Approximation of a ratio

Question

Is this approximation true? If so, why? $$\frac{1+x}{1+y}\approx 1+x -y$$

I think it has something to do with $x$ and $y$ being close to zero, so that the ratio of the two is approximately equal to zero, and therefore cancels out.

In advance thanks!

Answer 1

$$f(x,y) = \frac{1+x}{1+y}$$

Let's try to derive the 1st order Taylor expansion around $0$ of $f$;

$$f(x,y) \approx f(0,0) + f_x(0,0)x + f_y(0,0) y = 1 + x - y$$

where

$$f_x(x,y) = \frac{\partial f}{\partial x} = \frac{1}{1+y} \Rightarrow f_x(0,0) = 1$$

and

$$f_y(x,y) = \frac{\partial f}{\partial y} = -\frac{1+x}{(1+y)^2} \Rightarrow f_y(0,0) = -1$$

So, for $x$ and $y$ sufficiently small, this approximation is good.

Answer 2

You can check that $$(1+x-y)(1+y)=1+x-y+y+xy-y^2=1+x+xy+y^2\approx1+x.$$ The second order terms ($xy$ and $y^2$) are negligible for small $y$.

$x$ needn't be small.