Q117 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
$\lim _{x \rightarrow \infty}\left[\left(1+\frac{1}{n^3}\right)^{\frac{1}{n^3}}\left(1+\frac{8}{n^3}\right)^{\frac{4}{n^3}}\left(1+\frac{27}{n^3}\right)^{\frac{9}{n^3}} \ldots . .(2)^{\frac{1}{n}}\right]=$
- A$\log 2-\frac{1}{2}$
- B$e^{\left(\log 2-\frac{1}{2}\right)}$
- Correct answerC$e^{\left(\frac{2 \log 2-1}{3}\right)}$
- D$\frac{1}{3}(2 \log 2-1)$
Solution
$\begin{aligned}
& \quad \mathrm{I}=\lim _{x \rightarrow \infty}\left[\left(1+\frac{1}{n^3}\right)^{\frac{1}{n^3}}\left(1+\frac{8}{n^3}\right)^{\frac{4}{n^3}}\left(1+\frac{27}{n^3}\right)^{\frac{9}{n^3}} \ldots(2)^{\frac{1}{n}}\right] \\
& \Rightarrow \log \mathrm{I}=\lim _{n \rightarrow \infty}\left[\log \left(1+\frac{1}{n^3}\right)^{\frac{1}{n^3}}+\log \left(1+\frac{8}{n^3}\right)^{\frac{4}{n^3}}+\ldots+\log (2)^{\frac{1}{n}}\right] \\
& \log \mathrm{I}=\lim _{n \rightarrow \infty} \frac{1}{n^3} \log \left(1+\frac{1}{n^3}\right)+\frac{4}{n^3} \log \left(1+\frac{8}{n^3}\right)+\ldots+\frac{1}{n} \log (2) \\
& \log \mathrm{I}=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3} \log \left(1+\frac{r^3}{n^3}\right)=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{r^2}{n^2} \log \left(1+\frac{r^3}{n^3}\right) \\
& \therefore \log \mathrm{I}=\int_0^1 x^2 \log \left(1+x^3\right) d x
\end{aligned}$
Let $1+x^3=t \Rightarrow 3 x^2 d x=d t$ $\begin{aligned} & \therefore \log \mathrm{I}=\frac{1}{3} \int_1^2 \log t d t \mathrm{I}=\frac{1}{3}[t \log t-t]_1^2 \\ & \log \mathrm{I}=\frac{1}{3}[2 \log 2-2+1] \Rightarrow \mathrm{I}=e^{\left(\frac{2 \log 2-1}{3}\right)} \end{aligned}$
Let $1+x^3=t \Rightarrow 3 x^2 d x=d t$ $\begin{aligned} & \therefore \log \mathrm{I}=\frac{1}{3} \int_1^2 \log t d t \mathrm{I}=\frac{1}{3}[t \log t-t]_1^2 \\ & \log \mathrm{I}=\frac{1}{3}[2 \log 2-2+1] \Rightarrow \mathrm{I}=e^{\left(\frac{2 \log 2-1}{3}\right)} \end{aligned}$
Hence, the correct answer is option: 'C'.