From a Finite Slope to Continuity
Differentiability and continuity describe different features of a function near a point. Continuity says that nearby function values approach the value at the point. Differentiability says that the change in function value, divided by the change in input, approaches a finite slope.
The difference quotient connects these ideas directly. The change in function value is the input increment multiplied by the difference quotient. If the quotient approaches a finite number, it stays bounded near the point; multiplying that bounded quantity by an increment tending to zero forces the function's change to tend to zero as well.
Definitions at a Point
Let \(f:D\to\mathbb{R}\), where \(D\subseteq\mathbb{R}\), and let \(a\in D\). We consider increments \(h\) for which \(a+h\in D\) and \(h\ne 0\). For the usual two-sided derivative, the allowed inputs must approach \(a\) from both sides. At a domain endpoint, the corresponding one-sided derivative uses increments from the side that lies in the domain.
In particular, to speak of a derivative by this definition, \(a\) must be a limit point of the domain on the side or sides under consideration. The value \(f(a)\) must be defined. The word finite matters: the limit must be a real number, not an infinite value.
Equivalently, in terms of allowed increments, continuity at \(a\) means that \(f(a+h)-f(a)\to 0\) as \(h\to 0\). This formulation makes the relationship with the difference quotient especially clear.
The Main Result
The theorem applies to an ordinary two-sided derivative at an interior point and to a finite one-sided derivative at an endpoint. Its conclusion uses the same approach to \(a\) as the derivative definition.
Proof from the Difference Quotient
For every allowed \(h\ne 0\), rearrange the difference quotient identity:
The first factor, \(h\), tends to zero. The second factor tends to the finite real number \(L\). We can justify the conclusion using boundedness, without assuming in advance that \(f\) is continuous.
For allowed \(h\ne 0\), set \(Q(h)=\dfrac{f(a+h)-f(a)}{h}\). By hypothesis, \(Q(h)\to L\), where \(L\in\mathbb{R}\).
Because \(Q(h)\to L\), there is a positive \(\delta\) such that \(0<|h|<\delta\) implies \(|Q(h)-L|<1\). The triangle inequality then gives \(|Q(h)|\le |L|+1\).
Using \(f(a+h)-f(a)=hQ(h)\), we obtain \(|f(a+h)-f(a)|\le (|L|+1)|h|\) for those increments.
The constant \(|L|+1\) is finite, and \(|h|\to 0\). Therefore \(|f(a+h)-f(a)|\to 0\), which is precisely continuity at \(a\), relative to the domain.
For completeness, given any \(\varepsilon>0\), choose a positive \(\eta\) smaller than \(\delta\) and small enough that \((|L|+1)\eta<\varepsilon\). Then every allowed \(h\) with \(0<|h|<\eta\) satisfies
This is the definition of continuity in increment form. The proof also pinpoints where finiteness enters: convergence to a finite \(L\) makes \(Q(h)\) bounded near zero. Without that bound, the product of a small increment and the quotient need not tend to zero.
Worked Example: A Polynomial at an Interior Point
Worked Example: A Polynomial at an Interior Point
Let \(f(x)=x^2+3x\), and examine the point \(a=1\). First compute the function value:
For a nonzero increment \(h\), expand the value at \(1+h\):
Thus the difference quotient is
Taking the limit gives \(f'(1)=\lim_{h\to0}(5+h)=5\), a finite derivative. To check continuity directly, use the same computed values:
Therefore \(\lim_{x\to1}f(x)=f(1)=4\). The difference quotient calculation gives the finite derivative, and the change in function value confirms continuity at \(1\).
Worked Example: A Rational Function at an Interior Point
Worked Example: A Rational Function at an Interior Point
Let \(f(x)=\dfrac{1}{x+1}\), with domain \(D=\mathbb{R}\setminus\{-1\}\), and take \(a=1\). Since \(1\in D\),
For sufficiently small \(h\), \(1+h\ne -1\), so the quotient is defined. Compute it exactly:
Consequently,
This is a finite real number. Substituting the quotient back into the identity for the change gives
As \(h\to0\), the denominator tends to \(4\), so this difference tends to \(0\). Hence \(f(1+h)\to \frac12=f(1)\), verifying continuity at \(1\).
Worked Example: A One-Sided Derivative at a Domain Endpoint
Worked Example: A One-Sided Derivative at a Domain Endpoint
Let \(f(x)=\sqrt{x+1}\), with domain \(D=[0,\infty)\), and consider the endpoint \(a=0\). Only increments \(h>0\) are allowed when approaching \(0\) from within this domain. The function value is
For \(h>0\), rationalize the numerator of the one-sided difference quotient:
Therefore the right-hand derivative is
It is finite. The corresponding change in function value is
Thus \(\lim_{x\to0^+}f(x)=1=f(0)\). This checks continuity at the endpoint relative to the domain, using the same one-sided approach as the derivative.
Why the Finite Limit Is Essential
The proof does not rely merely on the quotient being defined for each nearby nonzero increment. It relies on that quotient approaching a finite number, which ensures that its magnitude is bounded near zero. A quotient can be defined at every such increment and still become unbounded.
Common Mistake: “A Small Number Times Anything Is Small”
A tempting but invalid argument starts with \(f(a+h)-f(a)=hQ(h)\) and says: since \(h\to0\), the product must tend to zero. This conclusion does not follow unless there is control over \(Q(h)\). For instance, if \(Q(h)=1/h\), then
which does not tend to zero. This quotient can occur for the function defined by \(f(0)=0\) and \(f(x)=1\) for \(x\ne0\): its difference quotient at zero is \(1/h\). The correction is to establish a finite limit for \(Q(h)\), then use that limit to bound \(Q(h)\) near zero before estimating the product.
What the Theorem Does—and Does Not—Say
The result is a one-way implication: a finite derivative at a point guarantees continuity there. It does not assert that every continuous function has a derivative. Nor does the proof require a formula for the derivative away from \(a\), or continuity of the derivative. It uses only the difference quotient limit at the particular point and the algebraic identity relating that quotient to the change in function value.
The reasoning is useful because it converts a local slope condition into a precise statement about nearby function values. Whenever a derivative is computed from first principles, the same factorization explains why the function cannot have a jump or other failure of continuity at that point: the change equals a vanishing increment times a quotient that remains bounded.
Check Your Understanding
Answer these questions using the difference-quotient argument.
- In the proof, what quantity is shown to be bounded near zero, and what assumption gives that bound?
- Why must the derivative limit be a finite real number for this proof?
- If the derivative is one-sided at a domain endpoint, which increments are used to establish continuity?
- For \(f(x)=x^2\) at \(a=2\), compute the difference quotient and state the derivative.
- Explain why knowing that \(Q(h)\) is defined for every allowed \(h\ne0\) does not by itself show that \(hQ(h)\to0\).
