Why This Implication Matters
Differentiability describes how a function changes near a point through its difference quotients. Continuity describes whether the function’s values approach its value at that point. These ideas are closely related, but they are not equivalent: differentiability guarantees continuity, while continuity by itself does not guarantee differentiability.
The implication is useful when justifying that a function is continuous at a point. If its derivative exists there, continuity follows. The justification depends on the definition of the derivative, not on a graph looking smooth or on a derivative formula alone.
Definitions at an Interior Point
Let \(f\) be defined on an open interval containing \(a\). The point \(a\) is an interior point of the domain if values of \(f\) are defined both to the left and to the right of \(a\), at least for inputs sufficiently close to \(a\).
The function \(f\) is continuous at \(a\) if
Continuity asks whether nearby function values approach the actual value at \(a\). Differentiability asks whether the slopes of secant lines from \((a,f(a))\) to nearby points approach one finite number. The proof below shows why the second condition forces the first.
The Limit-Based Justification
For \(x\ne a\), the difference between \(f(x)\) and \(f(a)\) can be written as
If \(f\) is differentiable at \(a\), the first factor approaches the finite number \(f'(a)\) as \(x\to a\). The second factor approaches zero. Their product therefore approaches zero, which means \(f(x)\to f(a)\). Here is a complete justification using the definition of a limit.
Let \(L=f'(a)\). By differentiability, \(L\) is finite and the difference quotient approaches \(L\) as \(x\to a\).
Because the quotient approaches \(L\), there is a number \(\rho>0\) such that \(0<|x-a|<\rho\) implies \(\left|\frac{f(x)-f(a)}{x-a}-L\right|<1\). The triangle inequality then gives \(\left|\frac{f(x)-f(a)}{x-a}\right|<|L|+1\).
For any \(\varepsilon>0\), set \(C=|L|+1\), which is positive, and choose \(\delta=\min(\rho,\varepsilon/C)\). If \(0<|x-a|<\delta\), then \(|f(x)-f(a)|=\left|\frac{f(x)-f(a)}{x-a}\right||x-a|<C\delta\le\varepsilon\).
At \(x=a\), \(|f(x)-f(a)|=0<\varepsilon\). Thus all inputs sufficiently close to \(a\), including \(a\), have function values within \(\varepsilon\) of \(f(a)\). This is continuity at \(a\).
The same reasoning can be summarized using limit laws: the difference quotient tends to \(f'(a)\), while \(x-a\) tends to zero, so their product tends to zero. The detailed argument above explains why the quotient remains bounded near \(a\), which is the key fact behind that product limit.
Worked Example: A Polynomial at a Point
Worked Example: A Polynomial at a Point
Let \(f(x)=3x^2-2x+1\). Verify differentiability at \(a=1\) from the difference quotient, then identify the continuity conclusion.
First evaluate the function at the point:
For \(x\ne1\), form the difference quotient:
Therefore,
The derivative exists and is finite, so the theorem guarantees continuity at \(1\). Directly, \(\lim_{x\to1}f(x)=3(1)^2-2(1)+1=2=f(1)\). As a check, the polynomial derivative is \(f'(x)=6x-2\), and \(f'(1)=6-2=4\), matching the difference-quotient result.
Worked Example: A Rational Function at a Point
Worked Example: A Rational Function at a Point
Let \(g(x)=\frac{x+2}{x+1}\), and consider \(a=1\). Since the denominator is nonzero at \(1\), the function is defined in an interval around that point. Compute the derivative from its definition.
First,
For \(x\ne1\), simplify the difference quotient:
Thus,
This finite derivative guarantees continuity at \(1\). Checking directly, \(\lim_{x\to1}g(x)=\frac{1+2}{1+1}=\frac32=g(1)\). The quotient derivative also agrees with the derivative formula \(g'(x)=-\frac{1}{(x+1)^2}\), which gives \(g'(1)=-\frac14\).
Why the Converse Fails
Continuity does not imply differentiability. A function may have no jump or hole at a point and still fail to have a single limiting secant slope there. The absolute-value function gives a direct example.
Worked Example: Continuous but Not Differentiable
Let \(h(x)=|x|\), and examine \(a=0\). Since \(h(0)=0\), and \(\bigl||x|-|0|\bigr|=|x|\), we have \(\lim_{x\to0}h(x)=0=h(0)\). Thus \(h\) is continuous at \(0\).
Now compute the difference quotient for \(x\ne0\):
If \(x>0\), then \(|x|=x\), so the quotient is \(1\). If \(x<0\), then \(|x|=-x\), so the quotient is \(-1\). Consequently,
The one-sided limits differ, so the two-sided derivative limit does not exist. Therefore \(h\) is continuous but not differentiable at \(0\). The check is decisive: continuity holds because the function values approach zero, whereas differentiability fails because the secant slopes do not approach one common value.
A Common Mistake: Dividing by a Quantity That Tends to Zero
A frequent error is to argue that continuity should imply differentiability because \(f(x)-f(a)\) tends to zero. The derivative quotient, however, divides that difference by \(x-a\), which also tends to zero. The resulting limit cannot be determined from the numerator’s limit alone.
The reverse mistake is to assume that failure of differentiability means failure of continuity. The same example rules that out: \(|x|\) is continuous at zero even though it is not differentiable there.
What the Result Does and Does Not Say
| Condition at an interior point \(a\) | Conclusion |
|---|---|
| \(f\) is differentiable at \(a\) | \(f\) is continuous at \(a\) |
| \(f\) is continuous at \(a\) | A derivative at \(a\) is not guaranteed |
| The derivative quotient has unequal one-sided limits | The two-sided derivative does not exist |
Differentiability at a point is a local condition: it depends on behavior arbitrarily close to that point. The implication established here is also pointwise. It does not claim that differentiability at one point guarantees continuity everywhere, nor does continuity at a point give information about derivatives at other points.
For a justification, identify the exact claim and its hypothesis. If the claim is that \(f\) is continuous at \(a\), and you know \(f'(a)\) exists as a finite two-sided limit, cite the definition of the derivative and show that the difference \(f(x)-f(a)\) is the product of the quotient and \(x-a\). If instead you are asked whether continuity proves differentiability, test the difference quotient; the absolute-value example shows why a universal implication cannot be made.
Check Your Understanding
For each question, distinguish carefully between what the function values do and what the difference quotients do.
- State the limit definition of differentiability at an interior point \(a\).
- In the proof that differentiability implies continuity, which factor tends to zero in the product expression for \(f(x)-f(a)\)?
- Why does the finite value of \(f'(a)\) help show that the difference quotient is bounded near \(a\)?
- For \(h(x)=|x|\), what are the left- and right-hand limits of the difference quotient at zero?
- Does continuity at a point guarantee differentiability there? Give a reason for your answer.
