Prove Differentiability Implies Continuity

The Difference Quotient Connects Differentiability to Continuity

The previous tutorial established uniform continuity on a compact set by coordinating pointwise estimates across the whole domain. Here the question is local: what does differentiability at one point tell us about the function at that same point? The defining limit for the derivative gives exactly the control needed to prove continuity there.

The key is to express the change in function values as the product of the input change and the difference quotient. Near the point, differentiability makes that quotient bounded; the input change tends to zero. This argument is pointwise and does not require compactness.

Definition: Let \(E\subseteq\mathbb R\), let \(a\in E\) be an accumulation point of \(E\), and let \(f:E\to\mathbb R\). The function \(f\) is differentiable at \(a\) if the limit \(\lim_{x\to a,\ x\in E,\ x\neq a}\frac{f(x)-f(a)}{x-a}\) exists as a finite real number. Its value is denoted by \(f'(a)\). The function \(f\) is continuous at \(a\) relative to \(E\) if, for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(x\in E\) and \(|x-a|<\delta\) imply \(|f(x)-f(a)|<\varepsilon\).

The accumulation-point condition ensures that the difference quotient is tested at domain points arbitrarily close to \(a\), other than \(a\) itself. If \(a\) is isolated in \(E\), continuity at \(a\) holds automatically, but this limit definition of a derivative has no nearby points on which to impose a condition.

The Main Implication

Theorem (Differentiability Implies Continuity): Let \(E\subseteq\mathbb R\), let \(a\in E\) be an accumulation point of \(E\), and let \(f:E\to\mathbb R\) be differentiable at \(a\). Then \(f\) is continuous at \(a\) relative to \(E\).

Proof. Write \(L=f'(a)\), and let \(\varepsilon>0\). Since the difference quotient tends to the finite number \(L\), there is a \(\delta_1>0\) such that, whenever \(x\in E\), \(0<|x-a|<\delta_1\),

\left|\frac{f(x)-f(a)}{x-a}-L\right|<1.

The triangle inequality then gives \(\left|\frac{f(x)-f(a)}{x-a}\right|<|L|+1\). Set \(\delta=\min\{\delta_1,\varepsilon/(|L|+1)\}\), which is positive because \(|L|+1>0\). If \(x\in E\) and \(0<|x-a|<\delta\), then

|f(x)-f(a)| =|x-a|\left|\frac{f(x)-f(a)}{x-a}\right| <|x-a|(|L|+1) <\delta(|L|+1) \leq\varepsilon.

If \(x=a\), then \(|f(x)-f(a)|=0<\varepsilon\) as well. Thus every \(x\in E\) with \(|x-a|<\delta\) satisfies the required continuity estimate. This proves continuity at \(a\). \(\square\)

This proof separates two roles. Differentiability controls the quotient near \(a\), while the factor \(|x-a|\) becomes small as \(x\) approaches \(a\). The point \(x=a\) is handled separately because the difference quotient is not defined there.

A Quantitative Local Estimate

The proof gives more than continuity: it bounds the change in \(f\) by a constant times the distance to \(a\), provided \(x\) is sufficiently close. This is a local estimate centered at \(a\); it does not assert a bound for arbitrary pairs of points in a whole domain.

Theorem (Local Linear Bound at a Differentiability Point): Under the hypotheses of the Differentiability Implies Continuity theorem, there is an \(r>0\) such that for every \(x\in E\) with \(|x-a|<r\), \(|f(x)-f(a)|\leq (|f'(a)|+1)|x-a|\).

Proof. By differentiability at \(a\), there is an \(r>0\) such that for every \(x\in E\) with \(0<|x-a|<r\),

\left|\frac{f(x)-f(a)}{x-a}-f'(a)\right|<1.

The triangle inequality implies \(\left|\frac{f(x)-f(a)}{x-a}\right|<|f'(a)|+1\). Multiplying by the positive number \(|x-a|\) gives \(|f(x)-f(a)|<(|f'(a)|+1)|x-a|\), and hence the stated weak inequality. At \(x=a\), both sides of the claimed inequality are zero, so it holds there too. This proves the estimate for every specified \(x\). \(\square\)

The constant \(|f'(a)|+1\) is convenient, not optimal. The purpose is to obtain a fixed finite bound on the difference quotient in some neighborhood of \(a\). Any such bound would yield continuity after multiplication by \(|x-a|\).

Worked Examples

Worked Example: A Polynomial at a Specified Point

Let \(f(x)=x^3-2x+4\), and consider \(a=1\). First, \(f(1)=1-2+4=3\). For \(x\neq1\), factor the difference:

\frac{f(x)-f(1)}{x-1} =\frac{x^3-2x+1}{x-1} =x^2+x-1.

