Why Classify the Failure?
A derivative can fail to exist for several different reasons. A graph may have a jump or a hole, or it may be continuous but have a sharp corner, a cusp, or a vertical tangent. The difference quotient can also oscillate without approaching any limit.
These cases should not be treated as interchangeable. In particular, an infinite one-sided difference-quotient limit does not by itself prove that a graph has a cusp or a vertical tangent. First check whether the function is continuous at the point.
The Difference Quotient and the First Test
For a function defined at \(a\), its derivative is the finite limit
provided this two-sided limit exists. The values of \(h\) used in the limit must keep \(a+h\) in the domain. The left-hand difference quotient uses \(h<0\), and the right-hand difference quotient uses \(h>0\).
Continuity at \(a\) means that \(f(a)\) is defined and \(\lim_{x\to a}f(x)=f(a)\). A function that is differentiable at \(a\) must be continuous there. This fact explains why continuity is the first test, rather than a final check.
Why Differentiability Requires Continuity
Suppose the derivative limit exists and equals the finite number \(L\). For \(h\ne0\) near zero, factor the change in the function value as
As \(h\to0\), the first factor tends to zero and the second tends to \(L\). Therefore their product tends to \(0\), so \(f(a+h)-f(a)\to0\). Equivalently, \(f(a+h)\to f(a)\), which is continuity at \(a\). Thus a jump or a removable discontinuity rules out differentiability before any slope classification is attempted.
A Classification Guide
After checking continuity, use the one-sided difference quotients. The following names describe common graph-based situations. A derivative that tends to infinity is not a finite derivative; the terms “cusp” and “vertical tangent” describe the local graph behavior when the function is continuous.
| Behavior at \(a\) | Typical classification |
|---|---|
| Finite one-sided function limits differ | Jump discontinuity; not differentiable |
| A finite function limit exists but does not equal the defined value | Removable discontinuity; not differentiable |
| Function is continuous; finite one-sided difference-quotient limits differ | Corner |
| Function is continuous; one-sided difference quotients tend to opposite infinities | Cusp, in the usual graph-based classification |
| Function is continuous; both one-sided difference quotients tend to the same infinity | Vertical tangent, in the usual graph-based classification |
| Function is continuous; the difference quotient has no limit because it oscillates | Oscillating difference quotient |
This guide is not a substitute for checking the conditions. In particular, infinite difference-quotient limits only support the cusp or vertical-tangent classifications after continuity at the point has been established.
Worked Example: A Jump Discontinuity
Worked Example: A Jump Discontinuity
Let \(f(x)=-1\) for \(x<0\), \(f(x)=2\) for \(x\ge 0\). Classify the failure of the derivative at \(x=0\).
The function value is \(f(0)=2\). As \(x\to0^-\), \(f(x)=-1\), so \(\lim_{x\to0^-}f(x)=-1\). As \(x\to0^+\), \(f(x)=2\), so \(\lim_{x\to0^+}f(x)=2\). The one-sided limits differ, so the function has a jump discontinuity at zero.
The difference quotient also confirms that no finite derivative exists. For \(h<0\), \(f(h)=-1\), and therefore
As \(h\to0^-\), this tends to \(+\infty\). For \(h>0\), \(f(h)=2\), so
The right-hand quotient tends to \(0\). These one-sided behaviors do not combine into a finite two-sided derivative. The final classification is a jump discontinuity, not a corner: the function is not continuous at zero.
Check: The left and right function limits are \(-1\) and \(2\), respectively, so they are unequal; this verifies the jump classification.
Worked Example: A Removable Discontinuity
Worked Example: A Removable Discontinuity
Let \(f(x)=1\) for \(x\ne0\), and let \(f(0)=0\). Classify the failure at zero.
For every \(x\ne0\), \(f(x)=1\), so \(\lim_{x\to0}f(x)=1\). But \(f(0)=0\), and the limit is not equal to the function value. The discontinuity is removable: changing the value at zero to \(1\) would make the function continuous there.
For \(h\ne0\), the difference quotient is
As \(h\to0^+\), this tends to \(+\infty\); as \(h\to0^-\), it tends to \(-\infty\). This does not make the point a cusp. The function is discontinuous at zero, so it fails the necessary condition for differentiability and the continuous-graph cusp classification.
Check: The nearby values tend to \(1\), while the specified value is \(0\); that mismatch verifies the removable discontinuity.
A removable discontinuity depends on the value assigned at the point. If the value in this example were changed to \(f(0)=1\), the function would instead be constant and its difference quotient would be \(0\) for every nonzero \(h\). The filled point would then be differentiable. A hole that is not part of the function’s domain has no ordinary derivative at that point because the function is not defined there.
Worked Example: A Corner
Worked Example: A Corner
Let \(f(x)=|x|\). Determine the behavior at zero using the difference quotient.
First, \(f(0)=0\), and \(\lim_{x\to0}|x|=0\), so the function is continuous at zero. For \(h>0\), \(|h|=h\), and
The right-hand quotient tends to \(1\). For \(h<0\), \(|h|=-h\), and
The left-hand quotient tends to \(-1\). Both limits are finite but unequal, so the two-sided derivative does not exist. Since the function is continuous, the graph has a corner at zero.
Check: Substituting \(h>0\) and \(h<0\) into the quotient gives exactly \(1\) and \(-1\), confirming the unequal one-sided slopes.
