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.

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