Calculus 1

Discontinuities and how to remove them

When a function fails to be continuous at a point, is every failure the same kind of break, and which breaks can be repaired by choosing a better value at that one point?

You need

Continuity on an intervalCalculus 1 Infinite limits and vertical asymptotesCalculus 1

Unlocks

Differentiability: corners, cusps and piecewise functionsCalculus 1 Making a piecewise function continuousCalculus 1

Where you are 5 in Continuity

Calculus8

Limits11

Continuity5

Continuity at a point Continuity on an interval

Discontinuities and how to remove them

The extreme value theorem Making a piecewise function continuous

Why this exists

Continuity at a point gives a yes or no answer: either \lim_{x \to a} f(x) exists and equals f(a), or the function is not continuous at a. A “no” hides very different pictures. A graph can be missing a single point, step from one height to another, run off to infinity, or swing faster and faster without settling. These breaks behave differently when the function is used. A missing point can be filled in, and the result is continuous; a step or an asymptote cannot be filled in by any choice of value.

This lesson sorts the breaks into four kinds using only the two one-sided limits, so each kind follows from limits already computed in One sided limits and when a limit exists and Infinite limits and vertical asymptotes. Day 1 classifies. Day 2 repairs the breaks that can be repaired, explains why the others cannot be, and follows breaks through compositions and through functions built from pieces. Reading breaks off a picture is done in Continuous graphs, jumps and holes and telling a hole from an asymptote in Vertical asymptotes; here the test is the limits themselves.

Definition

Let f be defined at every number of some open interval around a, except possibly at a itself. Then f has a discontinuity at a when f is not continuous at a. Write L^- = \lim_{x \to a^-} f(x) and L^+ = \lim_{x \to a^+} f(x) when these exist as numbers. The discontinuity is

removable when L^- and L^+ exist and are equal, so \lim_{x \to a} f(x) exists, but f(a) is undefined or differs from it;
a jump when L^- and L^+ both exist as numbers and L^- \ne L^+. The size of the jump is \lvert L^+ - L^- \rvert;
infinite when at least one one-sided limit is \infty or -\infty;
oscillating in every remaining case: neither side runs off to infinity, and at least one one-sided limit does not exist because the values keep swinging. The swings may stay bounded, as for \displaystyle \sin\frac{1}{x} at 0, or grow without bound in both directions, as for \displaystyle \frac{1}{x}\sin\frac{1}{x} at 0, whose one-sided limits are neither numbers nor \pm\infty.

Some books call every discontinuity that is not removable “essential”; the word is used in more than one way, so this lesson does not use it.

A function that is removable at a has a continuous extension there: the function F with F(x) = f(x) for x \ne a and F(a) = \lim_{x \to a} f(x). Building F is called removing the discontinuity.

Intuition

Picture standing at a and looking along the graph to the left and to the right. Each side either settles toward a height, runs off upward or downward, or never settles. The four kinds are the four possible reports.

Both sides settle at the same height: the graph has one height to arrive at, and only the single point above a is missing or misplaced. Removable.
Both sides settle, at different heights: the graph steps. Jump.
A side runs off: there is a vertical asymptote at a. Infinite.
A side never settles and does not run off: oscillating.

The value f(a) plays no part in sorting. It only decides whether there is a break at all, and in a jump, which side the point sits on. In Figure 1, the same function shows two kinds at once.

Figure 1Ahiga’s function. At 1 both sides head for -2, a removable discontinuity; at 2 the two sides run off to -\infty and \infty, an infinite discontinuity.

Results

Every discontinuity is exactly one kind. Removable, jump, infinite and oscillating cover every discontinuity, and no discontinuity is of two kinds.

Removable means extendable. If f has a removable discontinuity at a with \lim_{x \to a} f(x) = L, its continuous extension F is continuous at a, and L is the only value at a that makes the function continuous there.

The other three cannot be removed. If f has a jump, infinite or oscillating discontinuity at a, then no choice of value at a makes f continuous at a.

Breaks in sums and products. Let g be continuous at a.

If f has a jump of size J at a, then f + g has a jump of size J at a.
If f has a jump at a, f(a) is defined, and g(a) = 0, then f g is continuous at a; if g(a) \ne 0, f g has a jump of size \lvert g(a) \rvert J.

Breaks in compositions. If \lim_{x \to a^-} g(x) = L and f is continuous at L, then \lim_{x \to a^-} f\big(g(x)\big) = f(L), and the same from the right. If instead f is not continuous at L but g(x) approaches L through values less than L, then f\big(g(x)\big) approaches \lim_{u \to L^-} f(u) when that limit exists, and through values greater than L, \lim_{u \to L^+} f(u).

