function approximation using series

Question

I was trying to approximate a function by another one using some kind of a series.

Let $$ f(x) = m\cdot \left(1-\sqrt{\frac x2\biggm/\left(1+\frac x2\right)}\right) $$

I'm trying to approximate this function for large $x$.

I was trying to use taylor series but it didn't work because I don't have a specific value a that I'm converging to.

Another attempt was to take the limit at infinity, but this won't help either, because lots of functions converges to the same limit.

What I'm asked to do is to approximate $f(x)$ by $m/x$ for $x$ large enough.

Thank you for your time, Best

Answer 1

$$\begin{align} 1-\sqrt{\frac{x}{2+x}}&=\frac{1-\frac{x}{2+x}}{1+\sqrt{\frac{x}{2+x}}}\\ &=\frac{1}{2+x}\,\frac{2}{1+\sqrt{\frac{x}{2+x}}} \end{align}$$ Now $$ \frac{1}{2+x}=\frac1x+O\Bigl(\frac{1}{x^2}\Bigr) $$ and $$ \frac{2}{1+\sqrt{\frac{x}{2+x}}}=1+O\Bigl(\frac{1}{x}\Bigr). $$

Answer 2

$$ \begin{align} 1-\sqrt{1-\frac1{1+x/2}} &=\frac{\frac1{1+x/2}}{1+\sqrt{1-\frac1{1+x/2}}}\\ &=\frac1{x+2}\frac2{1+\sqrt{1-\frac1{1+x/2}}}\\ &=\frac1{x+2}\left(1+\frac{1-\sqrt{1-\frac1{1+x/2}}}{1+\sqrt{1-\frac1{1+x/2}}}\right)\\ &=\frac1{x+2}\left(1+\frac{\large\frac1{1+x/2}}{\small\left(1+\sqrt{1-\frac1{1+x/2}}\right)^2}\right)\\ &=\frac1{x+2}\left(1+\frac12\frac1{x+2}\left(\frac2{\small1+\sqrt{1-\frac1{1+x/2}}}\right)^{\!\!\!2}\right)\\[9pt] &=\frac1{x+2}+\frac12\frac1{(x+2)^2}+O\left(\frac1{(x+2)^3}\right) \end{align} $$ Then, if we wish, we can use $$ \begin{align} \frac1{x+2} &=\frac1x\frac1{1+2/x}\\ &=\frac1x-\frac2{x^2}+O\left(\frac1{x^3}\right) \end{align} $$