<s>[INST] Using the number e to find the limit of a sequence 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.
$$
B_n = \left(  \frac{n^2 -1}{n^2 +2} \right)^{2n^2-3}
= \left(  \frac{n^2 +2 -3 }{n^2 +2} \right)^{2n^2-3}
= \left(1 +  \frac{-3 }{n^2 +2} \right)^{2n^2-3}
$$
 [/INST] Set $y=n^2+2$, to get:
$$
\lim_{n\to\infty} \left(1 +  \frac{-3 }{n^2 +2} \right)^{2n^2-3}= \lim_{y\to\infty} \left(1 +  \frac{-3 }{y} \right)^{2y-7}\\
=\lim_{y\to\infty} \bigg(\left(1 +  \frac{-3 }{y} \right)^y\bigg)^2\left(1 +  \frac{-3 }{y} \right)^{-7}\\=(e^{-3})^2\cdot 1=e^{-6}
$$
</s>