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

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(C) ધારો કે $L = \lim _{n \rightarrow \infty}\left[\prod_{r=1}^n \left(1+\frac{r^2}{n^2}\right)\right]^{1 / n}$.બંને બાજુ પ્રાકૃતિક લઘુગણક લેતા:$\log

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

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(C) ધારો કે $L = \lim _{n ightarrow \infty}\left[\prod_{r=1}^n \left(1+\frac{r^2}{n^2}ight)ight]^{1 / n}$.બંને બાજુ પ્રાકૃતિક લઘુગણક લેતા:$\log L = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\frac{r^2}{n^2}ight)$.આ એક રીમેન સરવાળો છે,જેને નિશ્ચિત સંકલન તરીકે દર્શાવી શકાય:$\log L = \int_0^1 \log(1+x^2) dx$.ખંડશઃ સંકલનનો ઉપયોગ કરતા,$u = \log(1+x^2)$ અને $dv = dx$ લેતા:$\int \log(1+x^2) dx = x \log(1+x^2) - \int x \cdot \frac{2x}{1+x^2} dx$.$= x \log(1+x^2) - 2 \int \frac{x^2+1-1}{1+x^2} dx = x \log(1+x^2) - 2 \int \left(1 - \frac{1}{1+x^2}ight) dx$.$= x \log(1+x^2) - 2x + 2 	an^{-1} x$.$0$ થી $1$ સુધી મૂલ્ય શોધતા:$\log L = [1 \cdot \log(2) - 2(1) + 2 	an^{-1}(1)] - [0 - 0 + 0] = \log 2 - 2 + 2(\frac{\pi}{4}) = \log 2 - 2 + \frac{\pi}{2}$.$\log L = \log 2 + \frac{\pi-4}{2} = \log 2 + \log e^{\frac{\pi-4}{2}} = \log \left(2 e^{\frac{\pi-4}{2}}ight)$.તેથી,$L = 2 e^{\frac{\pi-4}{2}}$.

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

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