Introduction
A function can be continuous at a point and still fail to have a finite derivative there. At a corner, the two one-sided slopes may approach different finite values. A cusp presents a different pattern: the slopes become unbounded, with opposite signs on the two sides.
The standard example is \(f(x)=|x|^{2/3}\) at \(x=0\). Its graph meets at the origin, but the difference quotients from the right grow without bound while those from the left decrease without bound. Calculating these limits carefully shows both what the cusp means and why no finite derivative can exist at its tip.
Difference quotients at a point
For a function \(f\) defined near \(a\), its difference quotient at \(a\) is
The derivative \(f'(a)\) exists as a finite real number when the two-sided limit of this quotient as \(h\to 0\) exists and is finite. The right-hand limit examines positive \(h\), and the left-hand limit examines negative \(h\). These one-sided limits may instead diverge to \(+\infty\) or \(-\infty\); such divergence does not give a finite derivative.
This usage describes the behavior of the difference quotients at the point. It does not mean that every function with an unbounded difference quotient has a cusp: the opposing signs on the two sides are part of the pattern discussed here.
The model cusp: \(f(x)=|x|^{2/3}\)
Set \(f(x)=|x|^{2/3}\). Since \(f(0)=0\), the difference quotient at zero is
To determine its signs and limits, first consider positive and negative \(h\) separately. If \(h>0\), then \(|h|=h\), so
As \(h\to 0^+\), the positive cube root \(\sqrt[3]{h}\) approaches zero, so its reciprocal grows without bound:
If \(h<0\), then \(|h|=-h\), and the quotient can be written as
Here \(|h|>0\) approaches zero as \(h\to 0^-\). Therefore \(|h|^{-1/3}\) grows without bound, while its negative decreases without bound:
The right-hand and left-hand quotients diverge in opposite directions. Thus the difference quotient has no finite two-sided limit, and \(f'(0)\) does not exist as a finite real number.
Why these limits rule out a finite derivative
The conclusion follows directly from what it means for a difference quotient to have a finite limit. If a function \(Q(h)\) tends to a finite number \(L\) as \(h\to 0\), then for sufficiently small nonzero \(h\), its values must be close to \(L\). In particular, they must remain bounded. A quotient that grows beyond every positive bound on one side cannot converge to a finite number.
For the model function, the unboundedness can be verified with an explicit inequality. On the right, for any \(M>0\), choosing \(0<h<M^{-3}\) gives \(h^{-1/3}>M\). On the left, if \(0<|h|<M^{-3}\), then \(-|h|^{-1/3}<-M\). Thus the quotient exceeds every positive bound on the right and falls below every negative bound on the left.
This reasoning is a complete check of the two infinite limits. It also shows why a single finite candidate for \(f'(0)\) cannot fit the quotient values: sufficiently close to zero, the quotient is not even bounded.
Worked Example: Checking the real-power form on both sides
Worked Example: Checking the real-power form on both sides
For \(h\ne 0\), rational powers of negative numbers can be interpreted using real odd roots. Under this convention, \(h^{2/3}=(\sqrt[3]{h})^2=|h|^{2/3}\). Check the quotient identity for both signs of \(h\), including \(h=-8\).
For any \(h\ne 0\), the real cube root satisfies \(\sqrt[3]{h}=\operatorname{sgn}(h)|h|^{1/3}\), where \(\operatorname{sgn}(h)\) is \(1\) for positive \(h\) and \(-1\) for negative \(h\). Hence
On the other hand, using \(h=\operatorname{sgn}(h)|h|\),
because \(1/\operatorname{sgn}(h)=\operatorname{sgn}(h)\). Therefore
provided the negative fractional power uses the real odd-root convention. At \(h=8\), both sides equal \(1/2\): \(|8|^{2/3}/8=4/8=1/2\), and \(8^{-1/3}=1/2\). At \(h=-8\), both equal \(-1/2\): \(|-8|^{2/3}/(-8)=4/(-8)=-1/2\), and \((-8)^{-1/3}=1/\sqrt[3]{-8}=-1/2\).
