Proof Differentiability Implies Continuity

The Proof in One Line—and What It Requires

The previous tutorial examined local bounds on function increments and the behavior of those increments along sequences. This tutorial focuses on the proof that makes the central implication precise: differentiability at a point forces continuity at that point. The idea is to write the change in the function as the product of the change in the input and the difference quotient.

That factorization alone does not settle the proof. As the input increment tends to zero, the difference quotient must not grow too quickly. Differentiability supplies exactly the needed control: a difference quotient with a finite limit is bounded for all sufficiently small nonzero increments. Multiplying that bounded quantity by an increment tending to zero gives a function increment tending to zero.

Recall (Differentiability at a Point): Let $f$ be defined on an interval $I$, and let $a$ be an interior point of $I$. The function $f$ is differentiable at $a$ if the finite limit $$ f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h} $$ exists.

The theorem that differentiability at an interior point implies continuity there was established earlier in the course. We give its epsilon-delta proof here, making explicit the bounded-factor step. The result is a proof of a familiar implication, not a claim that continuity and differentiability are equivalent.

A Bounded-Factor Lemma

First isolate the elementary mechanism. Suppose an increment can be written as $h q(h)$, where $q(h)$ remains bounded for small nonzero $h$. Then the increment tends to zero as $h\to0$. The function $q$ need not itself have a limit for this argument; local boundedness is enough.

Lemma (A Vanishing Factor Times a Locally Bounded Factor): Suppose $q$ is defined for all sufficiently small nonzero $h$, and there are constants $M\geq0$ and $\rho>0$ such that $|q(h)|\leq M$ whenever $0<|h|<\rho$. Define $r(0)=0$ and $r(h)=h q(h)$ for nonzero $h$ in this neighborhood. Then $r(h)\to0$ as $h\to0$.

Proof. Let $\varepsilon>0$. If $M>0$, choose $\delta=\min(\rho,\varepsilon/M)$. For $0<|h|<\delta$, the assumed bound gives

|r(h)|=|h||q(h)|\leq M|h|<M\delta\leq\varepsilon.

At $h=0$, $r(0)=0<\varepsilon$ as well. If $M=0$, then $q(h)=0$ whenever $0<|h|<\rho$, so $r(h)=0$ there; any choice $0<\delta\leq\rho$ proves the limit. Thus in both cases $r(h)\to0$. $\square$

The separate treatment of $h=0$ matters in an epsilon-delta proof: the quotient $q(h)$ may be undefined there, while the increment $r(0)$ is defined. The estimate using the quotient applies only to nonzero $h$; at zero the increment is checked directly.

From a Finite Derivative to Continuity

Now take $q(h)$ to be the difference quotient of $f$ at $a$. Its finite limit $L=f'(a)$ makes it bounded near zero. The bounded-factor lemma then explains the proof conceptually. The following direct argument also shows exactly how to choose a continuity radius for a given $\varepsilon$.

Theorem (Differentiability Implies Continuity): Let $I$ be an interval, let $a$ be an interior point of $I$, and suppose $f:I\to\mathbb{R}$ is differentiable at $a$. Then $f$ is continuous at $a$.

Proof. Write $L=f'(a)$, a finite real number. Since $a$ is an interior point of $I$, there is a number $\rho>0$ such that $a+h\in I$ whenever $|h|<\rho$. By the definition of the derivative, there is a number $\eta>0$ such that, for every $h$ satisfying $0<|h|<\eta$,

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

The triangle inequality yields a bound on the difference quotient:

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

Given any $\varepsilon>0$, choose $\delta=\min(\rho,\eta,\varepsilon/(|L|+1))$. The denominator $|L|+1$ is positive because $|L|\geq0$. If $0<|h|<\delta$, then $a+h\in I$, the difference-quotient bound applies, and

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

The proof depends on two distinct bounds. The derivative limit supplies a neighborhood where the quotient is bounded by $1+|L|$; the choice of $\delta$ then makes $|h|$ small enough to force the product below $\varepsilon$. A finite limit is what makes the first bound possible.