The factorization is verified by multiplying: \((x-1)(x^2+x-1)=x^3-2x+1\). As \(x\to1\), the quotient tends to \(1+1-1=1\), so \(f'(1)=1\). The theorem therefore gives continuity at \(1\). Directly, the factorization also shows the function-value difference tends to zero: \(|f(x)-f(1)|=|x-1||x^2+x-1|\). The second factor tends to \(1\), so it remains bounded near \(1\), while the first factor tends to zero.

Worked Example: A Function with an Oscillating Factor

Define \(f:\mathbb R\to\mathbb R\) by \(f(0)=0\) and \(f(x)=x^2\sin(1/x)\) for \(x\neq0\). The difference quotient at zero is

\frac{f(x)-f(0)}{x-0}=x\sin(1/x),\qquad x\neq0.

Since \(|\sin(1/x)|\leq1\), we have \(|x\sin(1/x)|\leq|x|\), and \(|x|\to0\) as \(x\to0\). The Squeeze Theorem therefore gives \(f'(0)=0\). Differentiability implies continuity at zero. In this example the continuity estimate can also be checked directly: \(|f(x)-f(0)|=|x^2\sin(1/x)|\leq x^2\), which tends to zero.

The sine factor oscillates increasingly rapidly near zero, but its size remains at most one. The factor \(x^2\) is enough to make the function values approach \(f(0)\), and the difference quotient still tends to zero.

Worked Example: Continuity Without Differentiability

Let \(g(x)=|x|\) on \(\mathbb R\). At zero, \(|g(x)-g(0)|=|x|\), so \(g\) is continuous there: given \(\varepsilon>0\), choosing \(\delta=\varepsilon\) ensures that \(|x|<\delta\) implies \(|g(x)-g(0)|<\varepsilon\).

However, for \(x\neq0\), the difference quotient is \(\frac{|x|-0}{x}=\frac{|x|}{x}\). It equals \(1\) when \(x>0\) and \(-1\) when \(x<0\). These two values do not approach a common limit as \(x\to0\), so \(g\) is not differentiable at zero. This example confirms that the theorem gives a necessary consequence of differentiability, not a converse.

Applying the Pointwise Result on an Interval

The theorem is stated at one point, but it immediately applies wherever a function is differentiable. In particular, no uniform choice of \(\delta\) is needed to establish continuity at every point.

Corollary: Let \(I\subseteq\mathbb R\) be an interval, and suppose \(f:I\to\mathbb R\) is differentiable at every interior point of \(I\). Then \(f\) is continuous at every interior point of \(I\).

Proof. Let \(a\) be any interior point of \(I\). Because \(a\) is interior, there are points of \(I\) distinct from \(a\) arbitrarily close to it, so \(a\) is an accumulation point of \(I\). The function is differentiable at \(a\) by hypothesis. The Differentiability Implies Continuity theorem therefore gives continuity at \(a\). Since \(a\) was arbitrary, \(f\) is continuous at every interior point of \(I\). \(\square\)

The conclusion concerns interior points because the corollary assumes differentiability only there. At an endpoint, continuity follows from differentiability if an appropriate one-sided derivative exists and the same difference-quotient argument is used with domain points approaching from within the interval. It should not be inferred at an endpoint from a hypothesis that says nothing about differentiability there.

Why the Hypothesis Matters

Differentiability gives a finite limit for the difference quotient, and therefore bounds that quotient near the point. This is the essential step; merely knowing that a function has values nearby is not enough. The absolute-value example shows that continuity alone does not force the quotient to have a limit.

A common proof error is to write \(f(x)-f(a)=(x-a)\frac{f(x)-f(a)}{x-a}\) and conclude that the product tends to zero without justifying the behavior of the quotient. A product with a factor tending to zero need not tend to zero if the other factor is unbounded. Differentiability resolves this issue: the quotient converges to a finite number and is consequently bounded sufficiently near \(a\). The local linear bound makes that reasoning explicit.

Start with the derivative limit.
Choose a neighborhood in which the difference quotient differs from \(f'(a)\) by less than a fixed positive amount.

Bound the quotient.
The triangle inequality gives a finite bound such as \(|f'(a)|+1\) near the point.

Control the function-value difference.
Write it as the input distance times the quotient, then choose the input distance small enough for the desired continuity tolerance.

Check Your Understanding

Use the definition, proof, and examples above to answer the following questions.

Why is the quotient in the differentiability definition restricted to \(x\neq a\), and how is \(x=a\) handled in the continuity proof?
Where does the proof use that \(f'(a)\) is finite?
For \(f(x)=x^3-2x+4\) at \(a=1\), what is the factored difference quotient and its limit?
Why does the bound \(|x\sin(1/x)|\leq|x|\) establish the derivative of \(x^2\sin(1/x)\) at zero?
What prevents \(|x|/x\) from having a limit at zero, even though \(|x|\) is continuous there?