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.

$$ \displaystyle\lim_{x \to \infty}\left (\sqrt{\frac{x^3}{x-1}}-x\right) $$
$$ =\displaystyle\lim_{x \to \infty}\left (x\sqrt{\frac x{x-1}}-x\right) $$
$$ =\displaystyle\lim_{x \to \infty}\left (\frac{x\left(\sqrt x-\sqrt{x-1}\right)}{\sqrt{x-1}}\right) $$
$$ =\displaystyle\lim_{x \to \infty}\left ( \frac{x\{ x-(x-1)\}}{\sqrt{x-1}(\sqrt x+\sqrt{x-1})}\right) $$
$$ =\displaystyle\lim_{x \to \infty}\left (\frac1{\sqrt{\left(1-\frac1x\right)}\left(1+\sqrt{1-\frac1x}\right)} \right) $$ (dividing the numerator & the denominator by $x$)
$$=\frac12$$
