Introduction
Differentiability describes how a function changes near a point: its change is approximated by a linear expression, with an error that is small compared with the distance to the point. Continuity asks for less: the function's values must simply approach the value at the point.
The remainder definition makes the relationship precise. If the linear approximation and its remainder both tend to zero as the input approaches the point, then the function's change tends to zero. This gives the implication differentiability at a point implies continuity at that point.
The Remainder Definition
Let \(U\) be an open subset of \(\mathbb{R}^n\), let \(\mathbf{a}\in U\), and let \(f:U\to\mathbb{R}\). For a vector \(\mathbf{h}\in\mathbb{R}^n\), write \(\|\mathbf{h}\|\) for its Euclidean norm. Differentiability at \(\mathbf{a}\) means that there is a vector \(\mathbf{c}\in\mathbb{R}^n\) and a remainder \(r(\mathbf{h})\) such that, for all sufficiently small \(\mathbf{h}\),
The decomposition is required for \(\mathbf{h}\) close enough to \(\mathbf{0}\) that \(\mathbf{a}+\mathbf{h}\in U\). The vector \(\mathbf{c}\) gives the linear part of the change; in the usual notation for a differentiable scalar-valued function, it is \(\nabla f(\mathbf{a})\). The condition on \(r\) says that the remainder is smaller than first order: its size divided by the distance \(\|\mathbf{h}\|\) tends to zero.
The difference in this definition is exactly the part of the differentiability decomposition other than \(f(\mathbf{a})\):
As \(\mathbf{h}\to\mathbf{0}\), the linear term tends to zero because its absolute value is at most \(\|\mathbf{c}\|\,\|\mathbf{h}\|\). The remainder tends to zero as well: it is \(\|\mathbf{h}\|\) multiplied by a quantity tending to zero. The proof below turns those observations into the required epsilon-delta statement.
Why Differentiability Implies Continuity
Here is a complete proof. The radius on which the differentiability decomposition is valid must be respected throughout: the estimates below are applied only to increments inside that radius.
By differentiability, there is a radius \(\delta_0>0\) such that the decomposition holds whenever \(0<\|\mathbf{h}\|<\delta_0\), with \(\mathbf{a}+\mathbf{h}\in U\). Since \(U\) is open and contains \(\mathbf{a}\), the radius can be chosen so these nearby points lie in \(U\).
Given \(\varepsilon>0\), the limit \(r(\mathbf{h})/\|\mathbf{h}\|\to 0\) gives a \(\delta_1>0\) such that \(0<\|\mathbf{h}\|<\delta_1\) implies \(\lvert r(\mathbf{h})\rvert/\|\mathbf{h}\|<\varepsilon/2\).
Set \(\delta=\min\{\delta_0,\delta_1,1,\varepsilon/[2(\|\mathbf{c}\|+1)]\}\). This choice ensures that the decomposition is valid, the remainder estimate applies, and the linear term is small whenever \(0<\|\mathbf{h}\|<\delta\).
For such \(\mathbf{h}\), the Cauchy-Schwarz inequality gives \(\lvert\mathbf{c}\cdot\mathbf{h}\rvert\le\|\mathbf{c}\|\,\|\mathbf{h}\|<\varepsilon/2\). Also, because \(\|\mathbf{h}\|<1\), \(\lvert r(\mathbf{h})\rvert<(\varepsilon/2)\|\mathbf{h}\|<\varepsilon/2\). Consequently the absolute change is less than \(\varepsilon\).
Indeed, using the decomposition, the triangle inequality, and the two estimates,
This is precisely continuity at \(\mathbf{a}\). The proof also displays the role of each part of the definition: the linear term vanishes because it is bounded by a constant times \(\|\mathbf{h}\|\), and the remainder vanishes because its ratio to \(\|\mathbf{h}\|\) tends to zero.
