<s>[INST] How can I prove this statment? I have this problem to solve:
Prove that when $$x\rightarrow 0$$  $$\sqrt{1+x}-1 \sim \frac{x}{2}$$
Can someone give me a tip? or show me the way?
Thanks in advance
 [/INST] $$\sqrt { 1+x } -1=\sqrt { 1+x+\frac { { x }^{ 2 } }{ 4 } -\frac { { x }^{ 2 } }{ 4 }  } -1=\sqrt { { \left( \frac { x }{ 2 } +1 \right)  }^{ 2 }-\frac { { x }^{ 2 } }{ 4 }  } -1\\$$For $x$ very small$$ \frac { { x }^{ 2 } }{ 4 } <<{ \left( \frac { x }{ 2 } +1 \right)  }^{ 2 },$$therefore you can ignore the term $\frac { { x }^{ 2 } }{ 4 } $ in $\sqrt { { \left( \frac { x }{ 2 } +1 \right)  }^{ 2 }-\frac { { x }^{ 2 } }{ 4 }  } -1$. Then you have $$\sqrt { { \left( \frac { x }{ 2 } +1 \right)  }^{ 2 }-\frac { { x }^{ 2 } }{ 4 }  } -1\sim  \sqrt { { \left( \frac { x }{ 2 } +1 \right)  }^{ 2 } } -1\sim  \frac { x }{ 2 } $$
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