Epsilon-Delta Proofs: Quadratic Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 2.5: “The Precise Definition of a Limit” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First: The Extra Factor
For the linear limit $\lim_{x \to 2} 3x = 6$, the scratch work gave $|3x - 6| = 3|x - 2|$ -- a single factor of $|x - 2|$.
For $\lim_{x \to 2} x^2 = 4$, the scratch work starts:
$|x^2 - 4| = |x-2||x+2|$.
Predict: now there are two factors. How does this change the problem? What is the challenge with $|x + 2|$?
The problem is that $|x + 2|$ depends on $x$, which is exactly what we are controlling. We need to bound it in terms of $\delta$ before we can choose $\delta$.
Quantity-First Framing
For linear functions, the error $|f(x) - L|$ simplifies to $|m||x - a|$: a constant times the input error. For quadratic functions, the error contains a factor like $|x + a|$ that involves $x$ itself -- a moving target. The key technique: restrict $|x - a| \leq 1$ (a convenient bound) to make $|x + a|$ a computable constant, then choose $\delta$ as the minimum of 1 and $\varepsilon/\text{(constant)}$.
Prerequisite Check
Quick Reference
Key technique for quadratic limits. To prove $\lim_{x \to a} x^2 = a^2$:
- Write $|x^2 - a^2| = |x - a||x + a|$.
- Assume $|x - a| < 1$ (tentatively). This bounds $|x + a|$: since $|x - a| < 1$ gives $a - 1 < x < a + 1$, hence $|x + a| < |a| + a + 1$ (or compute exactly).
- Choose $\delta = \min\!\left(1,\, \dfrac{\varepsilon}{|x+a|\text{-bound}}\right)$.
Key Concepts
1. The Scratch Work for $x^2$
Example 1. Prove $\lim_{x \to 2} x^2 = 4$.
Scratch. Need $|x^2 - 4| < \varepsilon$.
Factor: $|x^2 - 4| = |x-2||x+2|$.
The factor $|x+2|$ contains $x$. Bound it: assume $|x-2| < 1$, so $1 < x < 3$. Then $3 < x + 2 < 5$, giving $|x+2| < 5$.
Now: $|x^2 - 4| = |x-2||x+2| < 5|x-2|$.
Need $5|x - 2| < \varepsilon$: choose $|x-2| < \varepsilon/5$.
Combine: choose $\delta = \min(1, \varepsilon/5)$. This ensures:
- $|x + 2| < 5$ (from $\delta \leq 1$), and
- $5|x-2| < \varepsilon$ (from $\delta \leq \varepsilon/5$).
Formal proof.
Given $\varepsilon > 0$. Let $\delta = \min(1, \varepsilon/5)$.
Suppose $0 < |x - 2| < \delta$.
Since $\delta \leq 1$: $|x - 2| < 1$, so $1 < x < 3$, giving $3 < x + 2 < 5$, hence $|x+2| < 5$.
Therefore: \[ |x^2 - 4| = |x-2||x+2| < \delta \cdot 5 \leq \frac{\varepsilon}{5} \cdot 5 = \varepsilon. \quad \square \]
2. Why $\min(1, \varepsilon/5)$?
The minimum ensures BOTH conditions are satisfied simultaneously:
- $\delta \leq 1$ guarantees $|x + 2| \leq 5$.
- $\delta \leq \varepsilon/5$ guarantees $5|x-2| < \varepsilon$.
If we only chose $\delta = \varepsilon/5$ without the “$\leq 1$” constraint, the bound $|x+2| \leq 5$ might fail (if $\varepsilon$ is large and $\delta = \varepsilon/5 > 1$). The minimum handles both requirements at once.
3. Two Representations
Geometric picture. On the number line, the condition $|x - 2| < \delta$ puts $x$ in the interval $(2 - \delta, 2 + \delta)$. The additional constraint $\delta \leq 1$ keeps $x$ in $(1, 3)$, where $|x + 2|$ is bounded by 5.
Table. For $\lim_{x \to 2} x^2 = 4$:
| $\varepsilon$ | $\delta = \min(1, \varepsilon/5)$ | Works? |
|---|---|---|
| $0.5$ | $\min(1, 0.1) = 0.1$ | Yes: if $|x-2| < 0.1$ then $|x^2-4| < 5(0.1) = 0.5$ |
| $10$ | $\min(1, 2) = 1$ | Yes: if $|x-2| < 1$ then $|x+2| < 5$, $|x^2-4| < 5(1) = 5 < 10$ |
4. General Quadratic
For $\lim_{x \to a} x^2 = a^2$:
$|x^2 - a^2| = |x-a||x+a|$.
Assume $|x - a| < 1$: then $|x| < |a| + 1$, giving $|x + a| \leq |x| + |a| < 2|a| + 1$.
Choose $\delta = \min(1, \varepsilon/(2|a|+1))$.
5. Ask Why: Why the Bound on $|x - a| < 1$ Specifically?
The choice of 1 is conventional and convenient, not essential. Any positive constant $M$ would work in its place:
Assume $|x - a| < M$. Then $|x + a| \leq |x - a| + 2|a| < M + 2|a|$.
Choose $\delta = \min(M, \varepsilon/(M + 2|a|))$.
With $M = 1$: $\delta = \min(1, \varepsilon/(1 + 2|a|))$, slightly different from before. The choice $M = 1$ is standard because it is simple.
Named Misconception: limit-as-unreachable-barrier
The $\min$ construction sometimes triggers the misreading: “we’re assuming $|x - a| < 1$ as a constraint, so the limit only holds within a 1-unit interval.”
