<s>[INST] How can I evaluate: $\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$ I need to find this limit:
$$\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$$
The answer I got from using the limit laws is $\sqrt{\infty} - \sqrt{\infty}$.
How do I proceed now?
Added I took the conjugate of the function and I got a new and probably better function to work with: $$\lim_{x\to \infty}\frac{9x}{\sqrt{x^2 + 4x} + \sqrt{x^2-5x}}$$
 [/INST] $$\\ \lim _{ x\rightarrow \infty  }{ \sqrt { { x }^{ 2 }+4x } -\sqrt { { x }^{ 2 }-5x }  } =\lim _{ x\rightarrow \infty  }{ \frac { \left( \sqrt { { x }^{ 2 }+4x } -\sqrt { { x }^{ 2 }-5x }  \right) \left( \sqrt { { x }^{ 2 }+4x } +\sqrt { { x }^{ 2 }-5x }  \right)  }{ \sqrt { { x }^{ 2 }+4x } +\sqrt { { x }^{ 2 }-5x }  }  } =\\ =\lim _{ x\rightarrow \infty  }{ \frac { 9x }{ \sqrt { { x }^{ 2 }+4x } +\sqrt { { x }^{ 2 }-5x }  }  } =\lim _{ x\rightarrow \infty  }{ \frac { 9x }{ \sqrt { { x }^{ 2 }\left( 1+\frac { 4 }{ x }  \right)  } +\sqrt { { x }^{ 2 }\left( 1-\frac { 5 }{ x }  \right)  }  }  } =\lim _{ x\rightarrow \infty  }{ \frac { 9x }{ \left| x \right| \left( \sqrt { 1+\frac { 4 }{ x }  } +\sqrt { 1-\frac { 5 }{ x }  }  \right)  }  } =\\ \overset { if\quad x>0 }{ = } \lim _{ x\rightarrow \infty  }{ \frac { 9x }{ x\left( \sqrt { 1+\frac { 4 }{ x }  } +\sqrt { 1-\frac { 5 }{ x }  }  \right)  }  } =\lim _{ x\rightarrow \infty  }{ \frac { 9 }{ \left( \sqrt { 1+\frac { 4 }{ x }  } +\sqrt { 1-\frac { 5 }{ x }  }  \right)  }  } =\frac { 9 }{ \sqrt { 1+0 } +\sqrt { 1-0 }  } =\frac { 9 }{ 2 } $$
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