How to show that this limit is identical to.. Suppose we have expressions
$$f_1=\frac{x(y-1)}{x(2y-1)-y}$$
and
$$f_2=\frac{xw/y+(1-x)w/(1-y)+4}{8}-\sqrt{\left(\frac{xw/y+(1-x)w/(1-y)+4}{8}\right)^2-1/4(1+xw/y)}.$$
Moreover, we can assume $w>0$, $x,y\in[0,1]$ and $x\ge y$. As you can probably guess, $f_2$ is a solution to a quadratic equation. 
In numerical calculations I almost accidentally discovered that 
$$\lim_{w\to\infty} f_2\to f_1,$$
which is not at all obvious from the expressions. Is that true - what's the limit of $f_2$ and how to calculate it? 
 A: Your expression is of the form
$$a-\sqrt{a^2-b}$$
which, when $a$ is "much" larger than $b$, is approximately $b/(2 a)$.  In your case,
$$a=\frac18 \left (\left [ \frac{x}{y} + \frac{1-x}{1-y}\right] w + 4\right )$$
$$b = \frac14 \left ( 1+ \frac{x}{y} w\right )$$
Now, as $w \to \infty$:
$$\frac{b}{2 a} = \frac{ 1+ \frac{x}{y} w}{\left [ \frac{x}{y} + \frac{1-x}{1-y}\right] w + 4} \sim \frac{\frac{x}{y}}{\frac{1-x}{1-y}+\frac{x}{y}}$$
which, when simplified, produces the sought-after expression.  
What I didn't answer is, when is this true, and it should be valid for all $x,y \in [0,1]$ because $a$ remains much larger than $b$ as $w \to \infty$.
ADDENDUM
I made the assertion
$$a-\sqrt{a^2-b} \sim \frac{b}{2 a}$$
This comes from the expression for the Taylor series for the function
$$\sqrt{1+z} = 1+\frac12 z + O\left(z^2\right)$$
where the $O$ term is an error term for small $z$.  Then
$$a-\sqrt{a^2-b} = a-a \sqrt{1-\frac{b}{a^2}} \sim a - a \left (1 - \frac{b}{2 a^2}\right ) = \frac{b}{2 a}$$
