Steps to solve $\lim_{n \to \infty} (\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} ) = 4 $ $$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} \right) = 4 $$
I think $\sqrt{n^4+1} - \sqrt{n^4-1}$ is approaching to zero, but it is not correct. What steps can evaluate above limit to 4?

Hint. Folowing @David Mitra piece of advice, as $n \to \infty$, one may write:
$$
\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}}=\frac{ 2(2n+1)^2}{ \sqrt{n^4+1}+ \sqrt{n^4-1}}=\frac{ 8+\frac{8}{n}+\frac{2}{n^2}}{ \sqrt{1+\frac{1}{n^4}}+ \sqrt{1-\frac{1}{n^4}}}.
$$
