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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.

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Think about what happens to each of the polynomial separately as $x$ becomes large first..... – Simon Hayward Dec 9 '12 at 12:31

Since $$\alpha-\beta=\frac{\alpha^6-\beta^6}{\alpha^5+...+\beta^5}$$ we have that \begin{gather}\lim\limits_{x\rightarrow+\infty}({x^3+x^2+1})^{1/3}-{(x^2+x+1)}^{1/2}=\lim\limits_{x\rightarrow+\infty}x[{(1+1/x+1/{x^2})}^{1/3}-{(1+1/x+1/{x^2})^{1/2}}]= \lim\limits_{x\rightarrow+\infty}x\frac{(1+1/x+1/{x^2})^2-(1+1/x+1/{x^2})^3}{(1+1/x+1/{x^2})^{5/3}+...+{(1+1/x+1/{x^2})}^{5/2}} \end{gather} The denominator goes to $6$ and the numerator: \begin{gather}\lim\limits_{x\rightarrow+\infty}x[(1+1/x+1/{x^2})^2-(1+1/x+1/{x^2})^3]=\\ \lim\limits_{x\rightarrow+\infty}x(1+1/x+1/{x^2})^2[-1/x-1/{x^2}]=\\ \lim\limits_{x\rightarrow+\infty}-(1+1/x+1/{x^2})^2[1+1/x]=-1 \end{gather}

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Best answer for me +1 – Amr Dec 9 '12 at 12:23
+1 And for me:very nice. – DonAntonio Dec 9 '12 at 15:43

$$ (x^3+x^2+1)^{1/3} - (x^2+x+1)^{0.5} = x \cdot (1+x^{-1}+x^{-3})^{1/3} - x \cdot (1+x^{-1}+x^{-2})^{0.5} $$

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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Using the degree-one Taylor polynomials for $e^x$ and $\log(1+x)$, $$ (x^3+x^2+1)^{1/3} - (x^2+x+1)^{0.5} = x \left[ \left(1+\frac1x+\frac1{x^3}\right)^{1/3} - \left(1+\frac1x+\frac1{x^2}\right)^{1/2}\right]= x \left[ e^{\frac13\,\log\left(1+\frac1x+\frac1{x^3}\right)} - e^{\frac12\,\log\left(1+\frac1x+\frac1{x^2}\right)}\right]=x\left[e^{\frac13\,\left(\frac1x+O(\frac1{x^2})\right)}-e^{\frac12\,\left(\frac1x+O(\frac1{x^2})\right)} \right]\\=x\,\left[1+\frac1{3x}+O(\frac1{x^2})-(1+\frac1{2x}+O(\frac1{x^2}))\right] =-\frac16+O(\frac1x)$$

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$$\lim_{x\to \infty}(x^3+x^2+1)^\frac13-(x^2+x+1)^\frac12$$

$$=\lim_{x\to \infty}x\{\left(1+\frac{x^2+1}{x^3}\right)^\frac13-\left(1+\frac{x+1}{x^2}\right)^\frac12\}$$

$$=\lim_{x\to \infty}x\{(1+\frac13\frac{x^2+1}{x^3}+\cdots)-(1+\frac12\frac{x+1}{x^2}+\cdots)\}$$ 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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