Differentiability Implies Continuity
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.2: “The Derivative as a Function” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Key idea
The theorem “differentiability implies continuity” is a one-way street. If you know $f$ is differentiable at $a$, you get continuity at $a$ for free. But continuity does not give you differentiability: $|x|$ is continuous at 0 but not differentiable there.
This matters in practice because many theorems (Mean Value Theorem, Rolle’s Theorem, integration theorems) require differentiability. Knowing the implication direction means you can use differentiability as a stronger hypothesis that covers continuity automatically.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
Theorem. If $f$ is differentiable at $a$, then $f$ is continuous at $a$.
Equivalently (contrapositive). If $f$ is not continuous at $a$, then $f$ is not differentiable at $a$.
The converse is FALSE. $f$ can be continuous at $a$ without being differentiable at $a$. (Example: $|x|$ at $x = 0$.)
Logical structure: \[ \text{differentiable at }a \implies \text{continuous at }a \] \[ \text{continuous at }a \centernot\implies \text{differentiable at }a \]
Key Concepts
1. The Proof
Theorem. If $f$ is differentiable at $a$, then $f$ is continuous at $a$.
Proof. We need to show $\lim_{x \to a} f(x) = f(a)$.
Write $f(x) - f(a)$ as a product: \[ f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a) \qquad (x \neq a). \]
Take the limit as $x \to a$: \[ \lim_{x \to a} [f(x) - f(a)] = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \cdot \lim_{x \to a}(x - a) = f'(a) \cdot 0 = 0. \]
The first factor has limit $f'(a)$ (this is the alternate form of the derivative, which exists because $f$ is differentiable). The second factor has limit 0.
So $\lim_{x \to a}[f(x) - f(a)] = 0$, which means $\lim_{x \to a} f(x) = f(a)$. This is exactly continuity at $a$. $\square$
The key algebraic move. Writing $f(x) - f(a)$ as $\dfrac{f(x)-f(a)}{x-a} \cdot (x-a)$ multiplies and divides by $(x-a)$, which is nonzero for $x \neq a$ (the only values the limit cares about). The first factor converges to $f'(a)$; the second converges to 0.
2. The Converse is False: $|x|$ at $x = 0$
$f(x) = |x|$ is continuous at $x = 0$: $\lim_{x \to 0} |x| = 0 = f(0)$. Continuous.
But the one-sided derivatives are $+1$ (from the right) and $-1$ (from the left), so $f$ is not differentiable at 0. Continuous but not differentiable.
Other examples of continuous-but-not-differentiable:
- $f(x) = x^{2/3}$ at $x = 0$: continuous (value is 0, limit is 0), but cusp -- not differentiable.
- $f(x) = |x^2 - 1|$ at $x = \pm 1$: continuous, with corners.
3. Using the Contrapositive
The contrapositive “not continuous implies not differentiable” is logically equivalent to the theorem and often easier to apply:
If $f$ has a jump, removable discontinuity, or vertical asymptote at $a$, then $f$ is not differentiable at $a$.
This shortcut saves you from computing the derivative limit when you already know the function is discontinuous.
Example 1. Let $f(x) = \dfrac{1}{x}$. At $x = 0$: $f$ is not defined (vertical asymptote). Therefore $f$ is not differentiable at $x = 0$.
Example 2. Let $g(x) = \lfloor x \rfloor$ (floor function). At every integer $n$: $g$ has a jump discontinuity, so $g$ is not differentiable at any integer.
4. Summary of the Logic
| At $a$ | Continuous? | Differentiable? |
|---|---|---|
| Polynomial (e.g., $x^2$) | Yes | Yes |
| $|x|$ | Yes | No (corner) |
| Jump discontinuity | No | No |
| Removable discontinuity | No (in strict definition) | No |
| Vertical asymptote | No | No |
“Yes” in the differentiability column always implies “Yes” in the continuity column. “Yes” in the continuity column does not guarantee “Yes” in the differentiability column.
if a function is continuous, it must be differentiable.
This is the concept-image-conflicts-definition error. The mental image many readers carry is “a smooth unbroken curve is differentiable,” and they extend this to conclude that any continuous function must be smooth. The function $f(x) = |x|$ breaks that image: it is continuous at every real number (its graph has no jumps or holes), but it fails to be differentiable at $x = 0$ because the graph has a corner there. The left-side slope approaching $x = 0$ is $-1$ and the right-side slope is $+1$; these disagree, so no single tangent slope exists. Continuity rules out jumps; differentiability additionally rules out corners. The concept of differentiability is strictly stronger than continuity.
discontinuity and non-existence of the limit are the same thing.
