<s>[INST] Simple limit of a sequence 
Need to solve this very simple limit $$ \lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$

I know how to solve these limits: by using
$a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - very tedious, boring and tiring. I hope there is some artful and elegant solution. Thank you!
 [/INST] You can use the binomial theorem to expand this.
$$\lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right)$$
Here 1 and 2 are the smallest terms and they can be ignored.
\begin{align}
&\lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right)
\\=& \lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x}-\sqrt[3]{3x^2+9x}\right)
\\=& \lim _{x\to \infty \:}\left(\sqrt[3]{3x^2\left(1+\frac{4}{3x}\right)}-\sqrt[3]{3x^2\left(1+\frac3x\right)}\right)
\\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left(\sqrt[3]{1+\frac{4}{3x}}-\sqrt[3]{1+\frac3x}\right)
\\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left({1+\frac{4}{9x}}-{1-\frac1x}\right)
\\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left({\frac{4-9}{9x}}\right)
\\=& \lim _{x\to \infty \:}\sqrt[3]{3}x^{2/3}\left({\frac{-5}{9x}}\right)
\\=& \lim _{x\to \infty \:}\sqrt[3]{3}x^{-1/3}\left({\frac{-5}{9}}\right)
\\=& \,0
\end{align}
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