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!
 A: $$\lim _{ x\to \infty \:  } \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 }  \right) =$$
$\lim _{ x\to \infty \:  } \frac { \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 }  \right) \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right)  }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right)  } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right)  }^{ 2 } }  \right)  }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right)  }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right)  } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right)  }^{ 2 } }  \right)  } =$$$\ =\lim _{ x\to \infty :  } \frac { 3x^{ 2 }+4x+1-3x^{ 2 }-9x-2 }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right)  }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right)  } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right)  }^{ 2 } }  \right)  } =\lim _{ x\rightarrow \infty  }{ \frac { -5x-1 }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right)  }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right)  } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right)  }^{ 2 } }  \right)  }  } $$
now ,the power of denominator of polynomial is higher than the  numerator so the limit is equal to $0$
A: You have $$f(x)=\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}=\sqrt[3]{3x^2}\left(\sqrt[3]{1+\frac{4}{3x}+\frac{1}{3x^2}}-\sqrt[3]{1+\frac{3}{x}+\frac{2}{3x^2}}\right)$$ Using Taylor expansion at order one of the cubic roots $\sqrt[3]{1+y}=1+\frac{y}{3}+o(y)$ at the neighborhood of $0$, you get: $$f(x)=\sqrt[3]{3x^2}\left(\frac{4}{9x}-\frac{1}{x}+o\left(\frac{1}{x}\right)\right)$$ hence $$\lim\limits_{x \to \infty} f(x)=0$$
A: 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}
A: Actually, this question can be solved without calculation if you're familiar with power.
If the first term(in the square root), the biggest term is $3x^2$ and so does the second term. And any term has power below 2 are negligible. So you can just forget them and get your answers, which is $0$.
A: Can you help with O-symbols? It's all right here? 
$$f(x) = \sqrt[3]{3x^2}\left(1 + \frac{4}{9x} + O\left(\frac{1}{x^2}\right) - 1 - \frac{1}{x} -O \left(\frac{1}{x^2}\right)\right)= \sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1}{18x^2} \right). $$
Hence
$$\lim _{x\to \infty }\sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1}{18x^2} \right)= \lim_{x \to \infty} \sqrt[3]{3}{x^{-1/3}}^{\to 0} \left( - \frac{5}{9}+\frac{1}{18x}^{\to 0} \right) = 0.$$
