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

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(C) ધારો કે $I = \lim _{n \rightarrow \infty} \prod_{r=1}^n \left(1+\frac{r^3}{n^3}\right)^{\frac{r^2}{n^3}}$.બંને બાજુ પ્રાકૃતિક લઘુગણક લેતા:$\log I

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

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(C) ધારો કે $I = \lim _{n \rightarrow \infty} \prod_{r=1}^n \left(1+\frac{r^3}{n^3}\right)^{\frac{r^2}{n^3}}$.બંને બાજુ પ્રાકૃતિક લઘુગણક લેતા:$\log I = \lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3} \log \left(1+\frac{r^3}{n^3}\right)$.આને આ રીતે ફરીથી લખી શકાય:$\log I = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \left(\frac{r}{n}\right)^2 \log \left(1+\left(\frac{r}{n}\right)^3\right)$.નિશ્ચિત સંકલનની વ્યાખ્યાનો ઉપયોગ કરતા,$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$:$\log I = \int_0^1 x^2 \log(1+x^3) dx$.ધારો કે $t = 1+x^3$,તો $dt = 3x^2 dx$,અથવા $x^2 dx = \frac{1}{3} dt$.જ્યારે $x=0, t=1$. જ્યારે $x=1, t=2$.$\log I = \frac{1}{3} \int_1^2 \log t dt = \frac{1}{3} [t \log t - t]_1^2$.$\log I = \frac{1}{3} [(2 \log 2 - 2) - (1 \log 1 - 1)] = \frac{1}{3} [2 \log 2 - 2 + 1] = \frac{2 \log 2 - 1}{3}$.તેથી,$I = e^{\left(\frac{2 \log 2 - 1}{3}\right)}$.

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

Similar Questions

$\lim _{n \rightarrow \infty} \left\{ \frac{\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{2n-1}}{n^{3/2}} \right\}$ ની કિંમત શોધો.

નીચેના વિધાનો પૈકી:
$(S1): \lim _{n \rightarrow \infty} \frac{1}{n^2}(2+4+6+\ldots+2n)=1$
$(S2): \lim _{n \rightarrow \infty} \frac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+\ldots+n^{15})=\frac{1}{16}$

જો $a$ અને $b$ ધન પૂર્ણાંકો હોય કે જેથી $b > a$,તો $\lim_{n \to \infty} \left[ \frac{1}{na} + \frac{1}{na + 1} + \frac{1}{na + 2} + \dots + \frac{1}{nb} \right] = $

$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\left( {n + 1} \right)\left( {n + 2} \right) \ldots \left( {3n} \right)}}{{{n^{2n}}}}} \right)^{\frac{1}{n}}} = $

$\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}}\right]=$

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