Differentiability Implies Continuity: A Limit Proof

Why Differentiability Guarantees Continuity

A derivative describes how a function changes near a point. If the derivative exists there as a finite real number, the change in the function must become small as the input change becomes small. That is precisely the behavior required for continuity.

The connection comes directly from the difference quotient. It does not require a graph, a derivative formula, or a separate continuity calculation. The key is to keep track of what the derivative limit says—and what it does not say.

Definitions at a Point

Let \(f\) be defined on an open interval containing \(a\). The derivative of \(f\) at \(a\) is the finite real limit

f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h},

provided this two-sided limit exists. The increment \(h\) is nonzero in the quotient, and \(h\) approaches zero through values for which \(a+h\) is in the domain of \(f\).

The function \(f\) is continuous at \(a\) if

\lim_{x\to a} f(x)=f(a).

Equivalently, writing \(x=a+h\), continuity at \(a\) means that \(f(a+h)-f(a)\) approaches zero as \(h\to 0\). Differentiability and continuity use limits of different quantities: differentiability concerns the quotient of the change in output by the change in input, while continuity concerns the change in output itself.

Theorem: If \(f\) is differentiable at an interior point \(a\) of its domain, then \(f\) is continuous at \(a\). Here, differentiability means that the difference quotient has a finite real limit as \(h\to 0\).

The Difference-Quotient Argument

For every nonzero \(h\) sufficiently close to zero, the definition of the difference quotient gives the exact identity

f(a+h)-f(a) = h\left(\frac{f(a+h)-f(a)}{h}\right).

If the quotient approaches a finite number, then the product on the right approaches zero: one factor, \(h\), approaches zero, and the other approaches a finite limit. Therefore the function’s change approaches zero, which is the condition for continuity.

The finite-limit condition matters. The product-limit rule used here applies when both factor limits exist as finite real numbers. In particular, the quotient must approach a finite value; it is not enough to know only that the quotient has some nonfinite or unbounded behavior.

Complete Proof

Suppose \(f'(a)=L\), where \(L\) is a finite real number. Define the quotient \(Q(h)\) for nonzero \(h\) near zero by

Q(h)=\frac{f(a+h)-f(a)}{h}.

By the assumption that \(f\) is differentiable at \(a\), \(\lim_{h\to 0}Q(h)=L\). For every such nonzero \(h\),

$$ f(a+h)-f(a)=hQ(h). $$

As \(h\to 0\), the first factor \(h\) tends to zero and the second factor \(Q(h)\) tends to the finite number \(L\). The product limit law therefore gives

\lim_{h\to 0}\bigl(f(a+h)-f(a)\bigr) = \left(\lim_{h\to 0}h\right) \left(\lim_{h\to 0}Q(h)\right) = 0\cdot L = 0.

Adding the constant \(f(a)\) to both sides of the limiting expression yields

\lim_{h\to 0}f(a+h)=f(a).

Since \(x=a+h\) approaches \(a\) exactly when \(h\) approaches zero, this says \(\lim_{x\to a}f(x)=f(a)\). Thus \(f\) is continuous at \(a\), as claimed.

Why the Finite Limit Is Explicitly Needed

The proof uses the fact that \(Q(h)\) approaches a finite real number \(L\). One way to see its role is to note that a function with a finite limit is bounded sufficiently close to the limit point. In this case, for all sufficiently small nonzero \(h\), we can arrange that

|Q(h)-L|<1, \qquad\text{so}\qquad |Q(h)|<|L|+1.

Consequently, the change in the function is bounded by

|f(a+h)-f(a)| = |h|\,|Q(h)| < |h|(|L|+1).

As \(h\to 0\), this upper bound approaches zero. A finite limit gives the quotient a finite bound near zero, so multiplying it by a sufficiently small increment forces the output change to be small too.

If a quotient does not approach a finite real number, this particular argument does not establish continuity: the factor \(Q(h)\) may not have a finite limit to use in the product-limit law. The theorem’s hypothesis is specifically a finite derivative at the point.

Worked Examples

Worked Example: A Quadratic at \(a=2\)

Let \(f(x)=x^2+1\). Find the difference-quotient limit at \(a=2\), then use it to establish continuity there.

For \(h\ne 0\), first compute the function values:

f(2+h)=(2+h)^2+1=4+4h+h^2+1=5+4h+h^2, \qquad f(2)=2^2+1=5.

Therefore,

\frac{f(2+h)-f(2)}{h} = \frac{(5+4h+h^2)-5}{h} = \frac{4h+h^2}{h} = 4+h.

Taking the limit gives \(f'(2)=\lim_{h\to 0}(4+h)=4\), a finite number. The theorem now gives \(\lim_{x\to 2}f(x)=f(2)=5\).

