Approximation of $\frac{x}{\sqrt{x^2+R^2}}$

Question

How do you prove this statement?

If $x\gg R$ then $$\frac{x}{\sqrt{x^2+R^2}}\cong 1-\frac{1}{2}\left(\frac{R}{x}\right)^2$$

I have no ideas even how to start.

Answer 1

Basically, this can be seen by a Taylor series expansion around $0$. Note that $x\gg R$ "means" $\frac{R}{x} \approx 0$: so you would need to make this quantity appear, first.

$$\begin{align} \frac{x}{\sqrt{x^2+R^2}} &= \frac{x}{x\sqrt{1+\left(\frac{R}{x}\right)^2}} = \frac{x}{x}\left(1-\frac{1}{2}\left(\frac{R}{x}\right)^2 + o\left(\left(\frac{R}{x}\right)^2\right)\right) \\ &= 1-\frac{1}{2}\left(\frac{R}{x}\right)^2 + o\left(\frac{R^2}{x^2}\right) \end{align}$$ using the fact that $(1+u)^\alpha = 1+\alpha u + o(u)$ when $u\to0$ (for any fixed $\alpha \in\mathbb{R}$).

Answer 2

Compare these two functions at $x=\infty$. If their quotient $f/g$ tends to $1$, it means $f\sim g$:

$$\lim_{x\to\infty} \frac{\frac{x}{\sqrt{x^2+R^2}}}{1-\frac{1}{2}\left(\frac{R}{x}\right)^2}= \lim_{x\to\infty} \frac{\frac{x}{\sqrt{x^2+R^2}}}{\frac{2x^2-R^2}{2x^2}}= \lim_{x\to\infty} \frac{x\cdot 2x^2}{\sqrt{x^2+R^2}(2x^2-R^2)}= \lim_{x\to\infty} \frac{2}{\sqrt{1+R^2/x^2}(2-R^2/x^2)}=1$$