Epsilon-delta
A skeptic keeps tightening the output tolerance $\varepsilon$ around $L = 1$. What must happen to your input window $\delta$ for the claim $\lim_{x \to 1} x^2 = 1$ to survive?
Every challenge gets an answer. At $\varepsilon = 0.5$ the instrument responds $\delta \approx 0.225$; at $\varepsilon = 0.2$, $\delta \approx 0.095$; at $\varepsilon = 0.05$, $\delta \approx 0.025$. The vertical band narrows every round, but its width never reaches zero.
So $\delta$ shrinks — and the rules allow that. The skeptic moves first, naming a tolerance $\varepsilon$ around $L = 1$. You answer with a window $\delta$ around $a = 1$ such that every $x$ within $\delta$ of 1 (except $x = 1$ itself) sends $f(x) = x^2$ within $\varepsilon$ of 1. The instrument plays the largest safe answer, $\delta = \min(\sqrt{1+\varepsilon} - 1,\ 1 - \sqrt{1-\varepsilon})$: the min, because the window must hold on both sides of $a$.
One won round proves nothing. The limit claim is that you can win every round — for each $\varepsilon > 0$, some $\delta > 0$ exists. That is the definition:
In the $\varepsilon$–$\delta$ game, who moves first — and what counts as proving $\lim_{x \to a} f(x) = L$?
The skeptic plays $\varepsilon = 0.5$. The largest workable window is $\delta = \min(\sqrt{1.5} - 1,\ 1 - \sqrt{0.5})$. Roughly what value does the instrument report?
Transfer: now prove $\lim_{x \to 2} 3x = 6$. The skeptic plays an arbitrary $\varepsilon > 0$. What is the largest $\delta$ you can answer with?
A limit is a winnable game, not one measurement. $\delta$ may depend on $\varepsilon$ and may shrink without bound; the claim $\lim_{x \to a} f(x) = L$ holds exactly when every tolerance $\varepsilon > 0$ has some working window $\delta > 0$.