Single-Variable-Calculus · Unit 2 · Video 6 · Interactive Practice

Differentiable Implies Continuous: A One-Line Proof That Never Divides by Zero

IKey Formulas

Formula	Name	Role in the proof
$f'(x_0) = \lim\limits_{x \to x_0} \dfrac{f(x) - f(x_0)}{x - x_0}$	The hypothesis	This one limit is assumed to exist
$\lim\limits_{x \to x_0} f(x) = f(x_0) \iff \lim\limits_{x \to x_0}\big(f(x) - f(x_0)\big) = 0$	Continuity, restated	The statement that has to be checked
$f(x) - f(x_0) = \dfrac{f(x) - f(x_0)}{x - x_0} \cdot (x - x_0)$	Multiplying by one	Valid at every $x \neq x_0$
$\lim\limits_{x \to x_0}\big(f(x) - f(x_0)\big) = f'(x_0) \cdot 0 = 0$	The one-line proof	Differentiable $\implies$ continuous

Key Insight: The hypothesis is spent exactly once — on giving the first factor a limit. The second factor, $x - x_0 \to 0$ , needs nothing about $f$ at all.

IIVisualization 1 — The Product That Kills the Gap

The gap $f(x) - f(x_0)$ splits into a quotient heading for $f'(x_0)$ times a run heading for $0$ .

x − x₀ 2.000

IIIVisualization 2 — The Value the Limit Never Uses

However close $x$ comes to $x_0$ , the difference $x - x_0$ stays nonzero and the ratio stays exactly $1$ .

x − x₀ 0.50

💡 The same licence as $N = N \cdot \tfrac{10}{10} = N \cdot \tfrac{3}{3} = N \cdot \tfrac{1}{1}$ : a fraction with equal, nonzero numerator and denominator is the number $1$ in disguise.

IVVisualization 3 — Does the Arrow Reverse?

Differentiable forces continuous. The converse fails: a corner is continuous with no derivative, a jump has neither.

Function x − x₀ 1.00

💡 Read the theorem backwards: a function that is not continuous at $x_0$ cannot be differentiable there — the quickest way to rule a derivative out.

VQuiz Questions

Problem 1 · Continuity in the Form the Proof Checks

Given: continuity at $x_0$ says $\lim\limits_{x \to x_0} f(x) = f(x_0)$ . Subtracting the constant $f(x_0)$ from both sides and moving it inside the limit — which statement results?

✅ Correct! This is the target of the proof: the gap between

f(x)

and

f(x_0)

has limit

0

❌ That is a much stronger claim. Dividing by

x - x_0

turns the statement into

f'(x_0) = 0

— a claim about the slope, not about continuity.

❌ Not quite. Only the constant

f(x_0)

is subtracted from both sides; the right-hand side becomes

0

, and on the left the constant slides inside the limit.

Show solution

Start from the definition of continuity at $x_0$ and subtract $f(x_0)$ from both sides:

\lim_{x \to x_0} f(x) = f(x_0) \quad\Longrightarrow\quad \lim_{x \to x_0} f(x) - f(x_0) = 0

Since $f(x_0)$ is a constant, it can be taken inside the limit:

\lim_{x \to x_0}\big(f(x) - f(x_0)\big) = 0

This is exactly what the one-line proof will produce, so it is the form worth writing down. It is not the same as $f'(x_0) = 0$ : no division by $x - x_0$ has happened.

Problem 2 · The Factor That Looks Like Zero

Given: the proof rewrites $f(x) - f(x_0)$ as $\dfrac{f(x) - f(x_0)}{x - x_0} \cdot (x - x_0)$ , and at $x = x_0$ the factor $x - x_0$ is $0$ . What makes the rewrite legitimate?

✅ Correct!

x

comes closer and closer to

x_0

from either side without ever landing on it, so

x - x_0

is small but never zero.

❌ $f$ is defined at $x_0$ . The theorem's conclusion is a statement about

f(x_0)

, and the expression

f(x) - f(x_0)

uses that value. What is excluded is the input

x = x_0

, not the function's value there.

❌ Cancelling is the step that needs justifying.

\tfrac{a}{c} \cdot c = a

is only valid when

c \neq 0

— so this answer assumes exactly what has to be argued.

❌ Not quite.

\tfrac{0}{0}

is never assigned a value; the escape is that the quotient is never formed at

x = x_0

in the first place.

