Why is $1-Cx\approx \frac{1}{1+Cx}$? How is this approximation derived? It is from a physics book.

$C$ is a constant and for small $Cx$ we have
  $$1-Cx\approx \frac{1}{1+Cx}$$ 

Is it also true without $C$? 
I.e. for small $x$, $1-x\approx \frac{1}{1+x}$?
 A: This is because for $|t|<1$, we have $(1-t)^{-1} = 1+t+t^2+t^3+\cdots$. Then Set $t=-Cx$.
A: If you are familiar with geometric series, that is 
$$1+x+x^2+\cdots+x^n=\dfrac{1-x^{n+1}}{1-x}$$
and when $|x|<1$ as $n\to\infty$ this expression takes the form 
$$1+x+x^2+\cdots=\dfrac{1}{1-x}$$
if you set $x=Ct$ then for $|Ct|<1$ we can use this series. If $Cx$ be so small then all terms $(Ct)^n$ will be negligible and we have this approximation
$$1+Ct\approx\dfrac{1}{1-Ct}$$
this approximation will be better if we use three first terms, i.e.
$$1+Ct+C^2t^2\approx\dfrac{1}{1-Ct}$$
A: It is related to the geometric series. In order to derive the formula start with
$$S_n = 1+x+x^2+...+x^n$$
multiply by $x$ to obtain
$$xS_n= x + x^2 +...+x^{n+1}.$$
Subtraction will result in (note that we have a lot of chancellations):
$$(1-x)S_n=1-x^{n+1} \implies S_n = \frac{1-x^{n+1}}{1-x}=1+x+x^2+...+x^{n}$$
The series converges for $|x|<1$ to give:
$1+x+x^2+... = \frac{1}{1-x}$$
Now, use $x=-Cu$ and only use the first two terms on the left-hand side to obtain your approximation.