The limit holds for ALL $x$ near $a$, not just within 1 unit. The constraint $|x - a| < 1$ appears only in the proof technique to control the moving factor -- it is a device for finding $\delta$, not a restriction on the limit itself. Once we find $\delta = \min(1, \varepsilon/5)$, the proof works for all $x$ with $|x - 2| < \delta$ (which is automatically within 1 unit).
Common Errors
| Error | Specific example | Correction |
|---|---|---|
| Forgetting the $\min$ | Choosing $\delta = \varepsilon/5$ without the $\leq 1$ bound | Must bound $|x+2|$ first; without $\delta \leq 1$, the bound fails for large $\varepsilon$ |
| Loose bound on $|x+a|$ | Writing $|x+2| < \infty$ | Compute the actual bound: on $(1, 3)$, $|x+2| < 5$ exactly |
| Circular reasoning | Using $|x^2 - 4| < \varepsilon$ to conclude $|x - 2| < \delta$ | The proof assumes $|x-2| < \delta$ (the input condition) and derives $|x^2-4| < \varepsilon$ (the output conclusion); not the reverse |
Leveled Practice
Level 1 -- Finding $\delta$
Problem 1. For $\lim_{x \to 3} x^2 = 9$: carry out the scratch work and find $\delta$ in terms of $\varepsilon$.
Show answer
$|x^2 - 9| = |x-3||x+3|$. Assume $|x-3| < 1$: then $2 < x < 4$, so $5 < x+3 < 7$, giving $|x+3| < 7$.
$\delta = \min(1, \varepsilon/7)$.
Problem 2. Prove $\lim_{x \to 1} x^2 = 1$.
Show answer
$|x^2 - 1| = |x-1||x+1|$. Assume $|x-1| < 1$: $0 < x < 2$, so $|x+1| < 3$.
$\delta = \min(1, \varepsilon/3)$.
If $0 < |x-1| < \delta$: $|x^2-1| = |x-1||x+1| < \delta \cdot 3 \leq \varepsilon$. $\square$
Level 2 -- Polynomials
Problem 3. Prove $\lim_{x \to 0} x^2 = 0$.
Show answer
$|x^2 - 0| = x^2 = |x|^2 = |x| \cdot |x|$. If $|x| < 1$: $|x+0| = |x| < 1$.
$|x^2| = |x||x| < 1 \cdot |x| = |x|$. So need $|x| < \varepsilon$: choose $\delta = \min(1, \varepsilon)$.
Alternatively: $|x^2| = |x|^2 < \delta^2$; choose $\delta = \min(1, \sqrt{\varepsilon})$ so $\delta^2 \leq \varepsilon$.
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 4 (Extension).
(a) (Floor) Verify that for $\varepsilon = 0.25$ and $a = 2$, the choice $\delta = \min(1, 0.25/5) = 0.05$ guarantees $|x^2 - 4| < 0.25$ for all $x$ with $|x-2| < 0.05$.
(b) (Mid) Adapt the proof to prove $\lim_{x \to a}(x^2 + c) = a^2 + c$ for a constant $c$.
(c) (Ceiling) Prove $\lim_{x \to a} x^3 = a^3$. Factor $x^3 - a^3 = (x-a)(x^2 + ax + a^2)$ and bound $|x^2 + ax + a^2|$ when $|x - a| < 1$.
Show answer
(a) If $|x-2| < 0.05$: then $|x+2| \leq |x-2| + 4 < 5$. $|x^2-4| = |x-2||x+2| < 0.05 \cdot 5 = 0.25$. Confirmed.
(b) $|(x^2+c)-(a^2+c)| = |x^2-a^2|$. Same proof as before: $\delta = \min(1, \varepsilon/(2|a|+1))$.
(c) $|x^3 - a^3| = |x-a||x^2+ax+a^2|$. Assume $|x-a| < 1$: $|x| < |a|+1$.
$|x^2+ax+a^2| \leq x^2 + |a||x| + a^2 < (|a|+1)^2 + |a|(|a|+1) + a^2$.
Let $M = (|a|+1)^2 + |a|(|a|+1) + a^2$. Choose $\delta = \min(1, \varepsilon/M)$. Proof follows the quadratic pattern.
Common Misconceptions
the bound on $|x + a|$ can be skipped because it “obviously” stays finite. The key step in the quadratic proof is showing that $|x + a|$ is bounded by a constant $K$ whenever $|x - a| < 1$. Students sometimes omit this step on the grounds that $x$ is “near $a$” so both are finite. But the proof requires an explicit numerical bound to construct $\delta = \min\{1, \varepsilon/K\}$. Without a specific $K$, the choice of $\delta$ cannot be written down, and the proof is incomplete.
the restriction $|x - a| < 1$ is part of the problem’s hypothesis. The restriction $|x - a| < 1$ is introduced by the prover, not given in the problem. It is a strategic choice to obtain a bound on the auxiliary factor $|x + a|$. Any positive bound (0.5, 2, or any other number) would also work and would lead to a different but equally valid $K$. The choice of 1 is convenient, not required.
Mastery Checklist
Mental Model
The quadratic proof introduces a two-step difficulty: the error $|x^2 - a^2|$ has two factors. The first factor ($|x-a|$) is directly controlled by $\delta$. The second factor ($|x+a|$) moves with $x$. The solution is to pre-restrict $x$ to a neighborhood (say, within 1 of $a$), bound the second factor by a constant in that neighborhood, and then choose $\delta$ small enough to make the whole product less than $\varepsilon$.
The $\min$ is the key tool: it enforces both requirements simultaneously with a single choice of $\delta$.
Connections
Within MATH161
- Polynomial proofs: The same technique extends to $x^n$ for any $n$: factor, bound the non-linear factor, use $\min$.
- Continuity: The epsilon-delta proof of $\lim_{x \to a} x^2 = a^2$ shows that $f(x) = x^2$ is continuous at every $a$ -- the limit equals the function value.
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