AP Calculus AB and BC

Vertical Tangent vs Vertical Asymptote

Q: How do I tell a cusp from a vertical tangent?

Compare the one sided limits of f'. If both run to +\infty, or both to -\infty, the curve has a vertical tangent, as y = x^{1/3} does at 0. If they run to opposite infinities the curve has a cusp, as y = x^{2/3} does at 0. The size \lvert f' \rvert blows up in both cases, so the sign is the whole test.

A vertical tangent sits at a point where the function is defined and continuous but its derivative runs off to the same infinity from both sides; a vertical asymptote sits where at least one one sided limit of the function itself is plus or minus infinity.

Vertical tangent

Use when: The function has a value at the point and the graph is unbroken there, yet the slope grows without bound and the one sided limits of the derivative both run to positive infinity, or both to negative infinity; if they run to opposite infinities the point is a cusp, not a vertical tangent.

Vertical asymptote

Use when: At least one of the one sided limits of the function at the point is positive or negative infinity, so the values grow without bound nearby and the graph runs alongside the line.

Side by side

	Vertical tangent	Vertical asymptote
What is infinite	The slope only, with the same sign from both sides: $\lim_{x \to a} f'(x) = +\infty$ or $\lim_{x \to a} f'(x) = -\infty$ . Opposite signs give a cusp	The function itself: at least one of $\lim_{x \to a^{-}} f(x)$ and $\lim_{x \to a^{+}} f(x)$ is $\pm\infty$
Value of $f(a)$	Exists, and the graph passes through that point	Usually undefined, and in any case irrelevant, since the asymptote is a statement about the limits
Continuity at $x = a$	Continuous	Infinite discontinuity
Does the graph cross the line $x = a$	Yes, it passes through at exactly one point	Never, since a vertical line meets the graph of a function at most once
Standard example	$y = \sqrt[3]{x}$ at $x = 0$	$y = \frac{1}{x}$ at $x = 0$

Both features draw the eye to a vertical line, so decide which object is misbehaving. At a vertical tangent the function is perfectly well behaved: $f(a)$ exists, the graph is unbroken there, and it is the slope that runs to infinity, with the same sign from both sides. At a vertical asymptote the function is the thing that fails, since at least one of $\lim_{x \to a^{-}} f(x)$ and $\lim_{x \to a^{+}} f(x)$ is $\pm\infty$ , which is why $x = a$ is normally outside the domain.

\underbrace{\lim_{x \to 0} \sqrt[3]{x} = 0}_{\text{function is fine}}, \quad \underbrace{\lim_{x \to 0} \frac{1}{3x^{2/3}} = +\infty}_{\text{slope is not, and one signed}} \qquad \text{versus} \qquad \underbrace{\lim_{x \to 0^{+}} \frac{1}{x} = \infty}_{\text{function itself blows up}}

Work the two examples side by side. For $f(x) = x^{1/3}$ the derivative is $f'(x) = \frac{1}{3}x^{-2/3}$ , which is undefined at $x = 0$ and tends to $+\infty$ from both sides, so the tangent line at the origin is the vertical line $x = 0$ and the curve passes straight through it. Contrast $y = x^{2/3}$ , whose derivative $\frac{2}{3}x^{-1/3}$ runs to $-\infty$ from the left and $+\infty$ from the right: the size of the slope blows up in both cases, but the mismatched signs make that point a cusp. For $g(x) = \frac{1}{x}$ there is no value $g(0)$ for a tangent line to touch, and the graph never meets $x = 0$ at any height.

The error this confusion produces

A derivative that blows up gets reported as an asymptote. Writing " $x = 0$ is a vertical asymptote of $y = \sqrt[3]{x}$ " is wrong twice over: that function is continuous everywhere and every limit it has at $0$ is finite. Ask what is infinite, the function or its slope, before naming the feature, and if it is the slope, check that both one sided limits of $f'$ carry the same sign before calling it a vertical tangent.

Frequently asked questions

Is a function differentiable at a vertical tangent?

No. The derivative fails to exist there because the limit of the difference quotient is infinite rather than a number. The function is still continuous at that point, which makes it a standard example of continuity without differentiability.

How do I tell a cusp from a vertical tangent?

Compare the one sided limits of $f'$ . If both run to $+\infty$ , or both to $-\infty$ , the curve has a vertical tangent, as $y = x^{1/3}$ does at $0$ . If they run to opposite infinities the curve has a cusp, as $y = x^{2/3}$ does at $0$ . The size $\lvert f' \rvert$ blows up in both cases, so the sign is the whole test.

Can a graph cross a vertical asymptote?

No. A vertical line meets the graph of a function in at most one point, so the graph can never pass from one side of $x = a$ to the other. Horizontal asymptotes are different: a curve may cross one many times and still approach it.

In the CED: Unit 1: Limits and Continuity, Unit 2: Defining the Derivative