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
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$\ds{\root{1 + x} - 1 = {x \over \root{1 + x} + 1} \approx {x \over \root{1 + 0} + 1} =  {x \over 2}}$. Also
$$
\root{1 + x} = \pars{1 + x}^{1/2} = 1 + \color{#ff0000}{\large{1 \over 2}\,x}
+
{\pars{1/2}\pars{1/2 - 1} \over 2!}\,x^{2}
+
{\pars{1/2}\pars{1/2 - 1}\pars{1/2 - 2} \over 3!}\,x^{3} + \cdots
$$
A: I am not sure if this helps but....
$\sqrt{1+x}-1 \sim \frac{x}{2}$
i.e., $\sqrt{1+x} \sim 1+\frac{x}{2}$ 
i.e., $2\sqrt{1+x} \sim 2+x$
i.e., $4+4x \sim 4+4x =x^2$ 
i.e., $0 \sim x^2$ 
i.e., $0 \sim x$
A: $$\sqrt{x+1}
$$
using binomial expansion
$$
=1+\frac{\text{x}}{2}-\frac{\text{x}^2}{8}+\frac{\text{x}^3}{16}...$$
hence when x-> 0 
then $$\sqrt{x+1}-1==\frac{\text{x}}{2}-\frac{\text{x}^2}{8}+\frac{\text{x}^3}{16}...
$$
what you are doing is ignoring the rest of the terms because~ $$x^2 , x^3...$$ are nearly equal to zero when x->0
A: Here is a hint, from which the result follows easily:
$$\sqrt{1+x}-1  = \frac{x}{\sqrt{1+x}+1}$$
A: $$\lim_{x\to0} \frac{\sqrt{1+x}-1}{x} = \left.\frac{d}{dx}\right|_{x=0}\sqrt{1+x} = \left.\frac{1}{2\sqrt{1+x}}\right|_{x=0} = \frac12$$
A: $$\sqrt { 1+x } -1=\sqrt { 1+x+\frac { { x }^{ 2 } }{ 4 } -\frac { { x }^{ 2 } }{ 4 }  } -1=\sqrt { { \left( \frac { x }{ 2 } +1 \right)  }^{ 2 }-\frac { { x }^{ 2 } }{ 4 }  } -1\\$$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 $$\sqrt { { \left( \frac { x }{ 2 } +1 \right)  }^{ 2 }-\frac { { x }^{ 2 } }{ 4 }  } -1\sim  \sqrt { { \left( \frac { x }{ 2 } +1 \right)  }^{ 2 } } -1\sim  \frac { x }{ 2 } $$
