Introduction
A derivative at a point describes the limiting behavior of the function near that point, relative to its value there. This places an important restriction on where a derivative can exist: a function cannot have a finite derivative at an interior point where it is discontinuous.
This gives a useful test. Before trying to calculate a difference quotient, check whether the function is continuous at the point. If it is not, the differentiability question is already settled: the derivative does not exist there. This test is particularly effective for jump discontinuities and for removable discontinuities caused by a hole or a mismatched point value.
The continuity requirement
Let \(f\) be defined on an open interval containing \(c\). The function is continuous at \(c\) if the limit \(\lim_{x\to c}f(x)\) exists and equals \(f(c)\). In particular, continuity concerns both the values near \(c\) and the value assigned at \(c\).
The derivative at \(c\), when it exists, is the finite limit
The result established in the earlier material on why differentiability implies continuity is that differentiability at an interior point requires continuity there. We use its contrapositive: if \(f\) is discontinuous at \(c\), then \(f\) is not differentiable at \(c\).
This is a one-way test. Discontinuity rules out differentiability, but continuity alone does not establish that a derivative exists. If the function is continuous at \(c\), the test has not decided the derivative question; a separate analysis is needed.
Why the test works
The logic is the contrapositive of the differentiability-implies-continuity result. That result says that, under the stated interior-point condition, if the derivative exists, then the function is continuous. Negating both the hypothesis and conclusion gives: if the function is not continuous, then the derivative does not exist.
Confirm that \(f(c)\) is defined and that \(c\) is an interior point of the interval under consideration.
Compare the limit of \(f(x)\) as \(x\) approaches \(c\) with the actual value \(f(c)\). If needed, compare the left- and right-hand limits.
If continuity fails, conclude that \(f'(c)\) does not exist. There is no need to evaluate the difference quotient.
The conclusion does not depend on how simple a formula for \(f\) may be away from \(c\). Differentiability uses the function’s actual value at \(c\), as well as its nearby values. A formula that describes the function only for \(x\ne c\) cannot, by itself, establish differentiability at \(c\).
Jump discontinuities
A jump discontinuity occurs when the left-hand and right-hand limits at \(c\) both exist as finite numbers but are unequal. Since the two sides approach different values, the two-sided limit does not exist. The function is therefore discontinuous at \(c\), whatever finite value has been assigned there.
Worked Example: A piecewise function with a jump
Consider the function
Determine whether \(f'(0)\) exists.
First evaluate the function at the point. Since \(0\ge 0\), the second branch applies:
For \(x<0\), the formula is \(x+1\), so the left-hand limit is
For \(x>0\), the formula is \(2-x\), so the right-hand limit is
The one-sided limits are unequal: \(1\ne 2\). Thus \(\lim_{x\to 0}f(x)\) does not exist, and \(f\) is discontinuous at \(0\). By the discontinuity test, \(f'(0)\) does not exist. The assigned value \(f(0)=2\) agrees with the right-hand limit, but it cannot make the two-sided limit exist.
Check: The jump is confirmed by the left-hand value \(1\) and the right-hand value \(2\). The continuity requirement fails, so a derivative at \(0\) is ruled out.
Removable discontinuities and mismatched values
A removable discontinuity at \(c\) occurs when the finite limit \(\lim_{x\to c}f(x)\) exists, but the function is not continuous there. For a function defined at \(c\), this means that its assigned value differs from the limit. The nearby values approach one number, while the function’s value at the point is another.
Worked Example: A finite limit with the wrong point value
Define
Determine whether \(g'(1)\) exists.
For \(x\ne 1\), factor the numerator and simplify:
This simplification is valid only when \(x\ne 1\), which is exactly the condition on that branch. Therefore, the limit of \(g(x)\) as \(x\) approaches \(1\) is
But the actual value is \(g(1)=0\). Since \(2\ne 0\), the limit does not equal the function value, so \(g\) is discontinuous at \(1\). The discontinuity test gives \(g'(1)\) does not exist.
Check: Values near \(1\) are given by \(x+1\) and approach \(2\), while the defined value at \(1\) is \(0\). This verifies the mismatch responsible for the discontinuity.
