Help solving the limit of the sequence: $\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$ Given the following sequence,
$$\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$$
I am asked to determine to which $l \in \mathbb{R}$ the sequence converges.
This is what I have tried so far:
$$
\lim_{n} \left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( \frac{2n^2 + 1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) }  \implies 
\left( \frac{2n^2}{2n^2 - 1} + \frac{1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies 
\left( 1 + \frac{1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies ? $$
My objective was to algebraically modify the sequence such that I could be able to reduce it to the form: $$\lim _{x\to +\infty }\left(1+{\frac {1}{x}}\right)^{x}=e$$
but seem to be stuck at the ? point.
Any suggestion? Thank you.

Use$$\ln l=\lim_{n\to\infty}2n^2\underbrace{\frac{1-1/(2n^2)}{1+3/n}}_{\sim1}\underbrace{\ln\underbrace{\frac{1-1/(2n^2)}{1+1/(2n^2)}}_{1-1/n^2+o(1/n^2)}}_{\sim-1/n^2}=-2\implies l=e^{-2}.$$
