Testing a Linear Approximation
A function may have partial derivatives at a point without being well approximated there by a linear function. To test whether a proposed linear approximation is genuinely accurate to first order, compare its error with the distance from the point. The error must become negligible relative to that distance as the input approaches the point.
This test is useful in both directions. A bound that works for every nearby direction can prove differentiability. A single path on which the error remains a substantial fraction of the distance can show that differentiability fails.
The denominator is the Euclidean distance from the increment \((h,k)\) to the zero increment. The condition says that the remainder is smaller than the size of the increment by a factor that tends to zero. Equivalently, \(R(h,k)=o(\sqrt{h^2+k^2})\) as \((h,k)\to(0,0)\).
When the partial derivatives exist, they determine the only possible coefficients of a derivative: \(A=f_x(a,b)\) and \(B=f_y(a,b)\). Thus, in practice, first identify the candidate coefficients and then test the remainder ratio. The existence of those partial derivatives alone does not guarantee that the ratio tends to zero.
Why the Candidate Coefficients Are Forced
The definition explains why a successful candidate must agree with the coordinatewise rates of change. Suppose the remainder ratio tends to zero. Restrict the increments first to the \(x\)-axis by setting \(h=t\) and \(k=0\), where \(t\ne0\). Then $$ \frac{|f(a+t,b)-f(a,b)-At|}{|t|}\longrightarrow 0. $$ This implies that the difference quotient for the partial derivative in the \(x\)-direction tends to \(A\). The same argument along the \(y\)-axis forces the other coefficient.
Set \(k=0\) in the remainder ratio. Its denominator becomes \(|h|\), so the condition says the \(x\)-direction difference quotient tends to \(A\).
Set \(h=0\). The corresponding difference quotient tends to \(B\).
If the partial derivatives exist, then \(A=f_x(a,b)\) and \(B=f_y(a,b)\). These values must be tested in the full two-variable remainder ratio.
This argument also gives a useful preliminary check: if a proposed coefficient disagrees with an existing partial derivative, that candidate cannot pass the remainder test. Matching both partial derivatives is necessary, but the ratio must still be checked in all directions.
Proving the Ratio Tends to Zero
A standard strategy is to find a bound on the absolute remainder that holds for every sufficiently small \((h,k)\). Write \(r=\sqrt{h^2+k^2}\). If one can show that $$ |R(h,k)|\le C r^{1+\alpha} $$ for some constant \(C\) and some \(\alpha>0\), then $$ \frac{|R(h,k)|}{r}\le C r^\alpha\longrightarrow 0. $$ The crucial point is that the bound must hold for all directions of approach, not merely along a few selected lines.
Worked Example: A Polynomial at a Nonzero Point
Test the candidate linear approximation for \(f(x,y)=x^2+3xy-2y^2\) at \((1,-1)\).
First calculate the function value and partial derivatives: $$ f(1,-1)=1+3(1)(-1)-2(-1)^2=1-3-2=-4, $$ $$ f_x(x,y)=2x+3y,\qquad f_x(1,-1)=2-3=-1, $$ $$ f_y(x,y)=3x-4y,\qquad f_y(1,-1)=3+4=7. $$ Thus the candidate approximation at the increment \((h,k)\) is \(-4-h+7k\).
Expand the function at the incremented point: $$ \begin{aligned} f(1+h,-1+k) &=(1+h)^2+3(1+h)(-1+k)-2(-1+k)^2\\ &=1+2h+h^2-3+3k-3h+3hk-2+4k-2k^2\\ &=-4-h+7k+h^2+3hk-2k^2. \end{aligned} $$ Therefore the remainder is \(R(h,k)=h^2+3hk-2k^2\).
Since \(h^2+k^2=r^2\) and \(2|hk|\le h^2+k^2\), we have $$ \begin{aligned} |R(h,k)| &\le h^2+3|hk|+2k^2\\ &\le h^2+\frac{3}{2}r^2+2k^2\\ &\le \frac{7}{2}r^2. \end{aligned} $$ For \(r>0\), the remainder ratio is consequently at most \(\frac{7}{2}r\), which tends to zero. The candidate is therefore valid. As a direct check, subtracting \(-4-h+7k\) from the expanded function gives exactly \(h^2+3hk-2k^2\), the remainder used in the bound.
Using Polar Coordinates
Near the origin, polar coordinates write \(h=r\cos\theta\) and \(k=r\sin\theta\), with \(r>0\). The distance in the denominator is then exactly \(r\). This separates the size of an increment from its direction: \(r\) measures how close the input is to the base point, while \(\theta\) records its direction.
