<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] $$\lim_{x\to \infty}(x^3+x^2+1)^\frac13-(x^2+x+1)^\frac12$$
$$=\lim_{x\to \infty}x\{\left(1+\frac{x^2+1}{x^3}\right)^\frac13-\left(1+\frac{x+1}{x^2}\right)^\frac12\}$$
$$=\lim_{x\to \infty}x\{(1+\frac13\frac{x^2+1}{x^3}+\cdots)-(1+\frac12\frac{x+1}{x^2}+\cdots)\}$$ using Binomial Expansion for rational exponent which is legal as $\mid\frac{x^2+1}{x^3}\mid<1$ and $\mid\frac{x+1}{x^2}\mid<1$ for $x\to \infty$
$$=\lim_{x\to \infty}\left(\frac{x^2+1}{3x^2}-\frac{x+1}{2x}\right)=\frac13-\frac12=-\frac16$$
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