Introduction
A derivative at a point is defined by a limit of difference quotients. When that limit is a finite real number, it gives the slope of the tangent line, provided the function is defined on both sides of the point. But sometimes the difference quotients do not approach a finite number: their magnitudes grow without bound. In an appropriate geometric setting, this behavior signals a vertical tangent.
The cube-root function is a basic example. Its graph is continuous at zero, but its difference quotient there becomes arbitrarily large and positive from both sides. This tutorial develops the test carefully and explains why an unbounded quotient alone is not enough to guarantee a vertical tangent.
Difference Quotients and Finite Derivatives
Let \(f\) be defined on an open interval containing \(a\). For nonzero \(h\) sufficiently close to zero, the difference quotient at \(a\) is
It is the slope of the secant line through the graph points \((a,f(a))\) and \((a+h,f(a+h))\). If its limit as \(h\to 0\) exists as a finite real number, that limit is the derivative \(f'(a)\). A limit of \(+\infty\) or \(-\infty\) is not a finite real limit, so it does not mean that the derivative exists as a real number.
For a vertical tangent, the secant slopes become unbounded in magnitude as the second graph point approaches the point of tangency. To make this a geometric conclusion about the graph near \((a,f(a))\), continuity at \(a\) matters: it ensures that \((a+h,f(a+h))\) actually approaches \((a,f(a))\).
The signed limits in this criterion describe a consistent infinite slope. More generally, if the absolute value of the quotient tends to infinity from both sides, the secant directions approach the vertical direction. If the signed limits disagree, that can describe a cusp rather than the same kind of vertical tangent behavior on both sides.
Worked Example: The Cube-Root Function at Zero
Worked Example: The Cube-Root Function at Zero
Let \(f(x)=\sqrt[3]{x}\), and examine the difference quotient at \(a=0\). First, \(f(0)=0\), so for \(h\ne 0\),
Since \(h=(\sqrt[3]{h})^3\), and \((\sqrt[3]{h})^2=|h|^{2/3}\), this simplifies to
This quantity is positive for every \(h\ne 0\), on either side of zero. To verify that it tends to \(+\infty\), let \(M>0\). If \(0<|h|<M^{-3/2}\), then \(|h|^{2/3}<1/M\), and therefore
Thus the difference quotient tends to \(+\infty\) from both sides. Also, the cube-root function is continuous at zero, so the criterion applies. The graph has vertical tangent line \(x=0\), and \(f'(0)\) is not a finite real number.
Check: Substitution into the original quotient gives \(\sqrt[3]{h}/h\), and the identity \(h=(\sqrt[3]{h})^3\) reduces it to the positive expression \(1/|h|^{2/3}\). This confirms both the sign and the unbounded behavior.
Why the Secant Lines Become Vertical
The geometric conclusion follows directly from the slopes. Write \(Q(h)\) for the difference quotient at \(a\). The secant line through \((a,f(a))\) and \((a+h,f(a+h))\) has slope \(Q(h)\). A direction vector for that line is \((1,Q(h))\). When \(|Q(h)|\) grows without bound, the horizontal component is small relative to the vertical component: their absolute ratio is \(1/|Q(h)|\), which tends to zero.
Each quotient \(Q(h)\) is the slope of a secant line from \((a,f(a))\) to a nearby graph point.
The ratio of horizontal to vertical change in the secant direction is \(1/|Q(h)|\), which approaches zero when \(|Q(h)|\to\infty\).
Continuity at \(a\) gives \(f(a+h)\to f(a)\), so these secants are formed by graph points approaching the point \((a,f(a))\).
Consequently, the secant lines approach the vertical direction through \((a,f(a))\), whose equation is \(x=a\). This explains why unbounded difference quotients yield a vertical tangent under the stated continuity condition. The quotient itself is still not a finite derivative.