Show solution

The fraction $\dfrac{x - x_0}{x - x_0}$ equals $1$ for every $x \neq x_0$ , exactly as $\tfrac{10}{10}$ , $\tfrac{3}{3}$ and $\tfrac{1}{1}$ do. Multiplying by it changes nothing:

f(x) - f(x_0) = \big(f(x) - f(x_0)\big)\cdot\frac{x - x_0}{x - x_0} = \frac{f(x) - f(x_0)}{x - x_0}\cdot(x - x_0)

A limit as $x \to x_0$ is built from values of $x$ that approach $x_0$ from either side without ever equalling it. So throughout the calculation $x - x_0$ is a small nonzero number, and dividing by it is legal. The rule "never divide by zero" is never broken — it is never even tested.

Problem 3 · Run the Proof on a Concrete Function

Given: $f(x) = 4x^2 - 3x$ and $x_0 = 2$ , so $f(2) = 10$ . Evaluate the two limits in the proof's chain.

Step 1 — the difference quotient

Step 2 — the whole product

✅ Correct!

13 \cdot 0 = 0

, so

f

is continuous at

x_0 = 2

— and

13

f'(2)

❌ That is the objection, not the answer. The quotient is only

\tfrac{0}{0}

x = 2

, and the limit never evaluates there. Factor the numerator and cancel

x - 2

first.

❌ Check the first factor.

f(x) - f(2) = 4x^2 - 3x - 10 = (x - 2)(4x + 5)

, so the quotient simplifies before the limit is taken.

❌ Check the product. The second factor,

\lim\limits_{x \to 2}(x - 2)

, is

0

— and a finite number times

0

0

Show solution

Step 1 — the difference quotient. Factor the numerator, then cancel (legal, since $x \neq 2$ inside the limit):

\frac{f(x) - f(2)}{x - 2} = \frac{4x^2 - 3x - 10}{x - 2} = \frac{(x - 2)(4x + 5)}{x - 2} = 4x + 5

\lim_{x \to 2}\frac{f(x) - f(2)}{x - 2} = 4(2) + 5 = 13 = f'(2)

Step 2 — the product. Now take the limit factor by factor:

\lim_{x \to 2}\big(f(x) - f(2)\big) = \lim_{x \to 2}\left[\frac{f(x) - f(2)}{x - 2}\right]\cdot(x - 2) = 13 \cdot 0 = 0

Since $\lim\limits_{x \to 2}\big(f(x) - f(2)\big) = 0$ , the function is continuous at $x_0 = 2$ . Note where each number came from: $13$ needed the hypothesis (the derivative exists), while $0$ needed nothing about $f$ at all.

Problem 4 · Which Way Does the Arrow Point?

Given: $u(x) = |x - 4|$ and $v(x) = \begin{cases} x, & x \lt 4 \\ x + 2, & x \ge 4 \end{cases}$ , both examined at $x_0 = 4$ , where $u(4) = 0$ and $v(4) = 6$ . Classify each function.

Continuous at 4?

Differentiable at 4?

✅ Correct!

u

is continuous but has a corner,

v

jumps — so continuity does not imply differentiability, while the failure of continuity kills differentiability outright.

❌ Check $v$ at $x = 4$ . From the left

v(x) \to 4

, but

v(4) = 6

: the two numbers disagree, so

v

jumps there.

❌ Check the one-sided slopes of $u$ .

|x - 4|

has slope

-1

just left of

4

and

+1

just right of it, so the difference quotient has no limit — a corner, not a derivative.

❌ Check continuity value by value. Compare the limit of each function at

4

with its value there:

|x - 4| \to 0 = u(4)

, while

v(x) \to 4

x \to 4^{-}

and

v(4) = 6

❌ Use the theorem in reverse. Whatever is not continuous at

4

cannot be differentiable there — and a corner defeats the remaining candidate.

Show solution

$u(x) = |x - 4|$ . As $x \to 4$ , $|x - 4| \to 0 = u(4)$ , so $u$ is continuous at $4$ . Its difference quotient is

\frac{u(x) - u(4)}{x - 4} = \frac{|x - 4|}{x - 4} = \begin{cases} -1, & x \lt 4 \\ +1, & x \gt 4 \end{cases}

The one-sided limits are $-1$ and $+1$ , so the limit does not exist: $u$ is not differentiable at $4$ . Continuity does not imply differentiability.

$v$ . From the left $v(x) \to 4$ ; but $v(4) = 6$ . The limit and the value disagree, so $v$ is not continuous at $4$ . By the contrapositive of today's theorem it cannot be differentiable there either — and indeed $v(x) - v(4) \to -2 \neq 0$ from the left, so the quotient $\frac{v(x) - v(4)}{x - 4}$ blows up.

The theorem is a one-way street: differentiable $\implies$ continuous, never the reverse.

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