Evaluate $\lim\limits_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)$ 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.
 A: Using $$\alpha-\beta=\frac{\alpha^2-\beta^2}{\alpha+\beta}$$
yields
$$\lim_{x \to \infty} \sqrt{\frac{x^3}{x-1}}-x=\lim_{x \to \infty} \frac{\frac{x^3}{x-1}-x^2}{\sqrt{\frac{x^3}{x-1}}+x}=\lim_{x \to \infty} \frac{x^3-x^3+x^2}{(x-1)(\sqrt{\frac{x^3}{x-1}}+x)}=
\lim_{x \to \infty} \frac{x^2}{\sqrt{x^4-x^3}+x^2-x}=\lim_{x \to \infty} \frac{x^2}{x^2(\sqrt{1-\frac1x}+1-\frac1x)}=\lim_{x \to \infty} \frac{1}{\sqrt{1-\frac1x}+1-\frac1x}=\frac12$$
A: 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
$$\cdots \approx x\left(1+\frac{1}{2x-2}-1\right) = \frac{x}{2x-2} \to \frac{1}{2}$$
A: Multiply and divide by $\sqrt{x^3/(x-1)}+x$, simplify and take the limit.
A: $$ \displaystyle\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right) $$
$$ =\displaystyle\lim_{x \to \infty}\left (x\sqrt{\frac x{x-1}}-x\right) $$
$$ =\displaystyle\lim_{x \to \infty}\left (\frac{x\left(\sqrt x-\sqrt{x-1}\right)}{\sqrt{x-1}}\right) $$
$$ =\displaystyle\lim_{x \to \infty}\left ( \frac{x\{ x-(x-1)\}}{\sqrt{x-1}(\sqrt x+\sqrt{x-1})}\right) $$
$$ =\displaystyle\lim_{x \to \infty}\left (\frac1{\sqrt{\left(1-\frac1x\right)}\left(1+\sqrt{1-\frac1x}\right)} \right) $$ (dividing the numerator & the denominator by $x$)
$$=\frac12$$
A: 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.
A: $$
\begin{aligned}
\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right)
& =\lim _{t\to 0}\left(\sqrt{\frac{\frac{1}{t^3}}{\frac{1}{t}-1}}-\frac{1}{t}\right)
\\& = \lim _{t\to 0}\left(\frac{1}{t\sqrt{-t+1}}-\frac{1}{t}\right)
\\& = \lim _{t\to 0}\left(\frac{1}{\sqrt{-t+1}\left(\sqrt{-t+1}+1\right)}\right)
\\& = \color{red}{\frac{1}{2}}
\end{aligned}
$$
Solved with substitution $\color{red}{t = \frac{1}{x}}$ and rationalization of the numerator.