Worked Examples
Worked Example: A Polynomial with a Nonzero Linear Term
Consider \(f(x,y)=x^2+3xy+y^2\) at \(\mathbf{a}=(1,-1)\). Write \(\mathbf{h}=(u,v)\), so the nearby input is \((1+u,-1+v)\). First evaluate the function at the point:
Now expand the function at the shifted input:
Thus the linear coefficient vector is \(\mathbf{c}=(-1,1)\), and the remainder is \(r(u,v)=u^2+3uv+v^2\). To verify the remainder condition, use \(2|uv|\le u^2+v^2\):
Since \(\|(u,v)\|=\sqrt{u^2+v^2}\), for \((u,v)\ne(0,0)\) this gives \(\lvert r(u,v)\rvert/\|(u,v)\|\le(5/2)\|(u,v)\|\to0\). Meanwhile, \(\lvert\mathbf{c}\cdot\mathbf{h}\rvert=\lvert-u+v\rvert\le\sqrt{2}\|(u,v)\|\to0\). Therefore the function change tends to zero. The expansion also checks the value directly: at \(u=v=0\), it gives \(-1=f(1,-1)\).
Worked Example: An Exact Linear Approximation
Let \(g(x,y)=5+2x-3y\) and consider the point \(\mathbf{a}=(1,2)\). Its value there is
For an increment \(\mathbf{h}=(u,v)\), direct substitution gives
The linear coefficient is \(\mathbf{c}=(2,-3)\), and \(r(u,v)=0\) for every increment. Hence the remainder ratio is zero whenever \((u,v)\ne(0,0)\), so the remainder condition holds. The linear term satisfies \(\lvert2u-3v\rvert\le\sqrt{13}\sqrt{u^2+v^2}\), which tends to zero as \((u,v)\to(0,0)\). Thus \(g(1+u,2+v)\to1=g(1,2)\). Substituting \(u=v=0\) into the expansion also recovers the original value \(1\).
Worked Example: A Vanishing Linear Term
Define \(q(x,y)=(x^2+y^2)/(1+x^2+y^2)\), and examine the origin. Here \(q(0,0)=0\). For \(\mathbf{h}=(u,v)\), let \(s=\|\mathbf{h}\|=\sqrt{u^2+v^2}\). Then
The linear coefficient is the zero vector, and the remainder is \(r(u,v)=s^2/(1+s^2)\). For \(s>0\), its ratio to the distance is
Thus the remainder condition holds, while the linear term is zero for every increment. The function change itself also tends to zero, since \(0\le q(u,v)=s^2/(1+s^2)\le s^2\to0\). This agrees with the theorem and directly checks that \(q(u,v)\to q(0,0)=0\).
A Common Remainder Mistake
The condition \(r(\mathbf{h})/\|\mathbf{h}\|\to0\) does not say that the remainder is identically zero. For example, consider \(p(x)=x^2\) at \(a=0\). Its value is \(p(0)=0\), its linear term is zero, and its remainder is \(r(h)=h^2\). For \(h\ne0\),
The differentiability condition permits a nonzero remainder; it requires that the remainder become negligible relative to the input displacement. In this example, \(p(h)=h^2\to0=p(0)\), as the theorem predicts.
A second point of care in a proof is the radius where the decomposition holds. It is not enough to choose a radius that controls the linear term and the remainder: the chosen radius must also be no larger than a radius on which the differentiability decomposition is valid. The proof above includes \(\delta_0\) in its minimum for exactly this reason.
Why the Implication Matters
Continuity is a necessary consequence of differentiability, but the definitions express different amounts of information. Differentiability requires a linear approximation whose error is smaller than the input displacement; continuity requires only that the total function change tend to zero. The proof works by decomposing that total change and showing that each term tends to zero.
This theorem gives a useful one-way test: if a function is not continuous at a point, it cannot be differentiable there. Continuity alone, however, is not what the differentiability definition asserts; the linear approximation and its controlled remainder are essential.
Check Your Understanding
Use the decomposition and the estimates in this tutorial to answer the following.
- In the remainder definition, what does the condition \(r(\mathbf{h})/\|\mathbf{h}\|\to0\) mean?
- Why does \(\mathbf{c}\cdot\mathbf{h}\) tend to zero as \(\mathbf{h}\to\mathbf{0}\)?
- In the epsilon-delta proof, why must the final radius be no larger than \(\delta_0\)?
- For \(p(x)=x^2\) at zero, identify the linear term and the remainder, and compute the remainder ratio.
- What necessary condition for differentiability follows from the theorem?