Continuity and Differentiability Ask Different Questions
A function can approach its value smoothly in the sense of continuity without having a single finite slope at a point. Continuity asks whether nearby function values approach the function value. Differentiability asks whether nearby changes in the function are described, to first order, by one finite rate of change.
The difference quotient makes the distinction precise. At a point \(a\) in the interior of the domain, it compares the change in the function value with the change in input. Its limit, when that limit exists as a finite real number, is the derivative.
These conditions are related, but they are not equivalent. Differentiability implies continuity, as established in the earlier material on why differentiability implies continuity. The reverse implication fails. We will see two distinct ways it can fail: the one-sided quotients can approach different finite numbers, or their magnitudes can grow without bound.
Example One: The Absolute-Value Function
Consider \(f(x)=|x|\) at \(x=0\). First, the function is continuous there: \(f(0)=0\), and \(|f(x)-f(0)|=|x|\). Given any \(\varepsilon>0\), choosing \(\delta=\varepsilon\) ensures that \(|x|<\delta\) implies \(|f(x)-f(0)|<\varepsilon\).
Worked Example: A Continuous Corner at the Origin
Compute the difference quotient for \(f(x)=|x|\) at \(0\). Since \(f(0)=0\), for \(h\ne 0\) the quotient is $$ \frac{f(0+h)-f(0)}{h} = \frac{|h|-0}{h} = \frac{|h|}{h}. $$
If \(h>0\), then \(|h|=h\), so the quotient is \(h/h=1\). If \(h<0\), then \(|h|=-h\), so the quotient is \((-h)/h=-1\). Therefore, $$ \lim_{h\to 0^+}\frac{|h|}{h}=1 \qquad\text{and}\qquad \lim_{h\to 0^-}\frac{|h|}{h}=-1. $$
Both one-sided limits are finite, but they are unequal. The two-sided limit does not exist, so \(f'(0)\) does not exist. The check is direct: the quotient equals \(1\) on the positive side and \(-1\) on the negative side, and thus cannot approach a single real number.
This is a failure of agreement, not a failure of the quotients to stay bounded. From the right, the secant slopes approach \(1\); from the left, they approach \(-1\). The function is continuous because its values meet at the origin, yet its two sides approach that point with different finite slopes.
Example Two: The Cube-Root Function
Now consider \(g(x)=x^{1/3}\), using the real cube root, at \(0\). It is continuous at \(0\), since \(g(0)=0\) and \(|g(x)-g(0)|=|x|^{1/3}\). Given \(\varepsilon>0\), choose \(\delta=\varepsilon^3\). If \(|x|<\delta\), then \(|g(x)-g(0)|=|x|^{1/3}<\varepsilon\).
Worked Example: Unbounded Difference Quotients at Zero
For \(h\ne 0\), the difference quotient for \(g(x)=x^{1/3}\) at \(0\) is $$ \frac{g(0+h)-g(0)}{h} = \frac{h^{1/3}}{h}. $$
For positive or negative \(h\), \(h=(h^{1/3})^3\). Thus $$ \frac{h^{1/3}}{h} = \frac{h^{1/3}}{(h^{1/3})^3} = \frac{1}{(h^{1/3})^2} = \frac{1}{|h|^{2/3}}. $$
This quotient is positive on both sides. To verify that it grows without bound, let \(M>0\) be arbitrary and choose \(\delta=M^{-3/2}\). If \(0<|h|<\delta\), then $$ |h|^{2/3}<\left(M^{-3/2}\right)^{2/3}=\frac{1}{M}, \qquad\text{so}\qquad \frac{1}{|h|^{2/3}}>M. $$
Therefore the quotient tends to \(+\infty\) as \(h\to 0\). It has no finite real limit, so \(g'(0)\) does not exist as a finite derivative. The check is that the computed quotient is exactly \(1/|h|^{2/3}\), which exceeds any prescribed positive bound sufficiently close to zero.
Unlike the absolute-value example, there are not two unequal finite slopes here. The quotient is positive from both sides, but it does not approach a finite number. In particular, the fact that the quotient has the same sign on both sides does not make the derivative exist: a finite limit is required.
