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Evaluate $$ \lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right) $$

The answer is $\frac{1}{2}$, have no idea how to arrive at that.

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Using $$\alpha-\beta=\frac{\alpha^2-\beta^2}{\alpha+\beta}$$ yields $$\lim_{x \to \infty} \sqrt{\frac{x^3}{x-1}}-x=\lim_{x \to \infty} \frac{\frac{x^3}{x-1}-x^2}{\sqrt{\frac{x^3}{x-1}}+x}=\lim_{x \to \infty} \frac{x^3-x^3+x^2}{(x-1)(\sqrt{\frac{x^3}{x-1}}+x)}= \lim_{x \to \infty} \frac{x^2}{\sqrt{x^4-x^3}+x^2-x}=\lim_{x \to \infty} \frac{x^2}{x^2(\sqrt{1-\frac1x}+1-\frac1x)}=\lim_{x \to \infty} \frac{1}{\sqrt{1-\frac1x}+1-\frac1x}=\frac12$$

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  • $\begingroup$ I realized the necessity to apply square difference identity but just fail to apply that. Thanks! $\endgroup$ – Paul Smith Dec 15 '12 at 11:45
  • $\begingroup$ @PaulSmith I made a mistake before. Look at the corrected answer $\endgroup$ – Nameless Dec 15 '12 at 11:47
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Alternatively:

$$ \sqrt{\frac{x^3}{x-1}} - x = x\sqrt{\frac{x}{x-1}} - x = x\left(\sqrt{1+\frac{1}{x-1}}-1\right) $$

For small $\alpha$ we have $\sqrt{1+\alpha}\approx 1+\alpha/2$, so for large $x$ we get

$$\cdots \approx x\left(1+\frac{1}{2x-2}-1\right) = \frac{x}{2x-2} \to \frac{1}{2}$$

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  • $\begingroup$ $\sqrt{1+\alpha}-1=\frac{\alpha}{\sqrt{1+\alpha}+1}$ helps to add some rigor to this. $\endgroup$ – robjohn Dec 15 '12 at 13:54
  • $\begingroup$ @robjohn, or just from calculus: $\sqrt{1+\alpha}=1+\alpha/2+o(\alpha)$ due to the derivative of the square root, and then carry the $o(\alpha)$ through the rest of the computation to see that it is harmless -- we end up with $\frac{x}{2x-2}+o(\frac{x}{x-1})$. $\endgroup$ – Henning Makholm Dec 15 '12 at 14:19
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Multiply and divide by $\sqrt{x^3/(x-1)}+x$, simplify and take the limit.

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  • $\begingroup$ This is essentially what Nameless did, with the details. This makes a nice hint. $\endgroup$ – robjohn Dec 15 '12 at 13:52
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$$ \displaystyle\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right) $$

$$ =\displaystyle\lim_{x \to \infty}\left (x\sqrt{\frac x{x-1}}-x\right) $$

$$ =\displaystyle\lim_{x \to \infty}\left (\frac{x\left(\sqrt x-\sqrt{x-1}\right)}{\sqrt{x-1}}\right) $$

$$ =\displaystyle\lim_{x \to \infty}\left ( \frac{x\{ x-(x-1)\}}{\sqrt{x-1}(\sqrt x+\sqrt{x-1})}\right) $$

$$ =\displaystyle\lim_{x \to \infty}\left (\frac1{\sqrt{\left(1-\frac1x\right)}\left(1+\sqrt{1-\frac1x}\right)} \right) $$ (dividing the numerator & the denominator by $x$)

$$=\frac12$$

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Method I
By simply applying l'Hôpital's rule, we have $$\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x \right)=\lim_{x \to \infty}\frac{\displaystyle\sqrt{\frac{x}{x-1}}-1}{\left (\displaystyle \frac{1}{x}\right)}=\lim_{x \to \infty}\frac{x^2}{2(x-1)^2\displaystyle\sqrt{\frac{x}{x-1}}}=\frac{1}{2}.$$ Done.

Method II
Using the same start, we resort to the elementary limit $\lim_{y\to1} \displaystyle \frac{\sqrt{y}-1}{y-1}=\frac{1}{2}.$ Then $$\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x \right)=\lim_{x \to \infty}\frac{\displaystyle\sqrt{\frac{x}{x-1}}-1}{\left (\displaystyle \frac{1}{x}\right)}=\lim_{x \to \infty}\frac{\displaystyle\sqrt{\frac{x}{x-1}}-1}{\left (\displaystyle \frac{x}{x-1}-1\right) }\cdot \frac{x}{x-1} = \frac{1}{2}.$$

Done.

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$$ \begin{aligned} \lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right) & =\lim _{t\to 0}\left(\sqrt{\frac{\frac{1}{t^3}}{\frac{1}{t}-1}}-\frac{1}{t}\right) \\& = \lim _{t\to 0}\left(\frac{1}{t\sqrt{-t+1}}-\frac{1}{t}\right) \\& = \lim _{t\to 0}\left(\frac{1}{\sqrt{-t+1}\left(\sqrt{-t+1}+1\right)}\right) \\& = \color{red}{\frac{1}{2}} \end{aligned} $$ Solved with substitution $\color{red}{t = \frac{1}{x}}$ and rationalization of the numerator.

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