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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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