Convert $x/y \cdot \frac{1-\delta_1-\delta_3}{(1-\delta_2)}$ to $x/y \cdot (1-\delta)$ using calculus

Question

Want expression in the form $x/y \cdot (1-\delta)$

• $\delta_3 \delta_1$ and anything $\delta^2$ can be excluded since they are too small

• $=x/y \cdot \frac{(1-\delta_1)}{(1-\delta_2)}(1-\delta_3)\\=x/y \cdot \frac{(\delta_3 \delta_1 - \delta_1 - \delta_3 + 1)}{(1-\delta_2)}\\\approx x/y \cdot \frac{1-\delta_1-\delta_3}{(1-\delta_2)}$

As you can see its not in the correct form I'm not sure how to simplify the denominator in this case.

Here is an example with multiplication/addition from my textbook

Answer 1

Suppose $\max_i(|\delta_i|)\leqslant\epsilon$: $$\frac{1-\delta_1-\delta_3}{1-\delta_2} \leqslant 1-\delta \iff 1-\delta_1-\delta_3 \leqslant (1-\delta_2)(1-\delta) = 1-\delta_2-\delta+\delta_2\delta \iff$$ $$\iff \delta \leqslant \delta_1-\delta_2+\delta_3 \iff |\delta| \leqslant |\delta_1-\delta_2+\delta_3| \leqslant |\delta_1|+|\delta_2|+|\delta_3| \leqslant 3\epsilon$$ So if you set $|\delta| \leqslant 3\epsilon$, you get: $$\frac{x}{y} . \frac{1-\delta_1-\delta_3}{1-\delta_2} \leqslant x/y.(1-\delta)$$