JEE practice!

$\displaystyle \begin{aligned} P_n & = \Pi _{r=1} ^n \dfrac {6r^2+r-1} {6r^2+7r-5} \\ & = \Pi _{r=1} ^n \dfrac {(3r-1)(2r+1)} {(3r+5)(2r-1)}\\ & = \dfrac {2 \times 3}{8 \times 1} \times \dfrac {5 \times 5}{11 \times 3} \times \dfrac {8 \times 7}{14 \times 5} \times \dfrac {11 \times 9}{17 \times 7} \times ... \times \dfrac {(3n-1)(2n+1)} {(3n+5)(2n-1)} \\ & = 2\times 5 \times \dfrac {1} {3n+2} \times \dfrac {2n+1} {3n+5} \\ & = \dfrac {10(2n+1)} {(3n+2)(3n+5)} \end{aligned}$

$\displaystyle \begin{aligned} \lim _{n \rightarrow \infty} \dfrac {9}{10}(n+1)P_n & = \lim _{n \rightarrow \infty} \dfrac {9}{10}(n+1) \dfrac {10(2n+1)} {(3n+2)(3n+5)} \\ & = \lim _{n \rightarrow \infty} \dfrac {9(n+1)(2n+1)} {(3n+2)(3n+5)} \\ & = \lim _{n \rightarrow \infty} \dfrac {9(2n^2+3n+1)} {9n^2+21n+10}\\ & = \lim _{n \rightarrow \infty} \dfrac {2(9n^2+13.5n+4.5)} {9n^2+21n+10}\\ & = \boxed{2} \end{aligned}$

$\begin{aligned} &\begin{aligned} P_n&=\prod_{r=1}^n\frac{6r^2+r-1}{6r^2+7r-5}\\ &=\prod_{r=1}^n\frac{(3r-1)(2r+1)}{(3r+5)(2r-1)} \end{aligned}\\ &\text{Notice that} (2(r+1)-1)=(2r+1),\ (3(r+1)+5)=(3r-1)\\ &\therefore P_n=10\cdot\frac{2n+1}{(3n+2)(3n+5)}\\ &\begin{aligned} \therefore\text{The original formula}&=\lim_{n\to\infty}\frac{9}{10}\left(\frac{(n+1)(2n+1)}{(3n+2)(3n+5)}\right)\\ &=9\lim_{n\to\infty}\frac{2n^2+3n+1}{9n^2+21n+10}\\ \end{aligned}\\ &\begin{aligned} \text{According to L'Hospital's rule, the original formula}&=9\lim_{n\to\infty}\frac{(2n^2+3n+1)''}{(9n^2+21n+10)''}\\ &=9\lim_{n\to\infty}\frac{4}{18}\\ &=9\times\frac{2}{9}\\ &=2 \end{aligned} \end{aligned}$