Worked Examples: The Proof in Concrete Settings

Worked Example: A Polynomial at a Chosen Point

Let $f(x)=x^3-2x$ and $a=1$. First, $f(1)=1-2=-1$. For $h\ne0$, compute the difference quotient:

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

As $h\to0$, this quotient tends to $1$, so $f'(1)=1$. To see the continuity estimate directly, if $0<|h|<1$, then

|1+3h+h^2| \leq 1+3|h|+|h|^2 <5.

Consequently, for $0<|h|<1$,

|f(1+h)-f(1)| = |h|\,|1+3h+h^2| <5|h|.

Given $\varepsilon>0$, take $\delta=\min(1,\varepsilon/5)$. For $0<|h|<\delta$, the increment is less than $\varepsilon$. At $h=0$, it is exactly zero. Thus the calculation verifies continuity at $1$, with the quotient estimate providing an explicit bound.

Worked Example: An Oscillatory Function with a Finite Derivative

Define $g(0)=0$ and $g(x)=x^2\sin(1/x)$ for $x\ne0$. The difference quotient at zero is, for $h\ne0$,

\frac{g(h)-g(0)}{h} = \frac{h^2\sin(1/h)}{h} = h\sin(1/h).

Since $|\sin t|\leq1$ for every real $t$,

|h\sin(1/h)|\leq|h|\longrightarrow0 \qquad\text{as }h\to0.

Therefore $g'(0)=0$. The same estimate also displays the continuity mechanism directly:

|g(h)-g(0)| = |h|\,|h\sin(1/h)| \leq |h|^2\longrightarrow0.

Although the sine factor oscillates and does not settle to a limit as its argument grows, it remains bounded. The difference quotient itself tends to zero because of the additional factor $h$, and the function increment tends to zero as required.

Worked Example: A Derivative at a Point Does Not Control the Whole Domain

Define $u:\mathbb{R}\to\mathbb{R}$ by $u(x)=0$ for $x\leq2$ and $u(x)=1$ for $x>2$. At $a=0$, if $|h|<1$, then $h<2$, so $u(h)=0=u(0)$. For all such nonzero $h$,

\frac{u(0+h)-u(0)}{h}=0.

Thus $u'(0)=0$. In particular, the theorem guarantees continuity at $0$. But the function is not continuous at $2$: $u(2)=0$, while $u(x)=1$ for every $x>2$. The differentiability hypothesis concerns behavior near the specified point $0$, so it cannot imply continuity at an unrelated point.

Why the Finiteness and Scope Matter

A finite derivative is essential because it makes the nearby difference quotients bounded. For a function that is continuous but not differentiable, the quotient may fail to have a finite limit; the continuity conclusion may still hold, but it cannot be obtained by claiming that a finite derivative exists. Conversely, a difference quotient that becomes unbounded does not by itself prove discontinuity: the proof above uses boundedness as a sufficient mechanism, not as a necessary condition for continuity.

A common proof error is to multiply an estimate for the difference quotient by $h$ without first restricting to $h\ne0$. The quotient is undefined at zero. The correct procedure is to establish the estimate for $0<|h|<\delta$, and then check the zero increment separately. Another frequent error is to infer continuity everywhere from differentiability at a single point. The conclusion is local: differentiability at $a$ proves continuity at $a$.

The reusable strategy is to identify a vanishing factor and a locally bounded factor. Here the vanishing factor is $h$, and the bounded factor is the difference quotient. In other problems, the factorization may look different, but the same question is useful: can one control one factor while the other tends to zero?

Check Your Understanding

Use the bounded-factor lemma and the epsilon-delta proof to answer the following.

Why is the difference quotient considered only for nonzero increments $h$?
How does a finite derivative limit guarantee that the difference quotient is bounded near the point?
In the theorem’s proof, why is $|L|+1$ used when choosing the continuity radius?
For the oscillatory example, which estimate establishes that the difference quotient tends to zero?
What does differentiability at one point imply about continuity elsewhere in the domain?