How the Two Examples Differ
For an interior point, the two-sided limit of the difference quotient exists as a finite real number exactly when both one-sided limits exist as finite real numbers and are equal. This gives a useful way to organize the test. First ask whether each side has a finite limit; then ask whether those limits agree. The absolute-value example fails the second test. The cube-root example fails the first.
| Function at the point | Continuity there? | Behavior of the difference quotient | Finite derivative? |
|---|---|---|---|
| \(|x|\) at \(0\) | Yes | One-sided limits are \(-1\) and \(1\) | No |
| \(x^{1/3}\) at \(0\) | Yes | Positive and unbounded on both sides | No |
These examples show why continuity cannot settle a question about differentiability. The continuity condition concerns \(f(a+h)-f(a)\) becoming small as \(h\) becomes small. The derivative condition concerns the ratio of that change to \(h\). A small numerator does not guarantee that the ratio approaches a finite number.
A Continuous Function Can Also Be Differentiable
The examples do not mean that continuity and nondifferentiability always occur together. A continuous function may have a finite derivative. The following calculation makes the contrast explicit.
Worked Example: A Continuous Function with a Finite Derivative
Let \(p(x)=x^2\), and examine \(x=0\). Since \(p(0)=0\), for any \(\varepsilon>0\), choosing \(\delta=\sqrt{\varepsilon}\) gives $$ |x|<\delta \quad\Longrightarrow\quad |p(x)-p(0)|=|x^2|=|x|^2<\delta^2=\varepsilon. $$ Thus \(p\) is continuous at zero.
For \(h\ne 0\), its difference quotient is $$ \frac{p(0+h)-p(0)}{h} = \frac{h^2-0}{h} = h. $$ As \(h\to 0\), this quotient tends to \(0\), so \(p'(0)=0\). Checking the limit from either side gives the same value, since the quotient is \(h\) for every nonzero \(h\). The derivative exists as a finite real number.
Thus continuity by itself neither guarantees nor rules out differentiability. It is compatible with a finite derivative, but does not force one to exist. To determine differentiability at a point, return to the difference quotient and examine its limit.
Why the Difference Quotient Is Decisive
The derivative is defined by a limit of ratios, not merely by how close the function values are to the function value at the point. The numerator and denominator can both approach zero while their ratio behaves in different ways. The calculations above display three possibilities:
- The quotient can approach a finite value, as for \(x^2\) at zero.
- The one-sided quotients can approach different finite values, as for \(|x|\) at zero.
- The quotient can grow without bound, as for \(x^{1/3}\) at zero.
Only the first possibility gives a finite derivative. This is why a continuity check and a differentiability check must not be confused. Continuity verifies that the function values fit together at the point; differentiability requires one finite limiting slope.
Common Pitfall: Treating Continuity as a Slope Condition
A common mistake is to reason, “The function is continuous, so the derivative exists.” That conclusion does not follow. For example, \(f(x)=|x|\) is continuous at zero, but the calculation $$ \frac{f(h)-f(0)}{h}= \begin{cases} 1,&h>0,\\ -1,&h<0 \end{cases} $$ shows that the quotient has no two-sided limit.
Another mistake is to see that the cube-root quotient is positive on each side and assume that matching signs are enough. They are not: its value is \(1/|h|^{2/3}\), which exceeds every fixed positive bound for sufficiently small nonzero \(h\). Positivity does not make it a finite limit.
The correction in both cases is the same: compute the difference quotient from its definition, simplify it, and check whether it approaches one finite real number from both sides. Do not infer differentiability from continuity, from a graph that looks connected, or from the sign of the quotient alone.
Check Your Understanding
Use the definition of the derivative and the behavior of the relevant difference quotient to answer these questions.
- Why is \(f(x)=|x|\) continuous at zero even though \(f'(0)\) does not exist?
- What are the right-hand and left-hand limits of the difference quotient for \(|x|\) at zero?
- Show that \(x^{1/3}\) is continuous at zero by giving a suitable choice of \(\delta\) for a given \(\varepsilon>0\).
- Why does \(1/|h|^{2/3}\) rule out a finite derivative for \(x^{1/3}\) at zero?
- What is the difference between the failure of the derivative for \(|x|\) and its failure for \(x^{1/3}\) at zero?
