Continuous but not differentiable: corners, cusps and vertical tangents

By Marius, private math tutor in Montreal · 2026-10-05

Differentiable implies continuous, but continuous does not imply differentiable. A graph can be unbroken at a point and still have no single slope there, and in a first calculus course that usually happens in one of three standard shapes: a corner, a cusp and a vertical tangent. Here is how one limit tells them apart, why the implication only runs one way, and a self-check with answers.

What each word asks for

Both words describe what happens as $x$ approaches a point $a$ , but they ask for different things.

Continuous at $a$ : the values of $f$ settle on the height of the point itself. ${lim}_{x \to a} f (x) = f (a)$
Differentiable at $a$ : the slopes of the secant lines through $(a, f (a))$ settle on one finite number. $f^{'} (a) = {lim}_{h \to 0} \frac{f (a + h) - f (a)}{h}$ That limit exists only when the two one-sided limits, $h \to 0^{-}$ and $h \to 0^{+}$ , are both finite and equal.

So an unbroken graph can still fail the second test, most often in one of two ways: the one-sided slopes are finite but different, or they grow without bound. The three graphs below are the standard cases.

All three functions are continuous at

x = 0

, and none has a derivative there. What differs is how the one-sided slopes behave.

A corner: the two slopes disagree

For $f (x) = | x |$ , the difference quotient at $0$ is

\frac{| 0 + h | - | 0 |}{h} = \frac{| h |}{h}

which equals $1$ for every $h > 0$ and $- 1$ for every $h < 0$ . The right-hand limit is $1$ , the left-hand limit is $- 1$ , and $f^{'} (0)$ does not exist. The slope jumps from $- 1$ to $1$ at the origin, and no single number can stand for both.

A cusp: the slopes run to opposite infinities

For $f (x) = x^{2 / 3} = \sqrt[3]{x^{2}}$ ,

\frac{f (h) - f (0)}{h} = \frac{h^{2 / 3}}{h} = \frac{1}{h^{1 / 3}}

A cube root keeps the sign of $h$ , so the quotient tends to $+ \infty$ as $h \to 0^{+}$ and to $- \infty$ as $h \to 0^{-}$ . Both halves of the graph turn vertical at the origin, but they arrive from opposite directions, and the curve comes to a point.

A vertical tangent: the slopes run to the same infinity

For $f (x) = \sqrt[3]{x}$ ,

\frac{f (h) - f (0)}{h} = \frac{h^{1 / 3}}{h} = \frac{1}{h^{2 / 3}}

and $h^{2 / 3} = {(\sqrt[3]{h})}^{2}$ is positive on both sides of $0$ , so the quotient tends to $+ \infty$ from the left and from the right. The curve passes smoothly through the origin, and its tangent line there is the vertical line $x = 0$ . A vertical line has no slope, so $f^{'} (0)$ still does not exist.

One limit decides all three

For a function that is continuous at $a$ , compute the two one-sided limits of $\frac{f (a + h) - f (a)}{h}$ , then read the result:

the same finite number on both sides: $f$ is differentiable at $a$ ;
two different finite numbers: a corner;
$+ \infty$ on one side and $- \infty$ on the other: a cusp;
the same infinity on both sides: a vertical tangent.

Check continuity first. At a jump, or at a single point lifted off an otherwise unbroken curve, a one-sided quotient also runs off to infinity, and neither graph has a cusp or a vertical tangent.

These are the common shapes, not the only ones. Take $f (x) = x \sin (1 / x)$ with $f (0) = 0$ . It is continuous at $0$ , yet its difference quotient there is $\sin (1 / h)$ , which keeps swinging between $- 1$ and $1$ and has no limit of any kind. The test does not change: unless both one-sided limits are the same finite number, there is no derivative.

Why a differentiable function is always continuous

The theorem: if $f$ is differentiable at $a$ , then $f$ is continuous at $a$ . The proof takes three lines. For $x \neq a$ , write

f (x) - f (a) = \frac{f (x) - f (a)}{x - a} \cdot (x - a)

As $x \to a$ , the fraction tends to $f^{'} (a)$ , which is a finite number, and $x - a$ tends to $0$ . So

{lim}_{x \to a} (f (x) - f (a)) = f^{'} (a) \cdot 0 = 0

and that says exactly ${lim}_{x \to a} f (x) = f (a)$ .

Run the argument backwards and it breaks. Knowing that $f (x) - f (a)$ tends to $0$ says nothing about the fraction, which can approach two different numbers (a corner) or grow without bound (a cusp or a vertical tangent). That is why the implication only goes one way.

The direction it does go is worth using: if $f$ is not continuous at $a$ , then $f^{'} (a)$ does not exist. A jump, a hole or a vertical asymptote rules out a derivative before you compute anything.

The common mistake is running the arrow backwards. The correct statement is that differentiable at $a$ implies continuous at $a$ . The trap is reading it the other way, and concluding that a function has a derivative because its graph is unbroken, for instance because the pieces of a piecewise function meet. Continuity is necessary for a derivative, never sufficient: after checking it, you still have to compute the two one-sided limits of the difference quotient.

