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

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Question

(C) Let $I = \lim _{n \rightarrow \infty} \prod_{r=1}^n \left(1+\frac{r^3}{n^3}\right)^{\frac{r^2}{n^3}}$.Taking the natural logarithm on both sides:$

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$e^{\left(\frac{2 \log 2-1}{3}\right)}$

Vedclass · Answer

(C) Let $I = \lim _{n \rightarrow \infty} \prod_{r=1}^n \left(1+\frac{r^3}{n^3}\right)^{\frac{r^2}{n^3}}$.Taking the natural logarithm on both sides:$\log I = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3} \log \left(1+\frac{r^3}{n^3}\right)$.We can rewrite this as:$\log I = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \left(\frac{r}{n}\right)^2 \log \left(1+\left(\frac{r}{n}\right)^3\right)$.Using the definition of a definite integral as the limit of a sum,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$:$\log I = \int_0^1 x^2 \log(1+x^3) dx$.Let $t = 1+x^3$,then $dt = 3x^2 dx$,or $x^2 dx = \frac{1}{3} dt$.When $x=0, t=1$. When $x=1, t=2$.$\log I = \frac{1}{3} \int_1^2 \log t dt = \frac{1}{3} [t \log t - t]_1^2$.$\log I = \frac{1}{3} [(2 \log 2 - 2) - (1 \log 1 - 1)] = \frac{1}{3} [2 \log 2 - 2 + 1] = \frac{2 \log 2 - 1}{3}$.Therefore,$I = e^{\left(\frac{2 \log 2 - 1}{3}\right)}$.

$\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^3}\right)^{\frac{1}{n^3}}\left(1+\frac{8}{n^3}\right)^{\frac{4}{n^3}}\left(1+\frac{27}{n^3}\right)^{\frac{9}{n^3}} \ldots \left(1+\frac{n^3}{n^3}\right)^{\frac{n^2}{n^3}}\right]=$

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