Doesn’t matter eventually it’s the same

I have two glasses that can hold a maximum volume of 24 fluid ounces. Initially, one glass contains precisely 12 fluid ounces of coffee, while the other contains precisely 12 fluid ounces of tea.

Your goal is to dilute the amount of coffee in the “coffee cup” by performing the following steps:

• Pour some volume of tea into the coffee cup.
• Thoroughly mix the contents.
• Pour that same volume out of the coffee cup, so that precisely 12 fluid ounces of liquid remain.

After doing this as many times as you like, in the end, you will have 12 ounces of liquid in the coffee cup, some of which is coffee and some of which is tea. In fluid ounces, what is the least amount of coffee you can have in this cup?

Let $C_k(x)$ be the ounces of coffee in the coffee cup after $k$ successive iterations of moving $x$ ounces back and forth for some $x\in (0, 12],$ since there's literally no action when $x=0.$ Here we have $C_0(x)=12$ for any value if $x \in (0,12].$ Since after each iteration we have the total volume in the coffee cup as 12, and extrapolating from the worked in the Classic problem where $x=1$, we will always have that the amount of tea in the coffee cup is equal to the amount of coffee in the tea cup, so we see that the amount of coffee in the tea cup at the end of $k$ successive iterations is $12-C_k(x).$

When $x$ ounces is transferred from the coffee cup to the tea cup, a total of $x\frac{C_k(x)}{12}$ ounces of coffee is trabsferred to the tea cup. At this point there is a total of $12-C_k(x) + x\frac{C_k(x)}{12}$ ounces of coffee in the tea cup, out of a total of $12+x$ ounces of liquid. After mixing, when the $x$ ounces of liquid is transferred back to the coffee cup, a total of $x\frac{12-(1-x/12)C_k(x)}{12+x}$ ounces of coffee is transferred. Using this information, we can get the recursive formula $$C_{k+1} (x) = C_k(x) - \frac{x}{12}C_k(x) + \frac{x}{x+12} \left( 12 - \left(1-\frac{x}{12}\right) C_k(x) \right) = \frac{12-x}{12+x} C_k(x) + \frac{12x}{12+x}.$$

In general a linear sequence $a_{n+1}=ra_n+b$, for $n\in \mathbb{N},$ will converge to some number $a^*=\frac{b}{1-r}$, whenever $|r|\lt 1.$ Here we have r=r(x)=\frac{12-x}{12+x} \in [0,1) for x \in (0,12], so we should converge for any value of $x$ to $$C^*=C^*(x) = \frac{\frac{12x}{12+x}}{1-\frac{12-x}{12+x}} = \frac{\frac{12x}{12+x}}{\frac{2x}{12+x}}=6.$$

In addition, we can rearrange the recursive formula to show that $C_{k+1}(x)$ is always a convex combination of $C_k(x)$ and $6,$ so in particular we see that $C_k(x)$ is a nonincreasing sequence for each $x \in (0,12].$ So we must have that the minimal amount of coffee in the coffee cup after many, many iterations is $\inf_k C_k(x) = C^* = 6$ ounces. Now for the strict constructivists amongst you, we need not appeal to countably infinitely many mixings or ignore infinitissimal quantities obtain optimality. Instead, if we chose $x=12$ ounces, then after a single iteration, voila, we have the perfect 6 oz coffee / 6 oz tea concoction.