<s>[INST] 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?
 [/INST] 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 $h\to0^+$
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