Approximation of $\frac{1-x}{1+x}$ if $x \ll 1$?

Question

In an exercise in general relativity, I am trying to show that in the limit as $\hat{r}\rightarrow\infty$,

\begin{equation*} \frac{1-GM/2\hat{r}}{1+GM/2\hat{r}} \approx 1 - \frac {2GM}{\hat{r}},\ \ \text{or}\ \ \frac{1-\phi}{1+\phi} \approx 1-4\phi\ \text{if}\ \phi\ll 1 \end{equation*}

At least, this is the approximation I deduce I need to show. I don't see a path to it, however. I see ways to show through Taylor expansion that to first order it's approximately $1-2\phi$, but I need $1-4\phi$.

Answer 1

What dxiv said in the comment is correct. One more idea for calculating $\frac{1-\phi}{1+\phi}$ is to assume $\tan^2(x)=\phi$ and hence using the formula $\cos(2x)=\frac{1-\tan^2(x)}{1+\tan^2(x)}$ and hence we have $\frac{1-\phi}{1+\phi} = \cos(2 (\tan^{-1}(\sqrt{\phi})))$. This is exact expression and not an approximation.

Answer 2

\begin{align} & 1+x \approx1 + x + x^2+ x^3 + \cdots = \frac 1 {1-x}. \\[8pt] & \frac 1 {1+x} \approx 1-x. \\[8pt] & \frac{1-x}{1+x} \approx(1-x)^2 = 1 - 2x+x^2 \approx 1 - 2x. \end{align}

Answer 3

To demonstrate my comment (that this approximation works if both combinations containing $\hat{r}$ are $\frac{GM}{\hat{r}}$)...\begin{align*} \frac{1 - GM/\hat{r}}{1 + GM/\hat{r}} &= \frac{1 + GM/\hat{r} - 2GM/\hat{r}}{1 + GM/\hat{r}} \\ &= 1 - \frac{2 G M/\hat{r}}{1 +GM/\hat{r}} \cdot \frac{\hat{r}}{\hat{r}} \\ &= 1 - \frac{2 G M}{\hat{r} + GM} \\ &\approx 1 - \frac{2 G M}{\hat{r}} \text{,} \end{align*} for $\hat{r} \gg GM$.