calculus - Evaluating the limit of $lim_{xtoinfty}(sqrt{frac{x^3}{x+2}}-x)$.

How do I evaluate this limit:
$$\lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right)$$

I tried to evaluate this using rationalizing the denominator, numerator and L'Hospital rule for nearly an hour with no success.

Answer

Rationalizing, observe that:
$$\begin{align*} \lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right) &= \lim_{x\to\infty}\left(\sqrt{\frac{x^3}{x+2}}-x\right)\left(\dfrac{\sqrt{\dfrac{x^3}{x+2}} + x}{\sqrt{\dfrac{x^3}{x+2}} + x}\right) \\ &= \lim_{x\to\infty} \dfrac{\dfrac{x^3}{x+2} - x^2}{\sqrt{\dfrac{x^3}{x+2}} + x} \\ &= \lim_{x\to\infty} \dfrac{\dfrac{-2x^2}{x+2}}{\sqrt{\dfrac{x^3}{x+2}} + x} \cdot \frac{\dfrac{1}{x}}{\dfrac{1}{x}}\\ &= \dfrac{\lim\limits_{x\to\infty} \dfrac{-2x}{x+2}}{\lim\limits_{x\to\infty} \left(\sqrt{\dfrac{x}{x+2}} + 1\right)}\\ &= \frac{-2}{\sqrt{1}+1} \\ &= -1 \end{align*}$$