - \frac{1}{1+x^2}ight) dx$.$= \log 2 - 2 [x - 	an^{-1}(x)]_0^1$.$= \log 2 - 2 [(1 - 	an^{-1}(1)) - (0 - 0)]$.$= \log 2 - 2(1 - \frac{\pi}{4}) = \log 2 - 2 + \frac{\pi}{2} = \log 2 + \frac{\pi-4}{2}$.Since $\log A = \log 2 + \frac{\pi-4}{2}$,we have $A = e^{\log 2 + \frac{\pi-4}{2}} = 2 e^{\frac{\pi-4}{2}}$.

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

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