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How to evaluate $\lim\limits_{n\to\infty}(n^3\cdot(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}))$?

My though is, that since $\lim\limits_{n\to\infty}n^3=\infty$, the limit must be $\infty$, if the right hand side is greater than or equal to $1$. Otherwise it's a fraction, and there might be many possible solutions.

However, I'm far for being confident about the solution. What am I missing here? What's the "trick"?

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A basic strategy when you have limits involving radicals is to multiply by the conjugate over the conjugate. –  Hakim May 27 at 20:49

3 Answers 3

up vote 2 down vote accepted

Hint: \begin{align}\\&\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2} \\\\=&\dfrac{n^2+\sqrt{n^4+1}-2n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}} \\\\=&\dfrac{\sqrt{n^4+1}-n^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}} \\\\=&\dfrac{n^4+1-n^4}{\left(\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}\right)\cdot\left(\sqrt{n^4+1}+n^2\right)}\end{align}

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This hint was helpful, thanks! Also, if somebody is not familiar with the conjugate over conjugate multiplication, this link will be helpful: dummies.com/how-to/content/… –  kdani May 27 at 21:03

Recall that by the Taylor series we have $$(1+x)^\alpha\sim_01+\alpha x$$ so we apply it:

$$\sqrt{n^2+\sqrt{n^4+1}}=n\sqrt{1+\sqrt{1+\frac1{n^4}}}\sim_\infty n\left(1+1+\frac{1}{2n^4}\right)^{1/2}\\=n\sqrt2\left(1+\frac{1}{4n^4}\right)^{1/2}\sim_\infty n\sqrt2\left(1+\frac{1}{8n^4}\right)$$ hence we see that

$$\lim_{n\to\infty}n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt{2}\right)=\frac{\sqrt2}8$$

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Apply conjugation twice to get the numerator to be $n^3$, then:$L \sim \dfrac{n^3}{2\sqrt{2}n\times 2n^2} \to \dfrac{1}{4\sqrt{2}}$

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