How to expand $\sqrt{x+\epsilon}$ in the following way?

Question

How to expand $\sqrt{x+\epsilon}$ in the following:

$$\sqrt{x+\epsilon}=\sqrt{x}\big(1+\frac{\epsilon}{2x}-\frac{\epsilon^2}{8x^2}+\frac{\epsilon^3}{16x^3}-\cdots\big)$$
as $\frac{\epsilon}{x}\rightarrow 0$

I just know that the Taylor expansion of $\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\cdots$. I know that this is related to the perturbation methods; however, I have no idea how to do the first step. Could anyone give me a hint or the rough idea? thanks!

Answer 1

$$\frac{\sqrt{x+\epsilon}}{\sqrt{x}}=\sqrt{\frac{x+\epsilon}{x}}=\sqrt{1+\frac{\epsilon}{x}}$$

Now you can use the Taylor expansion of $\sqrt{1+y}$ where $y = \frac{\epsilon}{x}.$

Answer 2

Since $$\sqrt{\epsilon+x}=\sqrt x\cdot \sqrt{1+\frac{\epsilon}{x}}$$ we have $$\sqrt{1+\frac{\epsilon}{x}}=1+\frac{1}{2}\cdot\frac{\epsilon}x-\frac{1}{8}\left(\frac\epsilon x\right)^2+\cdots=1+\frac\epsilon{2x}-\frac{\epsilon^2}{8x^2}+\cdots$$ and the result follows.