Why it is true

Exactly one kind. If both one-sided limits exist as numbers, they are equal or not: removable or jump. If they are equal and also equal f(a), there is no discontinuity, so a removable one really is broken. If some one-sided limit fails to exist as a number, either a side runs off to \pm\infty (infinite) or not (oscillating). The four cases do not overlap.

Removable means extendable. F agrees with f at every x \ne a, and a limit at a never looks at the value at a, so \lim_{x \to a} F(x) = \lim_{x \to a} f(x) = L = F(a). Any value that makes the function continuous at a must equal the limit, and a limit has only one value, so L is the only choice.

The other three cannot be removed. Changing the value at a changes no one-sided limit, for the same reason. Continuity at a needs \lim_{x \to a} f(x) to exist as a number, which by One sided limits and when a limit exists needs both one-sided limits to exist as numbers and agree. In a jump they disagree; in an infinite or oscillating discontinuity one of them is not a number.

Sums and products. These are the one-sided sum and product limit laws. With \lim_{x \to a^\pm} g(x) = g(a), the sum f + g has one-sided limits L^- + g(a) and L^+ + g(a), whose difference is still L^+ - L^-. The product has one-sided limits L^- g(a) and L^+ g(a): both 0 when g(a) = 0, which is also the value f(a) g(a), and otherwise differing by \lvert g(a) \rvert \lvert L^+ - L^- \rvert.

Compositions are stated with a sketch, as in Continuity on an interval: as x approaches a from one side, the inner values approach L, and the outer function turns inputs close to L into outputs close to f(L). When the inner values stay on one side of L, only that side of f near L is ever used, which gives the second statement. A full proof uses the The epsilon delta definition of a limit.

Worked examples

Example 1

Day 1: Ahiga’s function with two bad inputs

Ahiga studies f(x) = \dfrac{x^2 - 1}{x^2 - 3x + 2}. Find every discontinuity of f and classify each one.

Solution

A rational function is continuous wherever its bottom is not 0. The bottom factors as (x - 1)(x - 2), so the only candidates are 1 and 2. At each, f is undefined but defined at every nearby number, so each is a discontinuity. The kind depends on the one-sided limits. For x \ne 1, 2,

f(x) = \frac{(x - 1)(x + 1)}{(x - 1)(x - 2)} = \frac{x + 1}{x - 2}

where canceling is allowed because x \ne 1 (Limits by algebra: factoring and rationalizing).

At 1. The simplified form is continuous at 1, so both one-sided limits equal

\frac{1 + 1}{1 - 2} = -2

The limit exists and f(1) does not: a removable discontinuity. A table agrees: f(0.99) \approx -1.9703 and f(1.01) \approx -2.0303.

At 2. The top approaches 3, and the bottom x - 2 approaches 0 with the sign of x - 2. So

\begin{aligned} \lim_{x \to 2^+} f(x) &= \infty \\ \lim_{x \to 2^-} f(x) &= -\infty \end{aligned}

an infinite discontinuity, with the vertical asymptote x = 2 of Figure 1.

Example 2

Day 1: Nizhoni’s step

Nizhoni defines

p(x) = \begin{cases} 4 - x^2, & x < 1 \\ 2x + 3, & x \ge 1 \end{cases}

Classify the discontinuity at 1, find its size, and say whether p is continuous from either side there.

Solution

Each one-sided limit uses only the rule in force on that side, and each rule is a polynomial:

\begin{aligned} \lim_{x \to 1^-} p(x) &= 4 - 1 = 3 \\ \lim_{x \to 1^+} p(x) &= 2 + 3 = 5 \end{aligned}

Both exist as numbers and differ, so this is a jump, of size \lvert 5 - 3 \rvert = 2. The value p(1) = 2(1) + 3 = 5 comes from the second rule, so p is continuous from the right at 1 and not from the left (Figure 2). Every other input lies inside an open interval where p is one polynomial, so 1 is the only discontinuity.

Figure 2Nizhoni’s function of Example 2. The one-sided limits at 1 are 3 and 5, a jump of size 2; the value p(1) = 5 sits on the right piece.

Setting p(1) = 3 instead would make p continuous from the left and not from the right. The jump stays: no value lies on both sides.

Example 3

Day 1: A swing that never settles

Let s(x) = \sin\dfrac{1}{x} for x \ne 0 and s(0) = 0. Show that s has a discontinuity at 0 of none of the first three kinds. Then classify the discontinuity of t(x) = x^2 \sin\dfrac{1}{x} at 0.

