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$$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} \right) = 4 $$

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?

canoe
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    Have you tried multiplying top and bottom by $\sqrt{n^4+1}+\sqrt{n^4-1}$? – David Mitra Mar 27 '16 at 14:23
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    That difference is approaching $0$, but it gets divided by $\frac{1}{(2n+1)^2}$, so it gets multiplied by $(2n+1)^2$. So we need fairly precise information about how fast $\sqrt{n^4+1}-\sqrt{n^4-1}$ is approacing $0$. – André Nicolas Mar 27 '16 at 14:29

4 Answers4

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Hint. Folowing @David Mitra piece of advice, as $n \to \infty$, one may write:

$$ \frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}}=\frac{ 2(2n+1)^2}{ \sqrt{n^4+1}+ \sqrt{n^4-1}}=\frac{ 8+\frac{8}{n}+\frac{2}{n^2}}{ \sqrt{1+\frac{1}{n^4}}+ \sqrt{1-\frac{1}{n^4}}}. $$

Olivier Oloa
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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.

jim
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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 $\infty$.

Claude Leibovici
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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 $h\to0^+$

lab bhattacharjee
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