# 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?

Here is a hint, from which the result follows easily: $$\sqrt{1+x}-1 = \frac{x}{\sqrt{1+x}+1}$$

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{\root{1 + x} - 1 = {x \over \root{1 + x} + 1} \approx {x \over \root{1 + 0} + 1} = {x \over 2}}$. Also $$\root{1 + x} = \pars{1 + x}^{1/2} = 1 + \color{#ff0000}{\large{1 \over 2}\,x} + {\pars{1/2}\pars{1/2 - 1} \over 2!}\,x^{2} + {\pars{1/2}\pars{1/2 - 1}\pars{1/2 - 2} \over 3!}\,x^{3} + \cdots$$

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., $0 \sim x$

$$\sqrt{x+1}$$ 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

$$\lim_{x\to0} \frac{\sqrt{1+x}-1}{x} = \left.\frac{d}{dx}\right|_{x=0}\sqrt{1+x} = \left.\frac{1}{2\sqrt{1+x}}\right|_{x=0} = \frac12$$

$$\sqrt { 1+x } -1=\sqrt { 1+x+\frac { { x }^{ 2 } }{ 4 } -\frac { { x }^{ 2 } }{ 4 } } -1=\sqrt { { \left( \frac { x }{ 2 } +1 \right) }^{ 2 }-\frac { { x }^{ 2 } }{ 4 } } -1\\$$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 $$\sqrt { { \left( \frac { x }{ 2 } +1 \right) }^{ 2 }-\frac { { x }^{ 2 } }{ 4 } } -1\sim \sqrt { { \left( \frac { x }{ 2 } +1 \right) }^{ 2 } } -1\sim \frac { x }{ 2 }$$