Approximation of $(x+cx^2)/(1+cx^2)$, when c is small

Question

I'm reading a paper and can't wrap my head around the following approximation:

$f(x) = \frac{x+cx^2}{1+cx^2}$ $,$ $0 \le x \le 1$

Assuming that $c$ is small, $(c << 1)$, the following approximation is done

$f(x) \approx x + cx^2(1-x)$

So how is this approximation done and where do the -$cx^3$ term come from?

Answer 1

I think they'll be using the binomial theorem,

$$(1 + cx^2)^{-1} = 1 - cx^2$$

Then after ignoring the $c^2 x^4$ term, as $c$ is small, we get the result.

Answer 2

Notice, given $$f(x)=\frac{x+cx^2}{1+cx^2}=(x+cx^2)(1+cx^2)^{-1}$$ since $0\leq x\leq 1$ & by assuming $c<<1$ we get $cx^2<<1$

Hence, we can use the Binomial expansion of $(1+cx^2)^{-1}=1-cx^2$ using approximation by neglecting higher power terms,

$$f(x)\approx (x+cx^2)(1-cx^2)=x-cx^3+cx^2-c^2x^4$$ Neglecting higher power term $c^2x^4<<1$ we get $$f(x)\approx x-cx^3+cx^2$$ $$\color{red}{f(x)\approx x+cx^2(1-x)}$$