Why use a limit at a singular point?
A gradient is a useful way to calculate directional rates when a function is differentiable and its partial derivatives are available. At a singular point, however, the gradient formula may be unavailable: a partial derivative may fail to exist, or the function may be defined by a formula that is not valid at the point itself.
The directional rate can still be studied directly. Fix a straight-line direction through the point, substitute the line into the function, and calculate a one-variable limit. The result answers a precise question about change along that line; it does not require the function to have a gradient there.
Definition along a specified line
Let \(f\) be a function of two variables, let \(\mathbf{a}=(x_0,y_0)\) be a point in its domain, and let \(\mathbf{u}=(u_1,u_2)\) be a unit vector. The line through \(\mathbf{a}\) in direction \(\mathbf{u}\) is parameterized by \(\mathbf{a}+t\mathbf{u}\). If the limit exists as a finite real number, the directional derivative of \(f\) at \(\mathbf{a}\) in direction \(\mathbf{u}\) is
This is a two-sided limit: \(t\) approaches zero through both positive and negative values. Consequently, the quotient must approach the same value from either side. A right-hand limit as \(t\to 0^+\) describes the rate when moving forward along the parameterized line, but it is not, by itself, the two-sided directional derivative.
The unit-length convention makes the parameter \(t\) measure signed distance along the line. If a nonzero vector \(\mathbf{v}\) is given instead, its associated unit direction is \(\mathbf{u}=\mathbf{v}/\|\mathbf{v}\|\). Using \(\mathbf{v}\) without normalizing it still gives a limit along that line, but the parameter then measures a different distance, so its rate is scaled accordingly.
Why the line substitution works
Define a one-variable function \(g(t)=f(\mathbf{a}+t\mathbf{u})\) for values of \(t\) near zero. Since \(g(0)=f(\mathbf{a})\), the directional-derivative quotient becomes
Thus the directional derivative exists exactly when this one-variable difference quotient has a finite two-sided limit. This argument explains why the method applies even if partial derivatives or a gradient formula are unavailable: it uses only the values of \(f\) on the specified line. It does not claim that behavior on one line determines behavior in every direction.
Worked calculations at the origin
Worked Example: A radial function with no two-sided directional derivative
Let \(f(x,y)=\sqrt{x^2+y^2}\), and calculate the directional limit at the origin in the unit direction \(\mathbf{u}=(3/5,4/5)\). First verify that the vector is a unit vector:
The function value at the origin is \(f(0,0)=0\). Substituting the line \(t\mathbf{u}\) into the original function gives
Therefore the difference quotient is
The right-hand limit is \(1\), while the left-hand limit is \(-1\). They disagree, so the two-sided directional derivative does not exist. As a check, substituting \(t>0\) into the original function gives the quotient \(1\), and substituting \(t<0\) gives \(-1\), exactly as the piecewise calculation shows.
Worked Example: A direction where the derivative does exist
Let \(f(x,y)=|x|+y\), with \(f(0,0)=0\), and use the unit direction \(\mathbf{u}=(0,1)\). Along this line, \(t\mathbf{u}=(0,t)\), so
Hence, for \(t\ne 0\), the quotient is
Both one-sided limits equal \(1\), so \(D_{\mathbf{u}}f(0,0)=1\). Check by substituting into the function: its change from the origin along the vertical line is exactly \(t\), and dividing that change by \(t\) gives \(1\) for every nonzero \(t\). The function has an absolute-value term, but that term is zero along this particular line.
Worked Example: A singular quotient with a finite line rate
Define \(f(0,0)=0\), and for \((x,y)\ne(0,0)\) let
Calculate the directional derivative at the origin in the unit direction \(\mathbf{u}=(3/5,4/5)\). The point \((t\mathbf{u})\) is nonzero whenever \(t\ne0\). Substitution into the quotient gives
Since \(f(0,0)=0\), the difference quotient for \(t\ne0\) is
This value is constant for every nonzero \(t\), so the two-sided limit exists and \(D_{\mathbf{u}}f(0,0)=36/125\). The substitution checks against the original formula: the denominator on the line is \(t^2\), and the numerator is \(36t^3/125\), producing the line value \(36t/125\).
Direction can change whether the limit exists
The second example has different behavior on different lines. For instance, in the direction \(\mathbf{u}=(3/5,4/5)\), its difference quotient is
For \(t>0\), this equals \(3/5+4/5=7/5\). For \(t<0\), it equals \(-3/5+4/5=1/5\). The two-sided directional derivative in this direction therefore does not exist. By contrast, along the vertical direction \((0,1)\), the derivative exists and equals \(1\), as calculated above.
These calculations illustrate why the direction must be stated and used in the limit. A function may have a finite directional derivative along one line and fail to have one along another. A value computed using only \(t>0\) must also be identified as a right-hand rate rather than a two-sided directional derivative.
A common mistake: replacing an absolute value too soon
In the first example, a tempting but incorrect calculation is to write \(\sqrt{t^2}=t\), conclude that the quotient equals \(1\), and declare a two-sided derivative. The equality \(\sqrt{t^2}=|t|\) holds for all real \(t\); replacing \(|t|\) by \(t\) is valid only when \(t\ge 0\). For negative \(t\), \(|t|/t=-1\), not \(1\). The correction is to keep the absolute value until the signs of \(t\) have been considered.
What the calculation does and does not establish
A directional derivative is a limit along one specified straight line. It can be meaningful at a point where the usual gradient formula cannot be used, as the rational example demonstrates. But a successful calculation establishes only the rate along that line. It does not provide a gradient formula, and a failed limit in one direction says nothing by itself about whether the limit exists along a different direction.
When applying the definition, use this sequence of checks:
- Confirm the function value at the base point, including any special value assigned there.
- Write the specified unit direction and the line \(\mathbf{a}+t\mathbf{u}\).
- Substitute the line into the original function before simplifying the difference quotient.
- Keep absolute values and other sign-sensitive expressions intact.
- Compare the limits from positive and negative \(t\), and state whether the two-sided limit exists.
Check Your Understanding
For each question, use the limit along the stated line and distinguish a two-sided limit from a one-sided rate.
- For \(f(x,y)=\sqrt{x^2+y^2}\) at the origin, what are the right- and left-hand limits of the difference quotient in direction \((0,1)\)? Does the two-sided directional derivative exist?
- For \(f(x,y)=|x|+y\) at the origin, calculate the directional derivative in the unit direction \((1,0)\), or explain why it does not exist.
- In the rational-function example, what is the value of the difference quotient along the unit direction \((0,1)\) for nonzero \(t\)?
- Why is \(\sqrt{t^2}=t\) not a valid simplification for all real \(t\)?
- If the right- and left-hand limits of a directional difference quotient are unequal, what conclusion follows about the two-sided directional derivative?