The same test handles a removable discontinuity described as a hole. However, a distinction about the domain matters: a derivative at \(c\) is defined only if \(f(c)\) exists. If the point is excluded from the domain, there is no derivative of that function at \(c\) to test. If a value has been assigned at the point, then continuity can be checked by comparing that value with the limit.
Worked Example: A hole versus an assigned point value
Let \(q(x)=\dfrac{x^2-4}{x-2}\) for \(x\ne 2\). The point \(2\) is not in the domain of \(q\), so \(q'(2)\) is not defined for this function. To see the behavior near the hole, factor and simplify for \(x\ne 2\):
Consequently,
Now define a new function \(r\) by keeping the same formula away from \(2\) and assigning \(r(2)=1\). Its limit at \(2\) is still \(4\), because its values for \(x\ne 2\) have not changed. But \(r(2)=1\ne 4\), so \(r\) is discontinuous at \(2\). Since \(2\) is now in its domain, the discontinuity test applies and shows that \(r'(2)\) does not exist.
Check: The hole in \(q\) means its derivative at \(2\) is not defined because \(q(2)\) is missing. For \(r\), the value is defined but mismatched with the limit; that discontinuity rules out a derivative.
A common mistake: differentiating only the nearby formula
In the second example, the expression for \(g(x)\) simplifies to \(x+1\) whenever \(x\ne 1\). A tempting but incorrect argument is: “The derivative of \(x+1\) is \(1\), so \(g'(1)=1\).” This uses a derivative calculation for a formula that does not give the function’s actual value at \(1\).
The difference quotient shows the failure directly. For \(h\ne 0\), the point \(1+h\) is not \(1\), so the simplified branch applies there. Thus
As \(h\to 0\), the term \(2/h\) has no finite limit. Hence this quotient cannot approach a finite derivative. The calculation agrees with the continuity test: since \(g\) is discontinuous at \(1\), a derivative there cannot exist.
What the test does and does not establish
At an interior point in the domain, any failure of continuity is enough to rule out a finite derivative. Two common ways continuity fails are:
- Jump: the left- and right-hand limits are unequal, so the two-sided limit does not exist.
- Removable mismatch: a finite two-sided limit exists, but it differs from the assigned value at the point.
There are other forms of discontinuity as well, but the same implication applies: discontinuity at an interior point prevents differentiability there. The test does not calculate a derivative, and continuity does not guarantee that a derivative exists. It is a quick way to settle the question when continuity already fails.
Keep the domain condition in view. If \(c\) is not in the domain, \(f'(c)\) is not defined for that function. If \(c\) is in the domain but the function’s value disagrees with its limit, the function is discontinuous there and the derivative cannot exist. If the value agrees with the limit, this test alone gives no conclusion about differentiability.
Summary
Differentiability at an interior point requires continuity at that point. Therefore, discontinuity is a decisive test for nondifferentiability: a jump, a mismatched value, or any other failure of continuity rules out a finite derivative.
When applying the test, compare the function’s actual value with its nearby behavior. For a jump, inspect the one-sided limits. For a removable mismatch, calculate the finite limit and compare it with \(f(c)\). For a missing point, first note that the derivative at that point is not defined unless the function’s domain includes it.
Check Your Understanding
Use the continuity test where it applies. For each function, identify what information settles whether the derivative at the specified point can exist.
- A function has left-hand limit \(3\), right-hand limit \(5\), and is defined at \(c\) with \(f(c)=5\). Can \(f'(c)\) exist? Explain.
- A function has \(\lim_{x\to c}f(x)=7\) and \(f(c)=7\). What does the discontinuity test establish about \(f'(c)\)?
- A function is defined near \(c\) except at \(c\), where it has a hole. Is its derivative at \(c\) defined for this function?
- For \(p(x)=(x^2-16)/(x-4)\) when \(x\ne 4\), with \(p(4)=3\), find the limit at \(4\) and decide whether \(p'(4)\) exists.
- Why is it not enough to differentiate a simplified formula that holds only for \(x\ne c\) when deciding whether a derivative exists at \(c\)?
