Poles and Removable Discontinuities

A rational function has a pole (a vertical asymptote) at a zero of the denominator that is not cancelled by the numerator, and a removable discontinuity (a hole) where a common factor cancels. In brief: f(x) = (x² − 4)/(x − 2) simplifies to x + 2 with a hole at (2, 4), while g(x) = 1/(x − 2) has a pole at x = 2. The type of the discontinuity is found by factoring the numerator and the denominator.

At every excluded value, the graph of a rational function has some kind of break. The break can be a vertical asymptote, where the values grow without bound, or a single missing point. The difference decides how to draw the graph and how limits behave. This guide explains how to tell them apart, how the multiplicity of a pole affects the sign near it, how to find the coordinates of a hole and gives a table of examples.

What types of discontinuities do rational functions have?

A rational function f(x) = P(x)/Q(x) is undefined where Q(x) = 0. Writing the fraction in lowest terms (after cancelling common factors) distinguishes two cases:

• Removable discontinuity (hole): a factor (x − a) appears in both the numerator and the denominator and cancels. The function has a finite limit at a, but is not defined there. The graph of f equals the graph of the simplified function, except for one missing point.
• Pole (vertical asymptote): a factor (x − a) remains in the denominator after cancelling. The values |f(x)| grow without bound as x approaches a.

How do you find a removable discontinuity?

1. Factor the numerator and the denominator.
2. Identify the common factors (x − a).
3. Cancel them. The hole is at x = a.
4. Compute the y-coordinate by substituting x = a into the simplified function.

Example 1

f(x) = (x² − 4)/(x − 2) = (x − 2)(x + 2)/(x − 2) = x + 2 for x ≠ 2. The hole is at (2, 4). The graph is the line y = x + 2 with an open circle at (2, 4). The limit as x → 2 is 4, but f(2) is undefined.

Example 2

g(x) = (x² − x − 6)/(x² − 9) = (x − 3)(x + 2)/((x − 3)(x + 3)) = (x + 2)/(x + 3). The factor (x − 3) cancels: a hole at x = 3 with y = 5/6. The factor (x + 3) remains: a pole at x = −3. So g has both a hole at (3, 5/6) and a vertical asymptote x = −3.

The idea that the function can be “repaired” by assigning the limit value is connected with continuity.

How do you find a pole and its behaviour?

A pole at x = a remains after cancelling. The behaviour near a depends on the multiplicity m of the factor (x − a) in the denominator.

• Odd multiplicity (1, 3, …): the function goes to +∞ on one side of the asymptote and to −∞ on the other.
• Even multiplicity (2, 4, …): the function goes to the same infinity on both sides.

Example 3: simple pole

h(x) = 1/(x − 2). As x → 2⁺, h → +∞. As x → 2⁻, h → −∞.

Example 4: double pole

k(x) = 1/(x − 2)². Both sides give +∞, since the square is positive.

Example 5: sign analysis

m(x) = (x − 1)/((x + 2)(x − 3)). The poles are x = −2 and x = 3 (simple). Near −2: x − 1 < 0 and x − 3 < 0, so the numerator over (x − 3) is positive. For x slightly greater than −2, (x + 2) > 0, so m is positive (+∞). For x slightly less than −2, m is negative (−∞). Near x = 3: x − 1 > 0, x + 2 > 0, so the sign is the sign of (x − 3): +∞ from the right and −∞ from the left.

Step-by-step procedure

1. Find the domain: zeros of the original denominator.
2. Factor and cancel common factors.
3. Cancelled factors give holes. Remaining zeros of the denominator give vertical asymptotes.
4. Compute the y-value of each hole from the simplified function.
5. Plot the asymptotes, holes, intercepts and a few points, and draw the graph.

Example 6

f(x) = (x² − 1)/(x² + x − 2) = (x − 1)(x + 1)/((x + 2)(x − 1)). Domain: x ≠ 1, −2. After cancelling (x − 1): f(x) = (x + 1)/(x + 2). Hole at x = 1 with y = 2/3: (1, 2/3). Vertical asymptote at x = −2. Horizontal asymptote y = 1. The zero is x = −1, and the y-intercept is 1/2.

Table of examples

Where do these ideas come from?

The terms pole and removable singularity belong to complex analysis, developed in the nineteenth century by Cauchy, Riemann and Weierstrass, who studied the behaviour of functions near points where they are not defined. The word “pole” was introduced for the points at which a function of a complex variable becomes infinite. In real analysis, the classification of discontinuities as removable, jump or infinite is part of the standard treatment of continuity.

Where are they used today?

In engineering, the poles of a transfer function determine the stability and resonance of a system, and the position of the poles decides whether a vibration decays or grows. In physics, the response of a circuit at its resonant frequency corresponds to a pole. Holes appear when a formula is simplified: the formula (x² − 4)/(x − 2) for a rate may be undefined at x = 2 but the physical quantity is still 4, which is why the limit is used.

Which mistakes are common?

Treating every excluded value as an asymptote

If the factor cancels, there is only a hole, not an asymptote.

Forgetting the hole after simplifying

The simplified function x + 2 is defined at x = 2, but the original function is not. Mark the hole as an open circle.

Wrong y-coordinate of the hole

Compute it from the simplified expression, not from the original.

Assuming the sign always changes at a pole

It changes only for odd multiplicity. A double pole has the same sign on both sides.