The value of $\lim_{n \to \infty} \frac{1}{n}\sum_{j=1}^{n}\frac{(2j-1)+8n}{(2j-1)+4n}$ is equal to :
Show Hint
To convert a limit of a sum to a definite integral, manipulate the expression into the form $\lim_{n \to \infty} \frac{1}{n} \sum f(\frac{r}{n})$. Then, replace $\frac{1}{n}$ with $dx$, $\sum$ with $\int$, $\frac{r}{n}$ with $x$, and set the limits of integration from 0 to 1.
This limit is in the form of a limit of a Riemann sum, which can be converted to a definite integral.
The formula is $\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_0^1 f(x) dx$.
Let's manipulate the given expression to fit this form.
$L = \lim_{n \to \infty} \frac{1}{n}\sum_{j=1}^{n}\frac{(2j-1)+8n}{(2j-1)+4n}$
Divide the numerator and denominator inside the summation by n:
$L = \lim_{n \to \infty} \frac{1}{n}\sum_{j=1}^{n}\frac{\frac{2j}{n}-\frac{1}{n}+8}{\frac{2j}{n}-\frac{1}{n}+4}$
As $n \to \infty$, the term $\frac{1}{n} \to 0$. We replace $\frac{j}{n}$ with x, and the summation $\frac{1}{n}\sum$ with $\int_0^1 dx$.
Note that the summation index is j, so we should replace $\frac{j}{n}$ with x.
$L = \int_0^1 \frac{2x+8}{2x+4} dx$.
We can simplify the integrand:
$\frac{2x+8}{2x+4} = \frac{(2x+4)+4}{2x+4} = 1 + \frac{4}{2x+4} = 1 + \frac{2}{x+2}$.
Now, integrate this simplified expression from 0 to 1.
$L = \int_0^1 \left(1 + \frac{2}{x+2}\right) dx$
$= [x + 2\ln|x+2|]_0^1$
$= (1 + 2\ln|1+2|) - (0 + 2\ln|0+2|)$
$= (1 + 2\ln 3) - (2\ln 2)$
$= 1 + 2(\ln 3 - \ln 2)$
$= 1 + 2\ln\left(\frac{3}{2}\right)$.
This matches option (B).
