f is continuous at a when for every ε > 0 some δ > 0 gives |f(x) − f(a)| < ε whenever |x − a| < δ; for x² the classic choice is δ = min(1, ε/(2|a| + 1)).
Continuity at a point is the ε–δ limit with the hole filled in: the limit exists and equals the value. Writing the proofs teaches one trick you will use constantly, the "min with $1$," which restricts $x$ before bounding the rest of the expression.
🔑 Key idea
$f$ is continuous at $a$ when for every $\varepsilon > 0$ there is a $\delta > 0$ with $|f(x) - f(a)| < \varepsilon$ whenever $|x - a| < \delta$. For $f(x) = x^{2}$: $|x^{2} - a^{2}| = |x - a|\,|x + a|$; if $|x - a| < 1$ then $|x + a| < 2|a| + 1$, so $\delta = \min\left(1, \frac{\varepsilon}{2|a| + 1}\right)$ works.
💡 Why it works
The factor $|x - a|$ will be small; the trouble is the other factor $|x + a|$, which depends on $x$. Restricting $x$ to within $1$ of $a$ caps it: $$|x + a| = |(x - a) + 2a| \le |x - a| + 2|a| < 1 + 2|a|.$$ Now both conditions together, $|x - a| < 1$ and $|x - a| < \frac{\varepsilon}{2|a| + 1}$, give $$|x^{2} - a^{2}| < (2|a| + 1)\cdot\frac{\varepsilon}{2|a| + 1} = \varepsilon.$$ Taking $\delta$ to be the smaller of the two numbers guarantees both. The $1$ is arbitrary; any fixed positive number would do, with a different bound.
Proof, written out. Let $\varepsilon > 0$ and $\delta = \min\left(1, \frac{\varepsilon}{2|a| + 1}\right)$. Suppose $|x - a| < \delta$. Then $|x - a| < 1$, so $|x + a| < 2|a| + 1$, and $|x^{2} - a^{2}| = |x - a||x + a| < \delta(2|a| + 1) \le \varepsilon$. $\blacksquare$
Continuity of $x^{2}$ at $a = 3$: set ε, then compare the recipe $\delta = \min\left(1, \frac{\varepsilon}{7}\right)$ with the largest δ the picture finds.
Worked examples
✏️ Worked example: Running the recipe
Use $\delta = \min\left(1, \frac{\varepsilon}{2|a| + 1}\right)$ for $x^{2}$ at $a = 3$ with $\varepsilon = 0.7$.
$2|a| + 1 = 7$ and $\frac{0.7}{7} = 0.1$.
$\delta = \min(1, 0.1) = 0.1$.
Answer: $\delta = 0.1$. (The largest δ is $\sqrt{9.7} - 3 \approx 0.1145$; the recipe is safe, not sharp.)
✏️ Worked example: A different quadratic
For $f(x) = x^{2} - 4x$ at $a = 3$, write $|f(x) - f(3)|$ as a product and bound the second factor when $|x - 3| < 1$.
$f(x) - f(3) = x^{2} - 4x + 3 = (x - 3)(x - 1)$.
If $2 < x < 4$, then $1 < x - 1 < 3$, so $|x - 1| < 3$.
Forgetting the restriction. "$\delta = \frac{\varepsilon}{5}$ works for $x^{2}$ at $2$" fails for large $\varepsilon$: with $\varepsilon = 100$ it allows $x = 21$, where $|x^{2} - 4| = 437$. The bound $|x + 2| < 5$ was only true for $|x - 2| < 1$, so $\delta$ must also be at most $1$. And never let δ depend on $x$; it is chosen once, from $\varepsilon$ and $a$.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.