After substituting these expressions, a successful proof must show that the remainder ratio tends to zero uniformly over all angles \(\theta\). A bound independent of \(\theta\) is especially useful. Merely seeing a factor of \(r\) is not enough if the remaining angular factor can become unbounded as \(r\) shrinks.
Worked Example: A Uniform Bound in Polar Coordinates
Define \(f(0,0)=0\) and, for \((x,y)\ne(0,0)\), let $$ f(x,y)=\frac{x^2y^2}{x^2+y^2}. $$ Test the zero linear map at the origin. Along either coordinate axis the function is zero, so both partial derivatives at the origin are zero; the candidate is \(L(h,k)=0\).
Set \(x=r\cos\theta\) and \(y=r\sin\theta\). Because \(x^2+y^2=r^2\), $$ f(x,y)=\frac{r^4\cos^2\theta\sin^2\theta}{r^2} =r^2\cos^2\theta\sin^2\theta. $$ The remainder is \(f(x,y)\), and its ratio to the distance is $$ \frac{|f(x,y)|}{r} =r\cos^2\theta\sin^2\theta. $$ For every \(\theta\), \(\cos^2\theta\sin^2\theta\le \frac14\), because \(2|\sin\theta\cos\theta|\le \sin^2\theta+\cos^2\theta=1\). Hence $$ 0\le \frac{|f(x,y)|}{r}\le\frac{r}{4}\longrightarrow 0. $$ The bound is independent of direction, so it proves differentiability at the origin with zero derivative. Substitution back into the definition confirms that the remainder is precisely \(f(x,y)-f(0,0)-0=f(x,y)\), the expression just bounded.
Using a Path to Disprove the Test
To disprove a limit, it is enough to find one path approaching the base point along which the ratio does not tend to zero. This is a decisive test for failure. The converse is different: checking a few paths, even many straight lines, does not establish a two-variable limit. To prove the limit, control the ratio for every sufficiently small increment, often by a uniform inequality or a uniform polar-coordinate bound.
Worked Example: Partial Derivatives Exist but the Ratio Fails
Define \(f(0,0)=0\) and, away from the origin, set $$ f(x,y)=\frac{x^2y}{x^2+y^2}. $$ Along the coordinate axes \(f\) is zero, so \(f_x(0,0)=0\) and \(f_y(0,0)=0\). The only possible candidate linear map is therefore zero.
Now approach the origin along the path \(x=t,\ y=t\), where \(t\ne0\). The function value and distance are $$ f(t,t)=\frac{t^2t}{t^2+t^2}=\frac{t}{2}, \qquad \sqrt{t^2+t^2}=\sqrt{2}|t|. $$ The remainder ratio along this path is $$ \frac{|f(t,t)|}{\sqrt{t^2+t^2}} =\frac{|t|/2}{\sqrt{2}|t|} =\frac{1}{2\sqrt{2}}. $$ This is a fixed positive number, not a quantity tending to zero. The zero candidate fails, and because the partial derivatives force that candidate, \(f\) is not differentiable at the origin. Direct substitution into the ratio confirms the failure along the selected path.
A Common Mistake: Treating Line Tests as a Proof
A common error is to test the ratio along several lines, observe that it tends to zero on each one, and conclude that the full limit is zero. Those calculations only describe the tested directions. The direction of approach can also vary with the distance to the base point, so a path that bends as it approaches the point may reveal behavior that fixed lines miss.
The correction is to seek a bound that applies to every nearby increment. In polar coordinates, this means controlling all values of \(\theta\) with the same estimate. For instance, a bound of the form \(|R(r\cos\theta,r\sin\theta)|/r\le Cr\) proves the limit because the right-hand side is independent of \(\theta\) and tends to zero. A single path with a nonzero limiting ratio, on the other hand, is already enough to disprove the limit.
Check Your Understanding
Answer these questions by referring to the definition and reasoning in the examples.
- What quantity must tend to zero for a candidate linear map to be the derivative at a point?
- Why must the coefficient of \(h\) in a valid candidate equal \(f_x(a,b)\), when that partial derivative exists?
- For \(R(h,k)=O(h^2+k^2)\), explain why the remainder ratio tends to zero.
- What does a path with a nonzero limiting remainder ratio establish?
- Why do successful checks along a finite collection of lines not, by themselves, prove differentiability?