Worked Example: A Cusp
Worked Example: A Cusp
Let \(f(x)=|x|^{2/3}\). Classify its behavior at zero.
We have \(f(0)=0\), and \(|x|^{2/3}\to0\) as \(x\to0\), so the function is continuous. For \(h>0\),
For \(h<0\), the numerator is \(|h|^{2/3}\), which is positive, while \(h\) is negative. Thus
The function is continuous, and the one-sided difference quotients diverge with opposite signs. Under the usual graph-based classification, this is a cusp. There is no finite derivative at zero.
Check: The magnitude of each quotient is \(1/|h|^{1/3}\), which grows without bound as \(h\to0\); its sign is positive on the right and negative on the left.
Worked Example: A Vertical Tangent
Worked Example: A Vertical Tangent
Let \(f(x)=\sqrt[3]{x}\). Determine whether the graph has a vertical tangent at zero.
The function is defined at zero with \(f(0)=0\), and \(\sqrt[3]{x}\to0\) as \(x\to0\), so it is continuous there. For \(h\ne0\),
The last expression is positive for both signs of \(h\), and it tends to \(+\infty\) as \(h\to0\). The function is continuous and the two one-sided quotient limits have the same infinite sign, so the graph has a vertical tangent at zero in the usual classification. The derivative is not a finite real number.
Check: Squaring the real cube root gives \((\sqrt[3]{h})^2=|h|^{2/3}\), so the quotient is positive on both sides as calculated.
Worked Example: An Oscillating Difference Quotient
Worked Example: An Oscillating Difference Quotient
Define \(f(0)=0\) and \(f(x)=x\sin(1/x)\) for \(x\ne0\). Decide whether \(f'(0)\) exists.
Since \(|\sin(1/x)|\le1\), we have \(|f(x)|\le|x|\). As \(x\to0\), the bound tends to zero, so \(f(x)\to0=f(0)\). The function is continuous at zero.
For \(h\ne0\), the difference quotient simplifies to
To show that this expression has no limit, consider the positive sequences
Both sequences tend to zero. But \(\sin(1/h_n)=1\), while \(\sin(1/k_n)=-1\). Therefore the quotient has different values along sequences approaching zero and cannot have a limit. The derivative does not exist because the difference quotient oscillates.
Check: Substituting the sequence definitions gives \(1/h_n=\pi/2+2\pi n\) and \(1/k_n=3\pi/2+2\pi n\), whose sines are \(1\) and \(-1\), respectively.
A Reliable Classification Process
Use this order whenever a derivative appears to fail at a point. It prevents a discontinuity from being mislabeled as a cusp or vertical tangent.
Confirm that \(f(a)\) is defined. If it is not, the ordinary derivative at \(a\) is not defined.
Compare \(\lim_{x\to a}f(x)\) with \(f(a)\), considering one-sided limits if needed. A jump or removable discontinuity already rules out differentiability.
Compute \(\frac{f(a+h)-f(a)}{h}\) for \(h\ne0\), simplifying separately for \(h<0\) and \(h>0\) when the formula changes by side.
For a continuous function, unequal finite limits indicate a corner; opposite infinite limits indicate a cusp; same-sign infinite limits indicate a vertical tangent in the usual graph-based classification. If the quotient oscillates, demonstrate that it has no limit.
Common Mistake: Classifying Infinite Limits Before Checking Continuity
A tempting but incorrect shortcut is: “Opposite infinite quotient limits mean a cusp, and matching infinite quotient limits mean a vertical tangent.” Those descriptions need the function to be continuous at the point.
For example, the removable-discontinuity function above has quotient \(1/h\), which tends to \(-\infty\) from the left and \(+\infty\) from the right. Calling that a cusp would be wrong because \(f(0)=0\) while \(\lim_{x\to0}f(x)=1\). The graph is discontinuous, not a continuous cusp.
Likewise, define \(g(0)=0\) and \(g(x)=1/x\) for \(x\ne0\). Its quotient at zero is
from both sides. Nevertheless, \(g\) is discontinuous at zero, so this calculation does not establish a vertical tangent there. The correction is to check continuity first, then classify the one-sided quotient behavior.
Why This Classification Matters
The classification tells you what information a graph or formula provides and which test to use next. A jump or hole is a failure of continuity. A corner, cusp, vertical tangent, or oscillating quotient can occur even when the function is continuous, but each has a different one-sided difference-quotient pattern.
For AP Calculus problems, showing the relevant limits is more convincing than naming a shape from a sketch alone. A graph may suggest a corner or cusp, but the difference quotient establishes the behavior at the specified point. Always state whether continuity holds and show the one-sided calculations supporting the classification.
Check Your Understanding
For each question, identify the continuity status before deciding what the difference quotient implies.
- If the left and right function limits at \(a\) are finite but unequal, what kind of discontinuity occurs, and can \(f'(a)\) exist?
- A continuous function has left and right difference quotients tending to \(-4\) and \(2\). What is the usual graph-based classification?
- Why is continuity required before classifying opposite-sign infinite quotient limits as a cusp?
- For \(f(0)=0\) and \(f(x)=1\) when \(x\ne0\), identify the discontinuity and calculate the difference quotient at zero.
- If the difference quotient has value \(1\) along one sequence approaching zero and \(-1\) along another, can it have a limit? Explain.
