Differentiability Implies Continuity

From One-Sided Slopes to Continuity

The previous tutorial examined how the difference quotient can approach a limit from the left and from the right. When both sides approach the same finite number at an interior point, the Two-Sided Derivative Criterion says that the ordinary derivative exists there. Differentiability carries another important consequence: a function differentiable at a point is continuous at that point.

This implication is local. A derivative at $a$ describes the function’s behavior as its input approaches $a$; it does not by itself give information about every other point in the domain. The implication also runs in only one direction. A function can be continuous at a point without having a derivative there.

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 in “The Derivative as a Limit.” Here we will use that result and focus on what the difference quotient reveals about the size of nearby changes. This quantitative view is useful when estimating a function near a point or tracking values along a sequence.

A Local Bound on Function Increments

If the difference quotient approaches $L=f'(a)$, it must eventually stay close to $L$. In particular, its absolute value is eventually no greater than $|L|+\varepsilon$, for any chosen $\varepsilon>0$. Multiplying by the size of the input increment gives a bound on the change in the function value.

Theorem (Local Increment Bound): Suppose $f$ is differentiable at an interior point $a$, with $f'(a)=L$. For every $\varepsilon>0$, there is a $\delta>0$ such that, whenever $0<|h|<\delta$, $$ |f(a+h)-f(a)|\leq (|L|+\varepsilon)|h|. $$

Proof. By the definition of the derivative, there is a $\delta>0$ such that, whenever $0<|h|<\delta$,

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

The triangle inequality then gives

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

Since $|h|>0$, multiplying both sides by $|h|$ yields

|f(a+h)-f(a)|< (|L|+\varepsilon)|h|.

The strict inequality implies the stated weak inequality. This proves the theorem. $\square$

The bound says that, sufficiently close to $a$, the change in the function value is at most a constant times the distance from $a$. The constant can be chosen just above $|f'(a)|$. When the derivative is zero, the function changes more slowly than any fixed positive multiple of the input distance, once the inputs are close enough.

Worked Example: An Explicit Increment Estimate

Let $f(x)=x^2+3x$, and examine the point $a=1$. We have $f(1)=4$. For an increment $h$,

\frac{f(1+h)-f(1)}{h} = \frac{(1+h)^2+3(1+h)-4}{h} = \frac{5h+h^2}{h} = 5+h, \qquad h\ne0.

Thus $f'(1)=5$. If $0<|h|<1$, then $|5+h|\leq 6$, because $-1<h<1$ implies $4<5+h<6$. Therefore

|f(1+h)-f(1)|=|5+h||h|\leq6|h|.

For instance, if $|h|<0.01$, then $|f(1+h)-f(1)|\leq0.06$. The derivative gives a practical estimate of how much the function can change near the point, not merely a limiting slope.

What Changes When the Derivative Is Zero?

A zero derivative means the difference quotient approaches zero. Consequently, the function’s increment is small compared with the input increment: their ratio in absolute value tends to zero. This is a stronger rate statement than merely knowing that the function values approach $f(a)$.

Worked Example: A Smaller-Order Increment

Consider $g(x)=2+(x-4)^2$ at $a=4$. Since $g(4)=2$, substituting $x=4+h$ gives

\frac{g(4+h)-g(4)}{h} = \frac{2+h^2-2}{h} =h, \qquad h\ne0.

As $h\to0$, this quotient tends to zero, so $g'(4)=0$. The increment itself is $g(4+h)-g(4)=h^2$, and hence

\frac{|g(4+h)-g(4)|}{|h|}=|h|\longrightarrow0.

For every $C>0$, choosing $0<|h|<C$ gives $|h|^2<C|h|$. Thus the function’s change near $4$ is eventually smaller than any prescribed positive constant times the distance to $4$.

The same rate information can be expressed for sequences. It allows us to compare how quickly function values approach $f(a)$ with how quickly the input values approach $a$.

