<s>[INST] How to find the limit $\lim\limits _{ x\to \infty } \left( \sqrt { x^2+3x } -\sqrt { x^2+x } \right) $? When I try to do this type of indeterminations I reach to this point:
$\lim\limits_{ x\to \infty  } \dfrac { 2x }{ \sqrt { x^ 2 +3x } +\sqrt { x^2 +x }  } $
but I don't know how to continue. Thanks.
 [/INST] $\lim\limits_{ x\to \infty  }{ \dfrac { 2x }{ \sqrt { { x }^{ 2 }+3x } +\sqrt { { x }^{ 2 }+x }  }  } $
$\lim\limits_{x\to \infty  }\frac {2}{\large{\sqrt \frac{x^2+3x}{x^2}+\sqrt \frac{x^2+x}{x^2}}}=\lim\limits_{x\to \infty  }\frac {2}{\large{\sqrt{ 1+\frac{3x}{x^2}}+\sqrt{ 1+\frac{x}{x^2}}}}=\lim\limits_{x\to \infty  }\frac {2}{\large{\sqrt{ 1+\frac{3}{x}}+\sqrt{ 1+\frac{1}{x}}}}=\frac {2}{\large{\sqrt{ 1+0}+\sqrt{ 1+0}}}=1$
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