# 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: $$\sqrt{1+x}-1 = \frac{x}{\sqrt{1+x}+1}$$


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