Theorem (Relative Increment Along a Sequence): Suppose $f$ is differentiable at an interior point $a$, with $f'(a)=L$. If $x_n\to a$ and $x_n\ne a$ for all sufficiently large $n$, then $$ \frac{|f(x_n)-f(a)|}{|x_n-a|}\longrightarrow |L|. $$ In particular, if $L=0$, then $|f(x_n)-f(a)|/|x_n-a|\to0$.

Proof. For every sufficiently large $n$, the quotient

q_n=\frac{f(x_n)-f(a)}{x_n-a}

is defined. Since $x_n\to a$ and the terms are eventually distinct from $a$, the definition of differentiability gives $q_n\to L$. The inequality

\bigl||q_n|-|L|\bigr|\leq |q_n-L|

then implies $|q_n|\to |L|$. Finally,

|q_n|=\frac{|f(x_n)-f(a)|}{|x_n-a|},

which proves the claimed limit. When $L=0$, the limit is zero as stated. $\square$

Worked Example: Checking the Rate Along a Sequence

For the function $g(x)=2+(x-4)^2$ at $a=4$, take $x_n=4+1/n$. These points are distinct from $4$ and converge to it. Direct calculation gives

\frac{|g(x_n)-g(4)|}{|x_n-4|} = \frac{|(1/n)^2|}{|1/n|} = \frac{1}{n} \longrightarrow0.

This agrees with $g'(4)=0$. The sequence calculation illustrates the general theorem: when the derivative is zero, function-value changes become negligible relative to input changes along every sequence approaching the point, provided the sequence terms are eventually different from the point itself.

Continuity Does Not Guarantee Differentiability

The differentiability-implies-continuity theorem should not be reversed. Continuity controls whether function values approach the value at the point; differentiability asks for the more specific condition that the difference quotient approach a finite slope. A function can meet the first requirement and fail the second.

Worked Example: Continuous but Not Differentiable

Define $u(x)=\sqrt{|x|}$ for real $x$, and examine $a=0$. Since $u(0)=0$ and

|u(x)-u(0)|=\sqrt{|x|}\longrightarrow0 \qquad\text{as }x\to0,

the function is continuous at $0$. For $h>0$, however, the difference quotient is

\frac{u(h)-u(0)}{h} = \frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}},

which grows without bound as $h\to0^+$. For $h<0$, it is

\frac{u(h)-u(0)}{h} = \frac{\sqrt{-h}}{h} = -\frac{1}{\sqrt{-h}},

which decreases without bound as $h\to0^-$. The two-sided difference quotient therefore has no finite limit. Continuity holds, but differentiability fails.

There is a different limitation in the other direction: differentiability at one point does not imply continuity at every point in the domain. For example, define $v(x)=0$ when $x\leq1$, and $v(x)=1$ when $x>1$. At $a=0$, every sufficiently small increment satisfies $v(h)=v(0)=0$, so the difference quotient is zero and $v'(0)=0$. Yet $v$ is discontinuous at $1$. The derivative at $0$ describes behavior near $0$, not behavior near $1$.

Using the Implication Carefully

When a problem gives differentiability at $a$, the established theorem immediately supplies continuity at $a$; no separate continuity calculation is needed. The Local Increment Bound adds a useful estimate, and the sequence result identifies the relative rate of change. These conclusions all concern the same point and rely on the derivative existing there.

A common pitfall is to treat continuity and differentiability as interchangeable. Continuity does not ensure a finite limiting slope, as the square-root example shows. Another is to infer global regularity from a derivative at a single point. The step-function example is differentiable at $0$ but discontinuous elsewhere. Keep the location and scope of each hypothesis explicit: differentiability at $a$ implies continuity at $a$, not continuity throughout the domain.

Check Your Understanding

Use the definitions, estimates, and examples in this tutorial to answer the following.

In the Local Increment Bound, why must the increment $h$ be nonzero before the difference quotient can be used?
If $f'(a)=0$, what limit describes the size of $|f(a+h)-f(a)|$ relative to $|h|$?
What does the Relative Increment Along a Sequence theorem say when $f'(a)=L\ne0$?
Why does continuity of $\sqrt{|x|}$ at zero not imply differentiability there?
Does differentiability at one point guarantee continuity at every point of the domain? Explain using the scope of the hypothesis.