Continuity Is Necessary, but Not Enough
A graph can have no break at a point and still fail to have a well-defined tangent slope there. This is the key distinction between continuity and differentiability: continuity describes how function values behave near a point, while differentiability requires a finite, consistent limiting slope.
When deciding whether a derivative exists, continuity is an important check, but it is not a complete test. A discontinuous function cannot be differentiable at the point; a continuous function may or may not be differentiable there. The difference quotient reveals what continuity alone cannot.
Definitions and the One-Sided Test
Let \(f\) be defined on an open interval containing \(a\). The function is continuous at \(a\) if
The function is differentiable at \(a\) if the following limit exists as a finite real number:
The expression inside the limit is the difference quotient. For an interior point, the two-sided limit exists exactly when the left-hand and right-hand limits both exist as finite numbers and are equal. These limits are called the left-hand and right-hand derivatives.
Thus, a function’s values can fit together continuously while its slopes from the two sides disagree. Or the difference quotient may grow without bound rather than approach a finite value. Either behavior prevents differentiability.
Why Differentiability Implies Continuity
This implication explains why a discontinuity is decisive evidence against differentiability. Suppose \(f'(a)\) exists. For every nonzero \(h\) sufficiently close to zero, rearrange the difference quotient:
As \(h\to 0\), the first factor on the right approaches \(0\), and the factor in parentheses approaches the finite number \(f'(a)\). By the product limit law, their product approaches \(0\). Therefore \(f(a+h)-f(a)\to 0\), which means \(f(a+h)\to f(a)\). This is continuity at \(a\).
The proof gives a useful order for checking a claim. First, if the function is discontinuous at the point, conclude that it is not differentiable there. If it is continuous, do not stop: evaluate the difference quotient or compare the one-sided slopes.
Worked Example: Continuous at a Join, but Unequal Slopes
Worked Example: Continuous at a Join, but Unequal Slopes
Consider the function
Determine whether \(f\) is differentiable at \(0\).
Step 1: Check continuity. Since \(f(0)=0\), the left-hand limit is
The right-hand limit is
Both one-sided limits equal \(f(0)\), so \(f\) is continuous at \(0\).
Step 2: Compute the one-sided difference quotients. For \(h<0\), \(f(h)=h\), so
For \(h>0\), \(f(h)=2h\), so
The left-hand derivative is \(1\), and the right-hand derivative is \(2\). They are unequal, so the two-sided derivative does not exist. The function is continuous but not differentiable at \(0\). As a check, the two branches meet at the same function value, but their slopes, \(1\) and \(2\), do not match.
Worked Example: A Continuous Function with an Unbounded Quotient
Worked Example: A Continuous Function with an Unbounded Quotient
Let \(g(x)=\sqrt{|x|}\). Test continuity and differentiability at \(0\).
Step 1: Check continuity. We have \(g(0)=0\). As \(x\to 0\), \(|x|\to 0\), so \(\sqrt{|x|}\to 0=g(0)\). Thus \(g\) is continuous at \(0\).
Step 2: Form the difference quotient. For \(h\ne 0\),
When \(h>0\), this becomes \(1/\sqrt{h}\), which grows without bound as \(h\to 0^+\). When \(h<0\), \(h=-|h|\), so the quotient becomes \(-1/\sqrt{|h|}\), which decreases without bound as \(h\to 0^-\). Neither one-sided limit is a finite real number. Therefore \(g'(0)\) does not exist.
Check: The function values approach \(0\), confirming continuity, but the quotients do not approach any finite slope. A limit of function values and a limit of slopes are different tests.
Worked Example: A Discontinuity Rules Out a Derivative
Worked Example: A Discontinuity Rules Out a Derivative
Define
Can \(q\) be differentiable at \(1\)?
Step 1: Check the function value and one-sided limits. Since \(1\geq 1\), \(q(1)=1\). But
The left-hand limit does not equal \(q(1)\), so \(q\) is discontinuous at \(1\).
Step 2: Apply the necessary condition. A function differentiable at \(1\) would have to be continuous there. Since \(q\) is not continuous at \(1\), it cannot be differentiable there.
Check using the quotient: For \(h<0\),
As \(h\to 0^-\), this expression grows without bound. For \(h>0\), it equals \((1-1)/h=0\). The difference quotient cannot have a finite two-sided limit, consistent with the discontinuity test.
A Common Mistake: Treating Continuity as a Differentiability Test
A tempting but invalid argument is: “The graph has no gap at \(a\), so the derivative exists at \(a\).” That conclusion confuses a condition that is necessary with one that is sufficient.
The correction is to calculate the quotient from the definition. For \(h>0\), \(|h|/h=1\); for \(h<0\), \(|h|/h=-1\). The one-sided limits are unequal, so \(f'(0)\) does not exist. Continuity verifies that the function value fits the nearby values; it does not guarantee a single limiting slope.
A Reliable Check for a Differentiability Claim
When someone claims that a function is differentiable at an interior point, separate the continuity question from the slope question.
Compare the limit of the function at the point with its actual value. If they differ, differentiability is impossible.
Use the definition of the derivative, paying attention to the left-hand and right-hand limits.
The two one-sided limits must both be finite and equal. Continuity alone does not meet this requirement.
This approach prevents two opposite errors: claiming a derivative where a discontinuity rules it out, and claiming a derivative merely because the function is continuous. It also clarifies what evidence is needed: continuity is the first requirement, while the difference quotient decides whether a finite slope exists.
Check Your Understanding
For each question, distinguish what continuity establishes from what the difference quotient must establish.
- State the implication connecting differentiability and continuity. Does the converse follow?
- A function is continuous at \(a\), but its left-hand and right-hand difference-quotient limits are \(3\) and \(5\). Does its derivative exist at \(a\)? Explain.
- Why does a discontinuity at an interior point rule out differentiability there?
- For \(r(x)=\sqrt{|x|}\), what happens to the difference quotient from the right as \(h\to 0^+\)?
- What additional test is needed after confirming that a function is continuous at a point?
