<s>[INST] 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.
 [/INST] 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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