Stage review: for every ε there exists δ

Lesson 2.1.1 said that $\lim_{x \to a} f(x) = L$ means $f(x)$ gets as close to $L$ as you like once $x$ is close enough to $a$. This stage review turns that sentence into quantifiers and proves things with it. Take $f(x) = 3x - 1$ near $x = 2$, where $f(2) = 5$.

"As close as you like" is a tolerance $\varepsilon > 0$ (epsilon), and "close enough" is a distance $\delta > 0$ (delta). For each $\varepsilon$ the cell tries two choices of $\delta$. For each choice it finds the largest gap $|f(x) - 5|$ over 999 points strictly within $\delta$ of 2.

With $\delta = \varepsilon$ the largest gap is about $3\varepsilon$: 2.994 for $\varepsilon = 1$. With $\delta = \varepsilon / 3$ it stays below $\varepsilon$ every time: 0.998 for $\varepsilon = 1$. So a good $\delta$ exists for each $\varepsilon$ tried, and it depends on $\varepsilon$.

Now a function where the game is lost. The step function $H$ gives $H(x) = 0$ for $x < 0$ and $H(x) = 1$ for $x \ge 0$, so $H(0) = 1$. Near 0, does $H(x)$ stay close to $H(0)$? For each $\delta$ the cell tries the point $x = -\delta / 2$, halfway to the left edge.

Every $\delta$ tried has a point within $\delta$ of 0 whose gap is 1. So none of them meets a tolerance such as $\varepsilon = \tfrac{1}{2}$, and making $\delta$ smaller does not help.