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Compute $\lim\limits_{x\to \infty} \left(\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3} \right) $.
I tried multiplying by the conjugate and I get to an ugly expression. What should I do?

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    $\begingroup$ You need to multiply and divide by the conjugate. But you don't need to evaluate the resulting ugly expression. You just need to divide the numerator and denominator by the appropriate power of $x$, and then take the limit. $\endgroup$ – Robert Shore Mar 5 at 18:14
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So, you arrived at

$$\begin{align} \sqrt{(x+1/\sqrt x)^3}-\sqrt{x^3}&=\frac{3x^{3/2}+3+x^{-3/2}}{\sqrt{(x+1/\sqrt x)^3}+\sqrt{x^3}}\\\\ &=\frac{3x^{3/2}+3+x^{-3/2}}{x^{3/2}\left(1+\sqrt{(1+1/x^{3/2})^3}\right)} \end{align}$$

Now let $x\to \infty$. Can you finish now?

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  • $\begingroup$ Yes, thank you! $\endgroup$ – MathEnthusiast Mar 5 at 18:40
  • $\begingroup$ You're welcome! My pleasure. $\endgroup$ – Mark Viola Mar 5 at 18:48
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hint :- it becomes very easy if you substitute x = 1/t² where as x tends to infinity t tends to 0 ( precisely from right hand side) this would simplify to easily solvable form using l hopital or other techniques

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  • $\begingroup$ this would make the rationalisation very easy $\endgroup$ – Aditya Garg Mar 5 at 18:27
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Hello Taylor my old friend...

$\begin{array}\\ \sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3} &=\left(x+x^{-1/2}\right) ^{3/2} - x^{3/2}\\ &=x^{3/2}\left(\left(1+x^{-3/2}\right) ^{3/2} - 1\right)\\ &=x^{3/2}\left(\left(1+\frac32x^{-3/2}+O(x^{-3})\right) - 1\right)\\ &=x^{3/2}\left(\frac32x^{-3/2}+O(x^{-3})\right)\\ &=\frac32+O(x^{-3/2})\\ \end{array} $

Actually, of course, it's the generalized binomial theorem, but that doesn't scan as well.

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