Worked example: one function, two kinds of failure

Find every point where $g (x) = \sqrt[3]{x^{3} - 2 x^{2}}$ is continuous but not differentiable, and name each one.

Continuity. $x^{3} - 2 x^{2}$ is a polynomial and the cube root is continuous everywhere, so $g$ is continuous at every real number.
Candidates. Factor the inside: $x^{3} - 2 x^{2} = x^{2} (x - 2)$ . The cube root has a derivative at every number except $0$ , so the only places that can go wrong are where the inside is $0$ : $x = 0$ and $x = 2$ .
At $x = 0$ . $g (0) = 0$ and $g (h) = \sqrt[3]{h^{2}} \sqrt[3]{h - 2}$ , so $\frac{g (h) - g (0)}{h} = \frac{\sqrt[3]{h - 2}}{h^{1 / 3}}$ The numerator tends to $\sqrt[3]{- 2} \approx - 1.26$ , a negative number, while the denominator tends to $0$ with the sign of $h$ . The quotient tends to $- \infty$ as $h \to 0^{+}$ and to $+ \infty$ as $h \to 0^{-}$ : opposite infinities, so a cusp.
At $x = 2$ . $g (2) = 0$ and $g (2 + h) = \sqrt[3]{(2 + h)^{2} h}$ , so $\frac{g (2 + h) - g (2)}{h} = \frac{\sqrt[3]{(2 + h)^{2}}}{h^{2 / 3}}$ The numerator tends to $\sqrt[3]{4}$ , a positive number, and the denominator is positive on both sides, so the quotient tends to $+ \infty$ from both sides: a vertical tangent.
Answer. $g$ is continuous everywhere, has a cusp at $x = 0$ and a vertical tangent at $x = 2$ , and is differentiable at every other number.

Both points come from the same factorization. The squared factor $x^{2}$ behaves like $x^{2 / 3}$ and makes the cusp; the single factor $x - 2$ behaves like $x^{1 / 3}$ and makes the vertical tangent.

Check your understanding

No calculator needed.

True or false: if $f$ is continuous at $a$ , then $f$ is differentiable at $a$ .
True or false: if $f$ is not continuous at $a$ , then $f^{'} (a)$ does not exist.
Is $f (x) = | x | + x$ differentiable at $0$ ? If not, which of the three shapes does its graph have there?
At $x = 0$ , does $f (x) = \sqrt{| x |}$ have a corner, a cusp or a vertical tangent?
Does $f (x) = x^{4 / 3}$ have a derivative at $0$ ? If it does, find $f^{'} (0)$ .
Find the error: “The graph of $y = \sqrt[3]{x}$ has the tangent line $x = 0$ at the origin, so $f^{'} (0) = \infty$ and $f$ is differentiable at $0$ .”
Find every $x$ at which $k (x) = | x + 2 | + \sqrt[3]{x - 4}$ is not differentiable, and name the shape of the graph at each one.

Show the answers

False. $| x |$ is continuous at $0$ and has a corner there. Continuity is necessary for a derivative, not sufficient.
True. It is the contrapositive of “differentiable implies continuous”: if $f^{'} (a)$ existed, $f$ would have to be continuous at $a$ .
No: a corner. For $h > 0$ the quotient is $\frac{h + h}{h} = 2$ ; for $h < 0$ it is $\frac{- h + h}{h} = 0$ . The one-sided slopes are $0$ and $2$ , finite and different.
A cusp. The quotient is $\frac{\sqrt{| h |}}{h}$ , which equals $\frac{1}{\sqrt{h}} \to + \infty$ for $h > 0$ and $- \frac{1}{\sqrt{- h}} \to - \infty$ for $h < 0$ . The absolute value suggests a corner, but the limits run to opposite infinities.
Yes, $f^{'} (0) = 0$ . The quotient is $\frac{h^{4 / 3}}{h} = h^{1 / 3}$ , which tends to $0$ from both sides. A power greater than $1$ flattens the graph at the origin instead of sharpening it.
A derivative has to be a real number. The quotient $\frac{1}{h^{2 / 3}}$ grows without bound, so the limit does not exist and $f$ is not differentiable at $0$ . The tangent line exists, but it is vertical, and a vertical line has no slope.
A corner at $x = - 2$ and a vertical tangent at $x = 4$ . At $- 2$ the cube-root term has the finite slope $\frac{1}{3 \sqrt[3]{36}}$ , while $| x + 2 |$ adds $- 1$ from the left and $1$ from the right, so the one-sided slopes are finite and differ by $2$ . At $4$ , $| x + 2 |$ has slope $1$ , but the cube-root term contributes $\frac{1}{h^{2 / 3}} \to + \infty$ on both sides, so the whole quotient does too. Everywhere else both terms are differentiable.

To practise the definition itself (one-sided difference quotients at a corner, a vertical tangent at the end of a domain, and why the domain of $f^{'}$ can be smaller than the domain of $f$ ), work through the free CEGEP Calculus I worksheet on the derivative.

Want one-on-one help with this? I offer CEGEP Science calculus and linear algebra tutoring in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.