Thus this identity is valid for either sign of \(h\). Splitting into positive and negative cases is useful because it makes the signs of the one-sided limits explicit, not because the identity fails for negative \(h\).
Worked Example: A positive multiple of the model cusp
Worked Example: A positive multiple of the model cusp
Let \(g(x)=2|x|^{2/3}\). Calculate the one-sided difference quotients at zero and determine whether \(g'(0)\) exists as a finite number.
First, \(g(0)=2|0|^{2/3}=0\). For \(h\ne0\),
For \(h>0\), this equals \(2h^{-1/3}\). For \(h<0\), it equals \(-2|h|^{-1/3}\). Therefore
To verify the divergence explicitly, let \(M>0\). If \(0<h<(2/M)^3\), then \(2h^{-1/3}>M\). If \(h<0\) and \(0<|h|<(2/M)^3\), then \(-2|h|^{-1/3}<-M\). Thus the limits have the claimed signs and magnitudes, so \(g'(0)\) is not finite.
As a direct numerical sign check, at \(h=1/8\) the quotient is \(2(1/8)^{-1/3}=2\cdot2=4\). At \(h=-1/8\) it is \(-2(1/8)^{-1/3}=-4\). The right quotient is positive and the left quotient is negative, consistent with the limiting calculation.
Worked Example: A translated cusp with a linear term
Worked Example: A translated cusp with a linear term
Consider \(p(x)=3+5(x-2)-4|x-2|^{2/3}\). Find the one-sided difference-quotient limits at \(x=2\), and decide whether \(p'(2)\) exists as a finite number.
At the point in question, \(p(2)=3+5(0)-4|0|^{2/3}=3\). Put \(h=x-2\). Then \(x=2+h\), and for \(h\ne0\),
For \(h>0\), the quotient is \(5-4h^{-1/3}\), which tends to \(-\infty\). For \(h<0\), it is \(5+4|h|^{-1/3}\), which tends to \(+\infty\). Hence
The finite constant \(5\) does not prevent either quotient from becoming unbounded; the term involving \(|h|^{-1/3}\) determines the divergence. For a sign check, take \(h=1/8\). The quotient is \(5-4(2)=-3\). For \(h=-1/8\), it is \(5+4(2)=13\). These values have the signs predicted by their respective limits. Since the one-sided quotients diverge, \(p'(2)\) does not exist as a finite real number.
A common sign mistake
A frequent error is to replace the negative-side quotient \(|h|^{2/3}/h\) with the positive quantity \(|h|^{-1/3}\). That drops the sign contributed by the negative denominator. For example, at \(h=-8\), the original quotient is \(4/(-8)=-1/2\), whereas the incorrect expression \(|h|^{-1/3}\) is \(1/2\).
The correct sign-explicit formula is
Equivalently, it is \(h^{-1/3}\) for every \(h\ne0\) when the real odd-root interpretation is used. Keeping track of the sign is essential: it is exactly what distinguishes the two opposing infinite limits.
Why the cusp pattern matters
At a corner with unequal finite one-sided slopes, the quotient has two finite limiting values that disagree. At the cusp examples here, the quotient does not approach finite values at all: it becomes unbounded, and its sign reverses across the point. The two cases therefore fail the finite-derivative test in different ways.
The method is direct: start from the difference quotient at the point, simplify separately for positive and negative increments, and evaluate each one-sided limit. Do not infer differentiability from the appearance of the graph alone. In particular, a graph that seems to meet smoothly from a distance may still have an unbounded difference quotient near its tip.
Check Your Understanding
Use the difference quotient at the specified point to answer each question.
- For \(f(x)=|x|^{2/3}\), write the difference quotient at zero separately for \(h>0\) and \(h<0\).
- What are the right- and left-hand limits of that quotient?
- Why does an infinite one-sided limit prevent the derivative from being a finite real number?
- For \(g(x)=-|x|^{2/3}\), what are the signs of the right- and left-hand infinite limits at zero?
- For \(p(x)=2+3(x-1)+|x-1|^{2/3}\), write the difference quotient at \(x=1\) using \(h=x-1\), and state its two one-sided limits.
