Continuity Does Not Guarantee Differentiability
A function can have no break or gap at a point and still fail to have a finite derivative there. Continuity concerns the values of the function near the point; differentiability concerns the behavior of the slopes of nearby secant lines. These are related ideas, but they are not equivalent.
The earlier result that differentiability implies continuity gives one direction: if a function is differentiable at a point, it must be continuous there. The reverse direction does not always hold. Absolute-value and fractional-power functions provide simple examples of continuous functions whose difference quotients do not approach a finite number.
Continuity and the Derivative at a Point
The expression \(\frac{f(a+h)-f(a)}{h}\), for \(h\ne 0\), is the difference quotient. It measures the slope of a secant line through the points with inputs \(a\) and \(a+h\). Differentiability asks whether these slopes approach one finite value as \(h\) approaches zero from either direction.
Continuity asks a different question. It asks whether \(f(a+h)\) approaches \(f(a)\). There is no requirement in the definition of continuity that the change in function value, divided by the change in input, settle to a finite number. The function values can approach \(f(a)\) while the difference quotients do not approach any finite limit.
The Absolute-Value Function at Zero
Worked Example: Continuous but Not Differentiable at the Origin
Consider \(f(x)=|x|\) at \(x=0\).
First, \(f(0)=|0|=0\). For any \(x\), \(\bigl||x|-0\bigr|=|x|\), and \(|x|\to 0\) as \(x\to 0\). Therefore \(\lim_{x\to 0}f(x)=0=f(0)\), so \(f\) is continuous at zero.
Now compute the difference quotient at zero. Since \(f(h)=|h|\),
If \(h>0\), then \(|h|=h\), so the quotient is \(1\). If \(h<0\), then \(|h|=-h\), so the quotient is \(-1\). These values do not approach a common finite number as \(h\to 0\). Thus \(f'(0)\) does not exist.
For an explicit check, take \(h=1/n\) and \(h=-1/n\), where \(n\) is a positive integer. Both inputs approach zero as \(n\) increases, but the corresponding quotients are \(1\) and \(-1\). The function is continuous at zero, yet it is not differentiable there.
The key point is not simply that the formula for the function contains absolute-value symbols. The test is the limit of the difference quotient at the point in question. For this example, the quotient produces incompatible values for positive and negative increments.
Fractional Powers Can Also Fail to Have a Finite Derivative
Worked Example: A Fractional Power at Zero
Consider \(g(x)=x^{2/3}\), interpreted for real \(x\) as \((\sqrt[3]{x})^2=|x|^{2/3}\). This function is defined for every real number. At zero, \(g(0)=0\). Because \(|x|^{2/3}\to 0\) as \(x\to 0\), \(g\) is continuous at zero.
The difference quotient at zero is
For \(h>0\), this equals \(h^{2/3}/h=h^{-1/3}=1/h^{1/3}\). As \(h\) approaches zero through positive values, this quotient grows without bound. For \(h<0\), it equals \(|h|^{2/3}/(-|h|)=-|h|^{-1/3}\), which decreases without bound in the negative direction.
For instance, if \(h=1/n\), the quotient is \(n^{1/3}\); if \(h=-1/n\), it is \(-n^{1/3}\). These values do not approach a finite real number. Therefore \(g'(0)\) does not exist, although \(g\) is continuous there.
Check the substitutions: \(g(0)=0\), and for either sign of \(h\), \(g(h)=|h|^{2/3}\). Dividing by \(h\) gives exactly the two behaviors above.
This example shows why it is important to check the whole difference quotient rather than infer differentiability from how gently the function values seem to approach the point. The function values approach zero, while the quotient of the change in function value by the change in input becomes unbounded.
How the Exponent Affects the Difference Quotient
The same calculation works for the family \(F(x)=|x|^\alpha\), where \(\alpha>0\). The value at zero is \(F(0)=0\), and \(|x|^\alpha\to 0\) as \(x\to 0\), so every function in this family is continuous at zero. Its difference quotient there is
For positive \(h\), this is \(h^{\alpha-1}\). For negative \(h\), writing \(|h|=-h\) shows it is \(-|h|^{\alpha-1}\). The exponent \(\alpha-1\) determines what happens to the quotient’s size as \(|h|\) approaches zero.
