The limit of a difference between a cube root and square root I've been trying to evaluate the limit of $$\lim_{x \to \infty}[(x^3+x^2+1)^{1/3}-(x^2+x)^{1/2}]$$ I've tried using the property $$x^6-y^6 =(x-y)(x^5+x^4y+\cdots+xy^4+y^5)$$
where $x = (x^3+x^2+1)^{1/3}$ and $y=(x^2+x)^{1/2}$. But I can't seem to 'clean up' the denominator to get in a position to take the limit. Many thanks in advance.
 A: Replace $x$ by $\frac{1}{t}$ where we think of $t$ as small positive. Our expression becomes
$$\lim_{t\to 0^+}\frac{(1+t+t^3)^{1/3}-(1+t)^{1/2}}{t}.$$
To find the limit as $t\to 0^+$ use L'Hospital's Rule.
If we don't want to use L'Hospital's Rule, we can rewrite as 
$$\lim_{t\to 0^+} \frac{\left((1+t+t^3)^{1/3}-1\right)-\left((1+t)^{1/2}-1\right)}{t},$$
and recognize the difference of two derivatives.
A: Using your property, you have 
\begin{align}
(x^3+x^2+1)^{1/3}-(x^2+x)^{1/2}&=\frac{(x^3+x^2+1)^2-(x^2+x)^3}{(x^3+x^2+1)^{5/3}+\cdots+(x^2+x)^{5/2}}\\ \ \\
&=\frac{-x^5-2x^4+x^3+2x^2+1}{(x^3+x^2+1)^{5/3}+\cdots+(x^2+x)^{5/2}}\\ \ \\
&=\frac{-1-2/x+1/x^2+2/x^3+1/x^5}{(1+1/x+1/x^3)^{5/3}+\cdots+(1+1/x)^{5/2}}\\ \ \\
&\to-\frac16
\end{align}
(where in the last equality you divide numerator and denominator by $x^5$, and then use the fact that there are six terms in the denominator).
Also here is the way to do this limit that comes naturally to me. From the binomial series, $(1+a)^b\simeq 1+\frac ab$ for small $a$. Then, for small $x$,
\begin{align}
(x^3+x^2+1)^{1/3}-(x^2+x)^{1/2}&=x\,\left[ (1+1/x+1/x^3)^{1/3}-(1+1/x)^{1/2} \right]\\ \ \\
&\simeq x\,\left[1+\frac1{3x}+\frac1{3x^3}-1-\frac1{2x} \right]\ \\ \ \\
&=x\,\left[-\frac{1}{6x}+\frac1{3x^3} \right]=-\frac16+\frac1{3x^2}\\ \ \\
&\to-\frac16.
\end{align}
