<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] 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 $\infty$.
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