Check the change directly using the quotient identity: \(f(2+h)-f(2)=h(4+h)=4h+h^2\), which tends to zero as \(h\to 0\). Thus the nearby function values approach \(5\), as required for continuity.

Worked Example: A Square Root at \(a=4\)

Let \(f(x)=\sqrt{x}\), with \(x\ge 0\). Find the difference-quotient limit at \(a=4\), and use the theorem to conclude continuity.

For \(h\ne 0\) sufficiently close to zero, \(4+h>0\), and

\frac{f(4+h)-f(4)}{h} = \frac{\sqrt{4+h}-2}{h}.

Multiply the numerator and denominator by the conjugate \(\sqrt{4+h}+2\):

\frac{\sqrt{4+h}-2}{h} \cdot \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2} = \frac{(4+h)-4}{h(\sqrt{4+h}+2)} = \frac{h}{h(\sqrt{4+h}+2)} = \frac{1}{\sqrt{4+h}+2}.

Thus

f'(4) = \lim_{h\to 0}\frac{1}{\sqrt{4+h}+2} = \frac{1}{2+2} = \frac14.

This is a finite derivative, so \(f\) is continuous at \(4\), and \(\lim_{x\to 4}\sqrt{x}=f(4)=2\).

Check the function change: the quotient identity gives \(\sqrt{4+h}-2=h/(\sqrt{4+h}+2)\). The denominator approaches \(4\), so this change approaches zero as \(h\to 0\). The function values therefore approach \(2\).

Worked Example: A Reciprocal at \(a=2\)

Let \(f(x)=1/x\), with \(x\ne 0\). Calculate the derivative limit at \(2\) and apply the theorem.

For nonzero \(h\) close enough to zero that \(2+h\ne 0\),

\frac{f(2+h)-f(2)}{h} = \frac{\frac{1}{2+h}-\frac12}{h} = \frac{\frac{2-(2+h)}{2(2+h)}}{h} = \frac{-h}{2(2+h)h} = -\frac{1}{2(2+h)}.

It follows that

f'(2) = \lim_{h\to 0}-\frac{1}{2(2+h)} = -\frac{1}{2\cdot 2} = -\frac14.

The derivative is finite, so \(f\) is continuous at \(2\), with \(\lim_{x\to 2}f(x)=f(2)=1/2\).

As a check, the output change is \(f(2+h)-f(2)=h[-1/(2(2+h))]\). The bracketed factor approaches \(-1/4\), so the whole change approaches zero. This agrees with the continuity conclusion.

A Common Mistake: Treating the Limit as an Exact Quotient

A derivative is a limit of quotients, not necessarily the value of the quotient for every nonzero increment. For instance, in the quadratic example the quotient at \(2\) is \(4+h\), not exactly \(4\) for every \(h\ne 0\). It is the limit of \(4+h\) that equals \(4\).

A mistaken calculation might say, “Since \(f'(2)=4\), \(f(2+h)-f(2)=4h\) for every \(h\).” For \(f(x)=x^2+1\), that would predict a change of \(4h\). But the exact calculation gives

$$ f(2+h)-f(2)=4h+h^2. $$

The extra \(h^2\) term shows that the claimed equality is wrong whenever \(h\ne 0\). The correct statement is that the quotient \(4+h\) approaches \(4\), and therefore the change \(h(4+h)\) approaches zero. Keeping the limit in the argument is what makes the proof valid.

Why the Result Matters

This theorem provides a useful one-way test: if a function is differentiable at a point, it must be continuous there. In particular, when continuity at a point is absent, differentiability there is impossible. The converse is not part of this theorem; continuity alone does not supply the finite difference-quotient limit required for differentiability.

For AP problems, the result can save work. Once a finite derivative at a point has been established, continuity at that point follows without a second, independent limit calculation. Conversely, when checking an argument that claims differentiability, verify that the derivative limit is finite and that the point is an interior point where the difference quotient is defined for nearby inputs.

Key takeaway: For nonzero \(h\), the change in output equals the increment times the difference quotient. If the quotient has a finite limit, this product tends to zero, so the function is continuous at the point.

Check Your Understanding

Use the difference-quotient idea and the theorem to answer these questions.

In the proof, what does \(Q(h)\) represent, and what is its limit as \(h\to 0\)?
Why must the derivative limit be finite for the product-limit argument used here?
If \(f'(a)=L\), what exact identity connects \(f(a+h)-f(a)\) with the difference quotient for \(h\ne 0\)?
For \(f(x)=x^2\) at \(a=1\), the quotient simplifies to \(2+h\). What is the derivative, and what continuity conclusion follows?
Does this theorem say that every function continuous at a point is differentiable there? Explain what the theorem does and does not establish.