To justify each case, begin with the quotient for \(h\ne 0\). Its magnitude is \(|h|^{\alpha-1}\).
Then \(\alpha-1<0\), so \(|h|^{\alpha-1}=1/|h|^{1-\alpha}\) grows without bound as \(|h|\to 0\). The quotient cannot approach a finite real number.
Then \(F(x)=|x|\). The quotient is \(1\) for positive \(h\) and \(-1\) for negative \(h\), so it has no limit.
Then \(\alpha-1>0\), so \(|h|^{\alpha-1}\to 0\). The absolute value of the quotient approaches zero from either direction, which proves the quotient itself approaches zero and \(F'(0)=0\).
This derivation identifies a threshold: all positive exponents give continuity at zero, but only exponents greater than one give a finite derivative there. An exponent between zero and one makes the quotient unbounded; an exponent of one leaves two different finite values.
Worked Example: Comparing Two Fractional Powers
Compare \(u(x)=|x|^{1/2}\) and \(v(x)=|x|^{3/2}\) at zero. Both have value zero there, and both are continuous at zero because their exponents are positive.
For \(u\), the difference quotient is \(|h|^{1/2}/h\). Its magnitude is \(|h|^{-1/2}=1/\sqrt{|h|}\), which grows without bound as \(h\to 0\). More explicitly, it is \(1/\sqrt{h}\) for \(h>0\) and \(-1/\sqrt{|h|}\) for \(h<0\). Therefore \(u\) has no finite derivative at zero.
For \(v\), the difference quotient is \(|h|^{3/2}/h\). Its magnitude is \(|h|^{1/2}\), which approaches zero. The quotient is positive for \(h>0\) and negative for \(h<0\), but its magnitude tends to zero from both directions. Thus \(v'(0)=0\).
As a check, the general exponent calculation gives \(\alpha=1/2<1\) for \(u\), predicting no finite derivative, and \(\alpha=3/2>1\) for \(v\), predicting derivative zero. The direct quotient calculations confirm both results.
A Common Mistake: Using Only Positive Increments
A frequent error is to compute the difference quotient for \(f(x)=|x|\) at zero using only \(h>0\), then claim that the derivative is \(1\). That calculation is correct for positive \(h\), but it does not establish the limit as \(h\to 0\), which allows \(h\) to approach zero from either direction.
The conclusion does not follow because negative increments have not been considered. For \(h<0\), \(|h|=-h\), so \(\frac{|h|}{h}=-1\). The corrected conclusion is that the difference quotient has no single limit, and \(f'(0)\) does not exist.
Keep the distinction clear: checking just one direction may reveal part of a limit’s behavior, but a derivative at an interior point requires one finite limit as \(h\) approaches zero from both directions.
Why the Distinction Matters
Continuity describes whether function values fit together at a point; differentiability describes whether a finite local rate of change exists there. A continuous function therefore need not have a well-defined instantaneous rate of change at every point. When testing a proposed derivative, start from the difference quotient and examine its limit rather than relying only on continuity or on the appearance of the function’s formula.
Check Your Understanding
Use the definitions of continuity and derivative to answer these questions.
- For \(f(x)=|x|\), what is \(f(0)\), and why is \(f\) continuous at zero?
- What values does the difference quotient for \(f(x)=|x|\) at zero take when \(h>0\) and when \(h<0\)?
- Why is \(x^{2/3}\), interpreted as \(|x|^{2/3}\), continuous at zero but not differentiable there?
- For \(F(x)=|x|^\alpha\) with \(0<\alpha<1\), what happens to the magnitude of the difference quotient as \(h\to 0\)?
- For \(F(x)=|x|^{3/2}\), what is the derivative at zero, and how does the difference quotient verify it?
