The value of lim _{n rightarrow infty} frac{1 | vedclass

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(C) We are given the limit $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}$.Dividing the numerator and de

Vedclass · Accepted Answer

$1+2 \log _{e}\left(\frac{3}{2}\right)$

Vedclass · Answer

(C) We are given the limit $L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}$.Dividing the numerator and denominator of the term inside the sum by $n$,we get:$L = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{2(\frac{j}{n}) - \frac{1}{n} + 8}{2(\frac{j}{n}) - \frac{1}{n} + 4}$.Using the definition of the definite integral as the limit of a Riemann sum,where $\frac{j}{n} \rightarrow x$ and $\frac{1}{n} \rightarrow dx$ as $n \rightarrow \infty$,the expression becomes:$L = \int_{0}^{1} \frac{2x + 8}{2x + 4} dx$.We can simplify the integrand as follows:$\frac{2x + 8}{2x + 4} = \frac{(2x + 4) + 4}{2x + 4} = 1 + \frac{4}{2x + 4} = 1 + \frac{2}{x + 2}$.Now,integrate with respect to $x$ from $0$ to $1$:$L = \int_{0}^{1} (1 + \frac{2}{x + 2}) dx = [x + 2 \ln|x + 2|]_{0}^{1}$.Evaluating the definite integral:$L = (1 + 2 \ln(3)) - (0 + 2 \ln(2)) = 1 + 2(\ln(3) - \ln(2)) = 1 + 2 \ln(\frac{3}{2})$.Thus,the correct option is $C$.

The value of $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{(2 j-1)+8 n}{(2 j-1)+4 n}$ is equal to:

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