Yes, a function is continuous at a cusp. In fact, continuity is a prerequisite for a cusp to exist. A cusp is one of the classic examples of a point where a function is continuous but not smooth, meaning you can draw the graph through it without lifting your pencil, but the curve comes to a sharp, pointed tip rather than flowing smoothly.
Why a Cusp Is Continuous
Continuity at a point means three things are true: the function has a value at that point, the limit of the function exists as you approach that point, and those two values match. A cusp satisfies all three conditions. The graph has no gap, no jump, and no hole at the cusp. Both sides of the curve meet at exactly the same point.
Think of the “pencil test” often used to describe continuity. If you can trace the entire graph without picking up your pencil, the function is continuous. At a cusp, your pencil stays on the paper the whole time. The curve simply changes direction in a dramatic way, forming a sharp point where two branches meet.
What Makes a Cusp Different From a Smooth Point
The distinction is about the derivative, not the function’s value. At a smooth point on a curve, the slope changes gradually and the derivative exists. At a cusp, the slope from the left side and the slope from the right side both shoot off toward infinity, but in opposite directions. For a function f at a cusp located at x = c, one of these two things happens:
- The derivative approaches positive infinity from one side and negative infinity from the other
- Or the reverse: negative infinity from the left and positive infinity from the right
This means the derivative does not exist at the cusp. The function is continuous there, but it is not differentiable. Continuity and differentiability are related but separate properties. Every differentiable function is continuous, but a continuous function is not necessarily differentiable. Cusps are one of the clearest illustrations of that gap.
The Classic Example: f(x) = x^(2/3)
The function f(x) = x^(2/3) has a cusp at x = 0. If you graph it, you’ll see a smooth U-shaped curve that pinches down to a sharp point at the origin. The function is perfectly continuous at x = 0 because the limit as x approaches 0 equals f(0) = 0. No gap, no jump.
But look at the derivative: f'(x) = (2/3)x^(-1/3), which equals 2 divided by 3 times the cube root of x. As x approaches 0 from the left, the derivative plunges toward negative infinity. As x approaches 0 from the right, it climbs toward positive infinity. The slopes on either side are racing in opposite directions, making it impossible to assign a single tangent line at that point. That’s the signature of a cusp.
Cusps vs. Corners vs. Vertical Tangents
Cusps are sometimes confused with corners and vertical tangents, but they’re distinct geometric features. All three are points where a continuous function fails to be differentiable, yet they fail for different reasons.
At a corner, the derivative has two different finite values on either side. Think of the absolute value function f(x) = |x| at x = 0. The slope is -1 coming from the left and +1 from the right. The slopes disagree, but they’re ordinary numbers, not infinite. The result is a sharp angle rather than a pointed tip.
At a vertical tangent, the derivative approaches the same infinity on both sides. Both sides shoot toward positive infinity, or both toward negative infinity. The curve passes through the point going straight up. The function f(x) = x^(1/3) at x = 0 is the standard example: the cube root function is continuous there, and the slope approaches positive infinity from both directions, producing a vertical tangent rather than a cusp.
At a cusp, the derivative approaches opposite infinities from each side. One side goes to positive infinity while the other goes to negative infinity. This creates the characteristic pointed shape where two branches of the curve meet with matching tangent directions but from opposite orientations. The key distinction is that opposition between the two sides.
The Formal Continuity Check
If you need to prove continuity at a cusp using the formal epsilon-delta definition, the process is the same as proving continuity at any other point. You need to show that for every small positive number epsilon, you can find a distance delta such that whenever x is within delta of the cusp point c, the function value f(x) stays within epsilon of f(c).
For x^(2/3) at x = 0, this works out cleanly. As x gets closer to 0, x^(2/3) gets closer to 0 as well, from both sides. There’s no ambiguity about the limit, no disagreement between left and right. The function value and the limit match perfectly. The cusp creates drama in the derivative, not in the function itself.
Why This Matters in Calculus
Understanding that cusps are continuous but not differentiable matters whenever you apply theorems that require differentiability. The Mean Value Theorem, for instance, requires that a function be differentiable on an open interval. If your function has a cusp inside that interval, the theorem doesn’t apply. Continuity alone isn’t enough.
Similarly, when finding local extrema using the first derivative, cusps need special attention. Because the derivative doesn’t exist at a cusp, that point becomes a critical point by default. You can’t set the derivative equal to zero and solve, since the derivative simply isn’t defined. Instead, you analyze the behavior of the derivative on either side to determine whether the cusp is a local maximum, a local minimum, or neither. For x^(2/3), the cusp at x = 0 is actually a local minimum, since the function decreases on the left and increases on the right.