<s>[INST] Calculating limit-help needed (difference of cubes) I'm stuck trying to find this limit:
$$\lim_{x\to \infty} (\sqrt[3]{(x-2)^2(x-1)}-x)$$
Thanks in advance!
 [/INST] If the problem had a square root instead of a cube root, you probably know how to proceed. For example,
\begin{align*}
\lim_{x\to\infty} \big( \sqrt{(x-2)(x-1)} - x \big) &= \lim_{x\to\infty} \frac{(\sqrt{(x-2)(x-1)} - x)(\sqrt{(x-2)(x-1)} + x)}{\sqrt{(x-2)(x-1)} + x} \\
&= \lim_{x\to\infty} \frac{(x-2)(x-1) - x^2}{\sqrt{x^2-3x+2} + x} \\
&= \lim_{x\to\infty} \frac{-3x+2}{\sqrt{x^2-3x+2} + x} \\
&= \lim_{x\to\infty} \frac{-3+2/x}{\sqrt{1-3/x+2/x^2} + 1} = -\frac{3}{2}.
\end{align*}
That approach uses the identity $a^2-b^2=(a-b)(a+b)$ (with $a=\sqrt{(x-2)(x-1)}$ and $b=x$).
Given the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$, can you use a similar method in your problem?
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