Worked Example: A Vertical Tangent with Negative Unbounded Slope
Worked Example: A Vertical Tangent with Negative Unbounded Slope
Consider \(g(x)=2-4\sqrt[3]{x-5}\) at \(a=5\). Since \(\sqrt[3]{0}=0\), we have \(g(5)=2\). For \(h\ne 0\),
For every \(h\ne 0\), this quotient is negative. To check that it tends to \(-\infty\), let \(M>0\). Whenever \(0<|h|<(4/M)^{3/2}\), we have \(|h|^{2/3}<4/M\), and hence
The quotient therefore tends to \(-\infty\) from both sides. The function is continuous at \(5\), because it is a constant plus a constant multiple of a cube-root function. The graph has vertical tangent line \(x=5\), and its derivative at \(5\) is not finite.
Check: The quotient was formed by subtracting \(g(5)=2\); the constant terms cancel, leaving \(-4\sqrt[3]{h}/h\). Its sign is negative for either sign of \(h\), as required for the stated two-sided limit.
Unbounded Does Not Mean a Finite Derivative
It is important not to call \(+\infty\) or \(-\infty\) the value of the derivative in the ordinary real-valued definition. A derivative, when it exists, is a real number. An unbounded quotient says instead that no finite real number is its limit. The geometric phrase “vertical tangent” describes the limiting direction of secant lines; it does not change the definition of derivative.
Worked Example: A Finite Derivative for Comparison
Let \(p(x)=x^2\) and take \(a=0\). Since \(p(0)=0\), for \(h\ne 0\),
As \(h\to 0\), this quotient tends to \(0\), a finite real number. Thus \(p'(0)=0\); its derivative is not unbounded. The result checks directly against the definition: substituting \(p(0+h)=h^2\) and \(p(0)=0\) gives the quotient \(h\), whose limit is zero.
The contrast is in the limit of the same kind of quotient. For \(x^2\) at zero it approaches a real number; for the cube-root function at zero it grows without bound. One behavior gives a finite derivative, while the other—together with continuity—gives a vertical tangent.
Why Continuity Cannot Be Omitted
Unbounded difference quotients alone do not ensure a vertical tangent to the graph at \((a,f(a))\). The quotient only records the slopes of lines from that point to other graph points. Without continuity, those other points might not approach the point of tangency at all.
Worked Example: An Unbounded Quotient Without a Vertical Tangent
Define \(r(0)=0\) and \(r(x)=1/x\) for \(x\ne 0\). At \(a=0\), the difference quotient is
For every \(M>0\), if \(0<|h|<1/\sqrt{M}\), then \(1/h^2>M\). The quotient tends to \(+\infty\) from both sides. However, \(r(h)=1/h\) does not tend to \(r(0)=0\); in fact, its magnitude grows without bound as \(h\to 0\). The graph points \((h,1/h)\) therefore do not approach \((0,0)\), and \(r\) is not continuous at zero.
So the quotient meets the unbounded-slope condition, but the geometric requirement fails: there are no nearby graph points approaching \((0,0)\) to provide the usual tangent there. This example shows precisely why continuity is part of the vertical-tangent criterion.
A Practical Test
When investigating a possible vertical tangent at an interior point \(a\), keep the algebraic and geometric checks distinct:
- Form the difference quotient using \(f(a+h)-f(a)\), with \(h\ne 0\).
- Find its behavior as \(h\to 0\), checking both sides.
- Determine whether the quotient has a finite real limit or tends to signed infinity.
- Check continuity at \(a\) before drawing a geometric conclusion about the graph.
If the quotient tends to \(+\infty\) or \(-\infty\) from both sides and \(f\) is continuous at \(a\), the graph has vertical tangent \(x=a\), while the finite derivative does not exist. If the quotient tends to a real number, that real number is the derivative. If the quotient is unbounded but continuity fails, do not infer a vertical tangent from the quotient alone.
Check Your Understanding
Use the difference quotient and the continuity condition in each question.
- For \(f(x)=\sqrt[3]{x}\), why is the quotient at zero positive for both positive and negative \(h\)?
- What does it mean for the difference quotient to tend to \(+\infty\), and why is that not a finite derivative?
- For \(g(x)=2-4\sqrt[3]{x-5}\), what is the vertical tangent line and what is the limit of the quotient?
- Why does continuity help turn unbounded secant slopes into a geometric statement about a tangent at \((a,f(a))\)?
- In the example \(r(0)=0\), \(r(x)=1/x\) for \(x\ne 0\), why does the quotient tending to \(+\infty\) not justify claiming a vertical tangent at the origin?
