<s>[INST] Not getting the correct answer for an infinite limit problem The problem:
$\lim _{ x\to -\infty  } \frac { \sqrt { 9x^{ 6 }-x }  }{ x^{ 3 }+8 }$
My answer: $3$
What I did:
$\\ \lim _{ x\to -\infty  } \frac { \sqrt { x^{ 6 }(9-\frac { 1 }{ { x }^{ 5 } } ) }  }{ x^{ 3 }(1+\frac { 8 }{ { x }^{ 3 } } ) } \\ \\ \lim _{ x\to -\infty  } \frac { { x }^{ 3 }\sqrt { 9-\frac { 1 }{ { x }^{ 5 } }  }  }{ x^{ 3 }(1+\frac { 8 }{ { x }^{ 3 } } ) } \\ \lim _{ x\to -\infty  } \frac { \sqrt { 9-\frac { 1 }{ { x }^{ 5 } }  }  }{ 1+\frac { 8 }{ { x }^{ 3 } }  } \\ $
And since
$\\ \lim _{x\to-\infty} \frac{1}{{x}^{5}}\quad =\quad 0\\ \\ \lim _{x\to -\infty} \frac{8}{{x}^{3}} \quad =\quad 0\\ \\ $
plugging in $0$ for above values...
$\\ \frac {\sqrt{9}}{1} \quad =\quad 3\\$
The correct answer is apparently $-3$, I don't understand why.
 [/INST] Hint:  use that $\sqrt{x^6}=|x^3|$
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