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.

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.
