Using the number e to find the limit of a sequence

Question

I have been trying to solve this problem for the past day and I just can't get my head around it, I know I need to use Euler's formula but how do I proceed after simplifying it. $$ B_n = \left( \frac{n^2 -1}{n^2 +2} \right)^{2n^2-3} = \left( \frac{n^2 +2 -3 }{n^2 +2} \right)^{2n^2-3} = \left(1 + \frac{-3 }{n^2 +2} \right)^{2n^2-3} $$

Answer 1

Set $y=n^2+2$, to get: $$ \lim_{n\to\infty} \left(1 + \frac{-3 }{n^2 +2} \right)^{2n^2-3}= \lim_{y\to\infty} \left(1 + \frac{-3 }{y} \right)^{2y-7}\\ =\lim_{y\to\infty} \bigg(\left(1 + \frac{-3 }{y} \right)^y\bigg)^2\left(1 + \frac{-3 }{y} \right)^{-7}\\=(e^{-3})^2\cdot 1=e^{-6} $$