<s>[INST] $\lim\limits_{x\rightarrow+\infty}\left[(x^3+x^2+1)^{1/3}- (x^2+x+1)^{0.5}\right] = -1/6$? Every method I use seem to get me to something to the extent of $0/0$, stuff I can't work with. Wolfram Alpha claims the answer to this is $-1/6$ but they offer no step by step solution.
Would appreciate any tips and help.
 [/INST] $$  (x^3+x^2+1)^{1/3} - (x^2+x+1)^{0.5} = x \cdot (1+x^{-1}+x^{-3})^{1/3} - x \cdot (1+x^{-1}+x^{-2})^{0.5} $$
then I suggest Taylor series for both terms. 
$$ x \cdot \left [ 1+\frac{1}{3} \cdot x^{-1}-\frac{1}{9} \cdot x^{-2} + ... - 1-\frac{1}{2}\cdot x^{-1}-\frac{3}{8} \cdot x^{-2} - ...\right] = x \cdot [ -\frac{1}{6} x^{-1} + ... ] = -\frac{1}{6} + o(x^{-1})$$
which has limit of $-\frac{1}{6}$.
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