mathop {lim }limits_{n to infty } {left( {fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) ધારો કે $L = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{\left( {n + 1} ight)\left( {n + 2} ight) \ldots \left( {3n} ight)}}{{{n^

Vedclass · Accepted Answer

$\frac{{27}}{{{e^2}}}$

Vedclass · Answer

(D) ધારો કે $L = \mathop {\lim }\limits_{n 	o \infty } {\left( {\frac{{\left( {n + 1} ight)\left( {n + 2} ight) \ldots \left( {3n} ight)}}{{{n^{2n}}}}} ight)^{\frac{1}{n}}}$.બંને બાજુ પ્રાકૃતિક લઘુગણક લેતા:$\ln L = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r=1}^{2n} \ln \left( {\frac{{n+r}}{n}} ight) = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r=1}^{2n} \ln \left( {1 + \frac{r}{n}} ight)$.આ એક રીમેન સરવાળો છે,જે નિશ્ચિત સંકલનમાં રૂપાંતરિત થાય છે:$\ln L = \int\limits_0^2 \ln(1+x) dx$.ખંડશઃ સંકલનનો ઉપયોગ કરતા,$\int \ln(1+x) dx = (1+x)\ln(1+x) - (1+x) + C$.નિશ્ચિત સંકલનનું મૂલ્ય શોધતા:$\ln L = \left[ (1+x)\ln(1+x) - x ight]_0^2 = (3\ln 3 - 2) - (1\ln 1 - 0) = 3\ln 3 - 2$.આમ,$L = e^{3\ln 3 - 2} = e^{\ln(3^3) - 2} = e^{\ln 27} \cdot e^{-2} = \frac{27}{e^2}$.

$\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}}} = $

Similar Questions

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{\left( {n + 2} \right)}^{1/3}}}}{{{n^{4/3}}}} + \dots + \frac{{{{\left( {2n} \right)}^{1/3}}}}{{{n^{4/3}}}}} \right)$ ની કિંમત શોધો.

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

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

$\mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 0}^n {\frac{n}{{{{\left( {2r + n} \right)}^2}}}} $ ની કિંમત શોધો.

$\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}}{n^{\frac{3}{2}}} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module