Question:
The value of $\lim _{\mathrm{n} \rightarrow \infty} \frac{1}{\mathrm{n}} \sum_{\mathrm{j}=1}^{\mathrm{n}} \frac{(2 \mathrm{j}-1)+8 \mathrm{n}}{(2 \mathrm{j}-1)+4 \mathrm{n}}$ is equal to :
Correct Option: , 4
Solution:
$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{j=1}^{n} \frac{\left(\frac{2 j}{n}-\frac{1}{n}+8\right)}{\left(\frac{2 j}{n}-\frac{1}{n}+4\right)}$
$\int_{0}^{1} \frac{2 x+8}{2 x+4} d x=\int_{0}^{1} d x+\int_{0}^{1} \frac{4}{2 x+4} d x$
$=1+\left.4 \frac{1}{2}(\ell \mathrm{n}|2 \mathrm{x}+4|)\right|_{0} ^{1}$
$=1+2 \ell \mathrm{n}\left(\frac{3}{2}\right)$
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