<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] $$\lim _{ x\to \infty \:  } \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 }  \right) =$$
$\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 $0$
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