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.

Hint
$$=\left(\left(1-\dfrac2{2n^2+1}\right)^{-(2n^2+1)/2}\right)^{-\frac{2(2n^3-n)}{(2n^2+1)(n+3)}}$$
Alternatively
$$=\left(\dfrac{(1-1/2n^2)^{2n^2}}{(1+1/2n^2)^{2n^2}}\right)^{\dfrac{2n^3-n}{2n^2(n+3)}}$$
Now the numerator converges to $e^{-1}$
What about the denominator & the exponent?
