# How to find asymptotic behavior of this function?

I've read in a book that this function: $$b(r) = \sqrt{\frac{r^3}{r - 2M}}$$ behaves approximately as $$b(r) \approx r + M$$ for $$r \gg M$$.

I checked it numerically, by plotting a graph of this function, and found that it is true. How to show this mathematically?

Let $$R >0.$$

$$b(R)=R(1-2M/R)^{-1/2}.$$

Binomial expansion for rational exponent $$\alpha$$:

$$(1+x)^{\alpha}=$$

$$1+\alpha x+(\alpha (\alpha-1)/2!)x^2+ O(x^3)$$

converges for $$|x|<1.$$

Hence for large $$R$$:

$$b(R)=R(1+M/R +O((M/R)^2))$$ https://brilliant.org/wiki/fractional-binomial-theorem/

This is just a Taylor expansion $$b(r) = \sqrt{\frac{r^3}{r - 2M}}=r \sqrt{\frac{r}{r - 2M}}$$ Use long division or Taylor series $$\frac{r}{r - 2M}=1+\frac{2 M}{r}+\frac{4 M^2}{r^2}+O\left(\frac{1}{r^3}\right)$$ $$\sqrt{\frac{r}{r - 2M}}=1+\frac{M}{r}+\frac{3 M^2}{2 r^2}+O\left(\frac{1}{r^3}\right)$$ $$b(r)=r+M+\frac{3 M^2}{2 r}+O\left(\frac{1}{r^2}\right)=r+M+O\left(\frac{1}{r}\right)$$

$$\frac{\sqrt{\frac{r^3}{r - 2M}}}{r+M}=\sqrt{\frac{r^3}{(r-2M)(r+M)^2}}=$$ $$\sqrt{\frac{r^3}{r^3\left(1-\frac{2M}r\right)\left(1+\frac Mr\right)^2}}=\sqrt{\frac{1}{\left(1-\frac{2M}r\right)\left(1+\frac Mr\right)^2}}\to 1$$ as $$r\to\infty$$.

• While this is true, it works for $r+K$ for any constant $K$, so I don't think it fundamentally addresses OP's question. – Mees de Vries Jul 19 at 8:22