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?
 A: Hint. Folowing @David Mitra piece of advice, as $n \to \infty$, one may write:
$$
\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}}=\frac{ 2(2n+1)^2}{ \sqrt{n^4+1}+ \sqrt{n^4-1}}=\frac{ 8+\frac{8}{n}+\frac{2}{n^2}}{ \sqrt{1+\frac{1}{n^4}}+ \sqrt{1-\frac{1}{n^4}}}.
$$
A: You could also write $\sqrt{n^4 \pm 1} = n^2 \sqrt{1 \pm \frac{1}{n^4}}$ and expand the square roots out using the Binomial Theorem (for large $n, \, \frac{1}{n^4}$ is small), and see what this leaves you. 
A: Using jim's answer, by Taylor $$\sqrt{n^4+1}=n^2+\frac{1}{2 n^2}-\frac{1}{8 n^6}+O\left(\frac{1}{n^{9}}\right)$$ $$\sqrt{n^4-1}=n^2-\frac{1}{2 n^2}-\frac{1}{8 n^6}+O\left(\frac{1}{n^{9}}\right)$$ $$\sqrt{n^4+1}-\sqrt{n^4-1}=\frac{1}{n^2}+O\left(\frac{1}{n^{9}}\right)$$ Now, multiplying by $(2n+1)^2$, you get for the whole expression $$4+\frac{4}{n}+\frac{1}{n^2}+O\left(\frac{1}{n^7}\right)$$ which shows the limit and how it is approached.
Let us try with $n=10$; the exact value is $\approx 4.4100000055$ while the above approximation gives $4.41$. This does not seem to be too bad being so far from $\infty$.
A: Set $1/n=h\implies h\to0^+, h>0$
$$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \dfrac1{(2n+1)^2}} \right) =\lim_{ h\to0^+}\dfrac{(\sqrt{1+h^4}-\sqrt{1-h^4})(2+h)^2}{h^4} $$
$$=\lim_{ h\to0^+}(2+h)^2\cdot\dfrac1{\lim_{ h\to0^+}(\sqrt{1+h^4}+\sqrt{1-h^4})}\cdot\lim_{ h\to0^+}\dfrac{1+h^4-(1-h^4)}{h^4}=?$$
Cancel out $h$ as $h\ne0$ as $h\to0^+$
