Q48 / 155 · WBJEE 2019 · Mathematics
$\lim _{n \rightarrow \infty} \frac{3}{n}\left\{1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+\sqrt{\frac{n}{n+9}}\right.$$+\ldots+\sqrt{\frac{1}{n+3}(n-1)}$
- Adoes not exist
- Bis 1
- Correct answerCis 2
- Dis 3
Solution
$\lim _{n \rightarrow \infty} \frac{3}{n}\left[1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+\sqrt{\frac{n}{n+9}}+\ldots\right.+\sqrt{\frac{n}{n+3(n-1)}}$
$=\lim _{n \rightarrow \infty} \frac{3}{n}\left[\sqrt{\frac{n}{n+3 \times 0}}+\sqrt{\frac{n}{n+3 \times 1}}+\sqrt{\frac{n}{n+3 \times 2}}\right.$
$+\sqrt{\frac{n}{n+3 \times 3}}+\ldots+\sqrt{\frac{n}{n+3(n-1)}}$
$=\lim _{n \rightarrow \infty} \frac{3}{n}\left[\sum_{r=0}^{n-1} \sqrt{\frac{n}{n+3 r}}\right]$
$=\lim _{n \rightarrow \infty} \frac{3}{n}\left[\sum_{r=0}^{n-1}\left[\frac{1}{1+3\left(\frac{r}{n}\right)}\right]\right.$
$=3 \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left(\sqrt{1+3\left(\frac{r}{n}\right)}\right)^{-\frac{1}{2}}$
$=3 \int_{0}^{1} \sqrt{\frac{1}{1+3 x}} d x$
$=3 \int_{0}^{1}(1+3 x)^{-1 / 2} d x$
$=3\left[\frac{2 \sqrt{1+3 x}}{3}\right]_{0}^{1}$
$=[4-2]$
$=2$
Hence, the correct answer is option: 'C'.