Differentiability Implies Continuity

Overview

One of the first structural theorems relating the two central notions of single-variable calculus is that differentiability is a strictly stronger property than continuity: if a function f is differentiable at a point a, then f must also be continuous at a. The proof is a short, purely algebraic manipulation of the difference quotient, but the result has real teeth — its converse is dramatically false. Continuous functions need not be differentiable at a point (the corner of |x| at 0 is the elementary example), and in one of analysis's most startling constructions, Weierstrass exhibited a function that is continuous everywhere on ℝ yet differentiable nowhere, showing just how much weaker continuity is than differentiability.

Intuition

Differentiability at a point means the function has a well-defined instantaneous rate of change there — zoomed in closely enough, the graph looks like a single straight line (the tangent line) through that point. If the graph looked like a straight line locally, it certainly could not have a jump or a hole there: a jump would mean the 'zoomed in' picture still shows two disconnected pieces, which can never resemble a single line no matter how far you zoom. So having a well-defined tangent line is a strong local straightness requirement, and mere continuity (no jump or hole) is a much weaker requirement — the graph could still have a sharp corner (like |x| at the origin), where it's connected with no jump but the 'zoomed in' picture never settles into one single line because the left and right slopes disagree.

Formal Definition

Definition

f is differentiable at a if the following limit exists (as a finite real number):

f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}

Derivative at a point

f \text{ differentiable at } a \implies f \text{ continuous at } a

Theorem (this concept)

f \text{ continuous at } a \ \not\!\!\implies\ f \text{ differentiable at } a

Continuity is strictly weaker; e.g. f(x)=|x| is continuous but not differentiable at x=0

Converse fails

Notation

Notation	Meaning
$f'(a)$	The derivative of f at a — the limit of the difference quotient, when it exists
$\frac{f(x)-f(a)}{x-a}$	The slope of the secant line through (a,f(a)) and (x,f(x)); its limit as x→a is the derivative
$f \in C^0,\ f \in C^1$	f is continuous (C⁰) or continuously differentiable (C¹); C¹⊊C⁰ reflects that differentiability (with a continuous derivative) is a strictly stronger condition

Derivation

The key algebraic trick: rewrite f(x) - f(a) as the difference quotient (which has a finite limit, by hypothesis) multiplied by (x-a) (which → 0), so their product → 0 by the product rule for limits.

f(x) - f(a) = \frac{f(x)-f(a)}{x-a} \cdot (x - a) \quad \text{for } x \neq a

Purely algebraic identity — multiply and divide by (x-a)

\lim_{x\to a} \frac{f(x)-f(a)}{x-a} = f'(a) \quad \text{(exists, finite, by hypothesis)}

The differentiability hypothesis

\lim_{x\to a}(x-a) = 0

Trivial limit

\lim_{x\to a}\big[f(x)-f(a)\big] = \lim_{x\to a}\left[\frac{f(x)-f(a)}{x-a}\right]\cdot \lim_{x\to a}(x-a) = f'(a)\cdot 0 = 0

Product rule for limits (both factors' limits exist, so the limit of the product is the product of the limits)

\implies \lim_{x\to a} f(x) = f(a)

Rearranging gives exactly the definition of continuity of f at a

Proofs

Differentiability at a implies continuity at a

$\text{Suppose } f'(a) = \lim_{x\to a} \frac{f(x)-f(a)}{x-a} \text{ exists (finite).}$ (Hypothesis: f is differentiable at a)
$\text{For } x \neq a: \quad f(x) - f(a) = \frac{f(x)-f(a)}{x-a}(x-a).$ (Algebraic identity, valid since x≠a lets us multiply/divide by (x-a))
$\lim_{x\to a}\left[\frac{f(x)-f(a)}{x-a}(x-a)\right] = f'(a) \cdot 0 = 0$ (Product of limits: the first factor's limit is f'(a) by hypothesis, the second factor's limit is 0 trivially)
$\therefore \lim_{x\to a} \big[f(x)-f(a)\big] = 0 \implies \lim_{x\to a} f(x) = f(a).$ (This is exactly the ε-δ definition of continuity of f at a)

f(x) = |x| is continuous at 0 but not differentiable at 0

$\lim_{x\to 0} |x| = 0 = |0|,$ (Continuity of |x| at 0 is immediate: as x→0 from either side, |x|→0)
$\text{Right-hand difference quotient: } \lim_{x\to 0^+} \frac{|x|-0}{x-0} = \lim_{x\to 0^+} \frac{x}{x} = 1.$ (For x>0, |x|=x)
$\text{Left-hand difference quotient: } \lim_{x\to 0^-} \frac{|x|-0}{x-0} = \lim_{x\to 0^-} \frac{-x}{x} = -1.$ (For x<0, |x|=-x)
$\text{Since } 1 \neq -1, \lim_{x\to 0}\frac{|x|-0}{x-0} \text{ does not exist.}$ (The two one-sided limits disagree, so the two-sided limit (the derivative) fails to exist)
$\therefore |x| \text{ is continuous at 0 but not differentiable at 0.}$ (Confirms continuity does not imply differentiability — the converse of this concept's theorem is false)

Properties

\text{If } f'(a) \text{ exists, then } \lim_{x\to a} f(x) = f(a).

