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Identify $$\lim\limits_{x \to +\infty } x^2 \left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)$$

My Try :

$$\sqrt{x^4+x+1}=\sqrt{x^4(1+\frac{1}{x^3}+\frac{1}{x^4})}=x^2\sqrt{(1+\frac{1}{x^3}+\frac{1}{x^4})}$$

Now : $$\frac{1}{x^3}+\frac{1}{x^4}=z$$

$$(1+z)^{\frac{1}{n}}= 1 + \frac1n x + \frac{1 - n}{2n^2}x^2 + \frac{2n^2 - 3n + 1}{6n^3}x^3 + O(x^4)$$ $$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac12 z - \frac{1}{8}z^2 + + O(z^3)$$

$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac12 (\frac{1}{x^3}+\frac{1}{x^4}) - \frac{1}{8}(\frac{1}{x^3}+\frac{1}{x^4})^2 + O(z^3)$$

$$(1+(\frac{1}{x^3}+\frac{1}{x^4}))^{\frac{1}{2}}= 1 + \frac{1}{2x^3}+\frac{1}{2x^4} - \frac{x^2+2x+1}{8x^8} + O(z^3)$$

And :

$$\sqrt{x^4+x+1}=\sqrt{x^4(1+\frac{1}{x^3}+\frac{5}{x^4})}=x^2\sqrt{(1+\frac{1}{x^3}+\frac{5}{x^4})}$$

So :

$$(1+(\frac{1}{x^3}+\frac{5}{x^4}))^{\frac{1}{2}}= 1 + \frac{1}{2x^3}+\frac{5}{2x^4} - \frac{x^2+10x+25}{8x^8}+ O(z^3)$$

$$\lim\limits_{x \to +\infty }=x^4( 1 + \frac{1}{2x^3}+\frac{1}{2x^4} - \frac{x^2+2x+1}{8x^8} + O(z^3)-( 1 + \frac{1}{2x^3}+\frac{5}{2x^4} - \frac{x^2+10x+25}{8x^8}+ O(z^3)))$$

$$\lim\limits_{x \to +\infty }=x^4(\frac{1}{2x^4}-\frac{5}{2x^4})=x^4(\frac{-4}{2x^4})=-2$$

is it right ?

Did
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Almot1960
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2 Answers2

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Write \begin{align} \sqrt{x^4+x+1}-\sqrt{x^4+x+5} =\frac{(\sqrt{x^4+x+1}-\sqrt{x^4+x+5})(\sqrt{x^4+x+1}+\sqrt{x^4+x+5})}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}} \end{align} and use \begin{align} \frac{1}{2\sqrt{x^4+x+5}}\leq\frac{1}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}\leq \frac{1}{2\sqrt{x^4+x+1}} \end{align} and \begin{align} \sqrt{x^4+x+1}&=\sqrt{x^4(1+x^{-3}+x^{-4})}=x^2\sqrt{1+x^{-3}+x^{-4}}\\ \sqrt{x^4+x+5}&=\sqrt{x^4(1+x^{-3}+5x^{-4})}=x^2\sqrt{1+x^{-3}+5x^{-4}} \end{align}

Fakemistake
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    The OP's question is whether their approach is sound, not how to compute the limit. – Did Mar 31 '17 at 12:06
  • @Did An 11 line approach to what can be done in one line is obviously not sound, although maybe (who cares?) correct. – uniquesolution Mar 31 '17 at 12:18
  • @uniquesolution Maybe he wants to do it by this method, he previously asked http://math.stackexchange.com/questions/2211574/what-is-binomial-series-of-1z-frac1n-1-frac1nzoz2. –  Mar 31 '17 at 12:20
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    @uniquesolution This is obviously quite short-sighted. If you compare the two approaches to solve the question, one (to multiply by the conjugate quantity) is anecdotal (and short) while the other (limited expansions) is hugely important in the whole domain of analysis (and yes, longer in the present case). You might want to think again... (And anyway, the OP asked for the soundness of their approach, not for the limit itself, didn't they?) – Did Mar 31 '17 at 12:24
  • His approach is not very good, so i decided to give him a better one. In addition, the given problem was wrong and was edited after DonAntonio remarked, he should check what he wrote. I already thougt that he had made an mistake. – Fakemistake Mar 31 '17 at 12:46
  • @Did I disagree. – uniquesolution Mar 31 '17 at 13:11
  • @uniquesolution OK. Care to elaborate? You know, with arguments and all that... – Did Mar 31 '17 at 13:37
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$$ \begin{aligned} \lim _{x\to \infty }\left(x^2\:\left(\sqrt{x^4+x+1}-\sqrt{x^4+x+5}\right)\right) & = \lim _{x\to \infty }\left(-\frac{4x^2}{\sqrt{x^4+x+1}+\sqrt{x^4+x+5}}\right) \\& = \lim _{x\to \infty }\left(-\frac{4x^2}{x^2\sqrt{1+\frac{x}{x^4}+\frac{1}{x^4}}+x^2\sqrt{1+\frac{x}{x^4}+\frac{5}{x^4}}}\right) \\& \approx_\infty \lim _{x\to \infty }\left(-\frac{4x^2}{2x^2}\right) \\& = \color{red}{-2} \end{aligned} $$

In this case you can solve it with just an approximation without using the Taylor expansion, anyway your procedure is right.

Amarildo
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