This is the continuity-vs-limit-existence error. Discontinuity at a point can happen in three distinct ways: the function is undefined there, the limit fails to exist there, or the limit exists but does not match the function value. Only the second case means the limit does not exist. For $f(x) = |x|$ at $x = 0$: the function IS defined ($f(0) = 0$), the limit EXISTS ($\lim_{x \to 0} |x| = 0$), and the function value equals the limit ($f(0) = 0$). So $f$ is actually continuous at $x = 0$. Its failure to be differentiable is a separate matter from continuity or limits.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Reversing the implication | “Since $f$ is continuous, it must be differentiable” | Only the forward direction holds; continuity does not imply differentiability |
| Forgetting the zero factor | Claiming $f'(a) \cdot 0 = f'(a)$ in the proof | Anything times zero is zero, provided the other factor is finite |
| Using the theorem backwards | “Since $f$ is differentiable at $a$, check if it’s continuous” | Differentiability gives continuity automatically; no separate check needed |
Leveled Practice
Level 1 -- Applying the Theorem
Problem 1. If $f$ is differentiable at every point in $(-3, 3)$, what can you conclude about the continuity of $f$ on $(-3, 3)$?
Show answer
By the theorem, $f$ is continuous at every point in $(-3, 3)$.
Problem 2. At $x = 5$, a function $g$ has a jump discontinuity. Can $g$ be differentiable at $x = 5$?
Show answer
No. By the contrapositive: if $g$ is not continuous at $x = 5$, then $g$ is not differentiable at $x = 5$.
Level 2 -- Identifying the Implication
Problem 3. For each pair, determine which implication (if any) holds.
(a) $f(x) = \cos x$ at $x = 0$: differentiable? continuous?
(b) $h(x) = |x - 1|$ at $x = 1$: differentiable? continuous?
(c) $r(x) = \frac{1}{x-3}$ at $x = 3$: differentiable? continuous?
Show answer
(a) $\cos x$ is differentiable everywhere (derivative $= -\sin x$), hence continuous everywhere. Differentiable AND continuous at $x = 0$.
(b) $|x-1|$ is continuous at $x = 1$ ($\lim_{x\to 1}|x-1| = 0 = h(1)$) but has a corner there (left slope $= -1$, right slope $= 1$). Continuous but NOT differentiable.
(c) $r(x) = 1/(x-3)$ is not defined at $x = 3$ (vertical asymptote). Neither continuous nor differentiable.
Level 3 -- Proof-Based
Problem 4. Complete the following argument: if $f$ is differentiable at $a$, why must $\lim_{x \to a} f(x)$ equal $f(a)$?
Show answer
Since $f$ is differentiable at $a$, the limit $f'(a) = \lim_{x \to a} \dfrac{f(x)-f(a)}{x-a}$ exists and is finite.
Write $f(x) - f(a) = \dfrac{f(x)-f(a)}{x-a} \cdot (x-a)$.
By the product rule for limits: \[ \lim_{x \to a}[f(x)-f(a)] = \lim_{x \to a}\frac{f(x)-f(a)}{x-a} \cdot \lim_{x \to a}(x-a) = f'(a) \cdot 0 = 0. \]
Therefore $\lim_{x \to a} f(x) = f(a)$: $f$ is continuous at $a$.
Problem 5. Give an example of a function $f$ that is continuous on all of $\mathbb{R}$ but differentiable nowhere.
Show answer
This is a deep example: the Weierstrass function is a classical construction of a function that is continuous everywhere but differentiable nowhere. Its graph has a “fractal” structure with corners at every point, so no tangent line exists anywhere.
For MATH161, the key takeaway is that such functions exist -- continuity and differentiability are genuinely different properties -- even though such extreme examples are not encountered in standard calculus.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
The implication runs downhill: differentiability is a stronger condition than continuity, so differentiability implies continuity, not the other way around.
Think of differentiability as requiring the curve to look like a straight line under sufficient magnification. Continuity is weaker: the curve just needs no jumps or holes. A curve can be free of jumps and holes (continuous) but still have corners or cusps (not differentiable). The converse direction -- asking that continuity imply differentiability -- would require no corners either, which is a strictly stronger demand.
Connections
Within Calculus I (MATH161)
- Rolle’s Theorem and the Mean Value Theorem: Both require $f$ to be continuous on $[a, b]$ and differentiable on $(a, b)$. The theorem here means the differentiability hypothesis automatically covers continuity on the interior.
- Chain Rule: The chain rule requires that the inner and outer functions be differentiable. The theorem guarantees they are also continuous, which is used in some chain-rule proofs.
- Critical points: At a corner (like $|x|$ at 0), the derivative does not exist, but the function is still continuous and may still have a local extremum. Such non-differentiable critical points must be included in optimization.
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