Connecting Differentiability and Continuity

A derivative is the limit of $\frac{f(x) - f(a)}{x - a}$ as $x \to a$. The denominator goes to $0$, so for the quotient to settle on a finite number, the numerator $f(x) - f(a)$ has to go to $0$ as well. That forces $f(x) \to f(a)$, which is continuity. Differentiability is the stronger condition: it demands continuity and more.

Differentiable implies continuous

Here is the proof. For $x \neq a$, multiply and divide by $x - a$:

As $x \to a$, the first factor approaches $f'(a)$, a finite number because $f$ is differentiable at $a$, and the second factor approaches $0$. The limit of a product is the product of the limits, so

That is the definition of continuity at $a$ (topic 1.11).

Two consequences follow:

• No function value, no derivative. If $a$ is not in the domain of $f$, it is not in the domain of $f'$. The difference quotient needs the value $f(a)$, so no derivative can exist where $f$ is undefined. For $f(x) = \frac{1}{x}$, both $f$ and $f'$ are undefined at $0$.
• Every discontinuity is a point of non-differentiability. At a jump, a hole, or a vertical asymptote, the derivative does not exist, and nothing more needs checking. At a removable discontinuity this is true even though the graph looks smooth on both sides of the hole: the quotient's numerator $f(x) - f(a)$ approaches a nonzero number while the denominator goes to $0$.

Continuous does not imply differentiable

The converse is false. A function can be continuous at $a$ and still have no derivative there, because the difference quotient can fail to have a finite limit even when $f(x) - f(a) \to 0$. Usually it fails in one of two ways: its one-sided limits disagree, or it grows without bound. (A third way exists: the quotient can oscillate forever. For $f(x) = x\sin\frac{1}{x}$ with $f(0) = 0$, $f$ is continuous at $0$, but the quotient $\frac{h\sin(1/h)}{h} = \sin\frac{1}{h}$ swings between $-1$ and $1$ infinitely often as $h \to 0$ and approaches nothing.)

Corners: one-sided slopes disagree

For $f(x) = |x|$ at $x = 0$, the difference quotient is $\frac{|h|}{h}$. For $h > 0$ it equals $1$, and for $h < 0$ it equals $-1$. The one-sided limits are $1$ and $-1$, so the two-sided limit does not exist. The graph is continuous at $0$ but has a corner: there is no single line that matches its direction from both sides.

Any graph made of two smooth pieces that meet at an angle has the same problem. If the slope from the left approaches one value and the slope from the right approaches another, there is no derivative at the meeting point.

Vertical tangents and cusps: slopes grow without bound

For $f(x) = \sqrt[3]{x}$ at $x = 0$, the difference quotient is

which is positive for every $h \neq 0$ and grows without bound as $h \to 0$. The secant lines through the origin get steeper and steeper from both sides and approach the vertical line $x = 0$. That vertical line is the tangent line, but a vertical line has no slope, so $f'(0)$ does not exist. The function is continuous at $0$, and its graph is perfectly smooth there; it just stands straight up.

For $g(x) = x^{2/3}$ at $x = 0$, the quotient is $\frac{h^{2/3}}{h} = \frac{1}{h^{1/3}}$, which goes to $+\infty$ from the right and to $-\infty$ from the left. The graph comes down steeply into the origin and goes back up steeply, forming a sharp point called a cusp. Again, no derivative exists.

Piecewise functions: continuity first, then slopes

A function defined in pieces, such as

is differentiable away from the break point wherever its pieces are. At the break point $x = 1$, check in this order.

First, continuity. $f(1) = 2$, and $\lim\limits_{x \to 1^+}(3x - 1) = 2$, so $f$ is continuous at $1$. If this had failed, you would be done: no continuity, no derivative.

Then, the one-sided limits of the difference quotient. From the left, with $f(1) = 2$:

From the right:

The left-hand slope is $2$ and the right-hand slope is $3$, so $f$ is continuous but not differentiable at $x = 1$: the graph has a corner there.

Once continuity is established, the one-sided limits of the difference quotient equal the slopes of the two pieces at the break point, so with pieces whose derivatives you know, comparing those slopes is a valid shortcut. Without continuity, it is not, as the first trap below shows.

Worked examples

Worked exampleExample 2: reading a graph

The graph of $f$ on $-4 \leq x \leq 5$ is shown below.

(a) At which values of $x$ in $-4 < x < 5$ is $f$ not continuous?

(b) At which values of $x$ in $-4 < x < 5$ is $f$ not differentiable? Give a reason for each.

Show solution

(a) $f$ is not continuous at $x = -2$, where the graph jumps (the left-hand limit is 2, the right-hand limit is 4), and at $x = 4$, where $\lim\limits_{x \to 4} f(x) = 4$ but $f(4) = 1$.

(b) $f$ is not differentiable at four values:

• $x = -2$ and $x = 4$: $f$ is not continuous there, and a function is differentiable only where it is continuous.
• $x = 0$: $f$ is continuous, but the graph has a corner. The slope from the left is $-1$ (the segment from $(-2, 4)$ to $(0, 2)$), and the slope from the right is a small positive number, so the one-sided limits of the difference quotient are not equal.
• $x = 2$: $f$ is continuous, but the tangent line there is vertical, so the difference quotient grows without bound and $f'(2)$ does not exist.

Everywhere else in the interval, the graph is a smooth unbroken curve or segment with a non-vertical tangent, so $f$ is differentiable there.

Where students lose points

On the exam

Questions on this topic ask you to decide whether a function is differentiable at a point and justify it. For a piecewise function, the justification states continuity with limits and then compares the one-sided slopes. Multiple-choice questions may show a graph and ask at how many points the function is not differentiable, or give a piecewise function with unknown constants. The fact "differentiable implies continuous" also appears as a step inside other justifications: when a question says $f$ is differentiable, you may cite it to conclude $f$ is continuous wherever it is differentiable. Check what that gives you against each theorem's hypotheses. The Intermediate Value Theorem needs $f$ continuous on the closed interval $[a, b]$, and the Mean Value Theorem needs $f$ continuous on $[a, b]$ and differentiable on $(a, b)$. If $f$ is differentiable at every point of $[a, b]$ (for example, differentiable for all real $x$), continuity on $[a, b]$ follows. If you are told only that $f$ is differentiable on the open interval $(a, b)$, that gives continuity inside the interval but not at $x = a$ or $x = b$, so continuity at the endpoints must be given or checked separately.