Condition: The converse is false; continuity is necessary but not sufficient for differentiability.

Existence of a corner blocks differentiability, not continuity

f(x)=|x| \text{ is continuous at every } x, \text{ but not differentiable at } x=0.

Example: The one-sided derivatives at 0 are +1 (from the right) and -1 (from the left), which disagree.

Weierstrass function

\text{There exist functions continuous everywhere on } \mathbb{R} \text{ but differentiable nowhere.}

Condition: Constructed via an infinite series of rescaled cosine waves; shows the gap between continuity and differentiability can be maximal, not just a finite set of 'bad' points.

Local linearity (equivalent characterization)

f \text{ is differentiable at } a \iff f(a+h) = f(a) + f'(a)h + o(h) \text{ as } h \to 0

Condition: The o(h) (little-o) error term vanishing faster than h itself is what forces continuity: as h→0, f(a+h)-f(a) = f'(a)h + o(h) → 0.

Applications

Physical trajectories with sharp corners (e.g. an idealized bouncing ball's height function at the instant of impact) are continuous but not differentiable there, signaling a discontinuity in velocity — a genuine physical event, not a modeling artifact.

Worked Examples

Compute the difference quotient and its limit.
$\lim_{x\to 2} \frac{x^2 - 4}{x-2} = \lim_{x\to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x\to 2}(x+2) = 4$
Since f'(2)=4 exists (finite), the theorem guarantees continuity at 2.
$f'(2) = 4 \implies \lim_{x\to 2} f(x) = f(2) = 4$

Answer: f'(2) = 4, and hence f is continuous at x=2 (as directly confirmed: lim_{x→2} x² = 4 = f(2))

Practice Problems

Difficulty 4/10

If g is differentiable at x=5, what can you immediately conclude about lim_{x→5} g(x)?

Difficulty 5/10

Which statement correctly describes the logical relationship between continuity and differentiability at a point?

They are equivalent conditionsDifferentiability implies continuity, but not converselyContinuity implies differentiability, but not converselyNeither implies the other

Difficulty 7/10

Prove that f(x) = x^(1/3) (real cube root) is continuous at x=0 but not differentiable at x=0.

Common Mistakes

Common Mistake

Assuming continuity implies differentiability (reversing the theorem's direction).

The theorem only goes one way. f(x)=|x| is continuous everywhere but fails to be differentiable at x=0, where the left- and right-hand difference quotients disagree (-1 vs +1).

Common Mistake

Believing non-differentiability can only occur at isolated 'corner' points, so a continuous function is 'differentiable almost everywhere' in some naive sense.

The Weierstrass function is a continuous function on all of ℝ that is differentiable at NO point whatsoever — the failure of differentiability can be total, not just confined to a few corners.

Common Mistake

Thinking that if f'(a) fails to exist, f cannot be continuous at a either.

This inverts the actual implication. Non-differentiability at a says nothing about continuity at a — f could still be perfectly continuous there (as with |x| at 0); it's only the CONVERSE direction (continuity ⟹ differentiability) that fails, not this one.

Quiz

Flashcards

1 / 4

Summary

If f is differentiable at a, then f is continuous at a: differentiability is strictly stronger than continuity.
Proof: f(x)-f(a) = [(f(x)-f(a))/(x-a)]·(x-a) → f'(a)·0 = 0 as x→a, using the product rule for limits.
The converse fails: f(x)=|x| is continuous at x=0 but not differentiable there (one-sided derivatives -1 and +1 disagree).
The Weierstrass function is continuous everywhere on ℝ yet differentiable nowhere, showing the gap between continuity and differentiability can be total, not just isolated corners.
Contrapositive form is a useful shortcut: if f fails to be continuous at a, it automatically fails to be differentiable at a.

References

WebsiteWikipedia — Differentiable function
WebsiteWikipedia — Weierstrass function
BookRudin, W. Principles of Mathematical Analysis, 3rd ed. Ch. 5.