Solution

For s. Take \displaystyle x_k = \dfrac{1}{\frac{\pi}{2} + 2\pi k} and \displaystyle y_k = \dfrac{1}{\frac{3\pi}{2} + 2\pi k} for k = 1, 2, 3, \ldots Both lists are positive and approach 0, yet

s(x_k) = \sin\left(\tfrac{\pi}{2} + 2\pi k\right) = 1, \qquad s(y_k) = -1

So inside every interval (0, \delta) the function takes both values 1 and -1, and no single number can be approached: \lim_{x \to 0^+} s(x) does not exist. The same happens from the left, since s(-x) = -s(x). Neither side runs off to infinity, because \lvert s(x) \rvert \le 1. So the discontinuity is not removable, not a jump, not infinite: it is oscillating. The picture is in What a limit is: graphs and tables.

For t. The swinging is now squeezed. For x \ne 0,

0 \le \left\lvert x^2 \sin\tfrac{1}{x} \right\rvert \le x^2

and x^2 \to 0, so \lim_{x \to 0} t(x) = 0 by the squeeze theorem of The squeeze theorem and the limit of sine x over x. The limit exists and t(0) is undefined: a removable discontinuity. Multiplying by a factor that tends to 0 changed the kind.

Example 4

Day 1: A jump with no pieces

Classify the discontinuity of v(x) = \dfrac{1}{1 + e^{1/x}} at 0, and show it is the only one.

Solution

For x \ne 0 the bottom is 1 plus a positive number, so it is never 0, and v is continuous at every x \ne 0 by the composition and quotient results. At 0 it is undefined. The one-sided limits of e^{1/x} were found in Infinite limits and vertical asymptotes: \infty from the right and 0 from the left. So

\lim_{x \to 0^+} v(x) = 0

because the bottom grows without bound, and

\lim_{x \to 0^-} v(x) = \frac{1}{1 + 0} = 1

Both exist as numbers and differ: a jump of size 1 (Figure 3). A single formula can jump; pieces are not needed. Already at x = 0.1 the value is \displaystyle \frac{1}{1 + e^{10}} \approx 0.0000454, and at x = -0.1 it is about 0.99995.

Figure 3The function of Example 4 is given by one formula, yet it jumps from 1 to 0 at x = 0.

Example 5

Day 2: Removing Ahiga’s hole

Remove the removable discontinuity of Ahiga’s f(x) = \dfrac{x^2 - 1}{x^2 - 3x + 2}, and show that no value at 2 makes f continuous there.

Solution

Example 1 found \lim_{x \to 1} f(x) = -2. The continuous extension is

F(x) = \begin{cases} \dfrac{x^2 - 1}{x^2 - 3x + 2}, & x \ne 1, 2 \\ -2, & x = 1 \end{cases}

and since \displaystyle \frac{x + 1}{x - 2} equals f(x) for x \ne 1, 2 and equals \displaystyle \frac{2}{-1} = -2 at 1, the same function has the one-line formula

F(x) = \frac{x + 1}{x - 2}, \qquad x \ne 2

That formula is rational with bottom 0 only at 2, so F is continuous on (-\infty, 2) and on (2, \infty). No other value at 1 would do, since continuity at 1 forces the value to be the limit.

At 2 suppose some number c were chosen as the value. The one-sided limits do not depend on c, and they are \infty and -\infty, so \lim_{x \to 2} f(x) is not a number and f cannot be continuous at 2 whatever c is. Simplifying the formula removed one break and left the other.

Example 6

Day 2: Doli’s floor of a square

Doli studies w(x) = \lfloor x^2 \rfloor on (-2, 2), the greatest integer at most x^2. Find every discontinuity and classify it. Is w continuous at 0, where the floor function itself jumps?

Solution

The outer floor is continuous at every non-integer, so by the composition result w is continuous wherever x^2 is not an integer. On (-2, 2), x^2 lies in [0, 4), so the candidates are where x^2 = 0, 1, 2, 3, that is \displaystyle x = 0, \pm 1, \pm\sqrt{2}, \pm\sqrt{3}. At each, ask from which side x^2 approaches its integer value.

At 1. As x \to 1^-, x^2 \to 1 through values below 1, where the floor is 0; as x \to 1^+, through values above 1, where it is 1. One-sided limits 0 and 1: a jump of size 1, and w(1) = 1 makes it continuous from the right.

At -1. The sides swap. As x \to -1^-, x is below -1 and x^2 is above 1, giving 1; as x \to -1^+, x^2 is below 1, giving 0. A jump of size 1, continuous from the left since w(-1) = 1.

At \displaystyle \pm\sqrt{2} and \displaystyle \pm\sqrt{3} the same reasoning gives jumps of size 1: from 1 to 2 and from 2 to 3 on the positive side, mirrored on the negative side.

At 0. From both sides x^2 approaches 0 through positive values, where the floor is 0. Both one-sided limits are 0, equal to w(0) = \lfloor 0 \rfloor = 0, so w is continuous at 0. The floor function jumps at 0, but x^2 never feeds it the inputs just below 0, where the other step lies.

So w has six discontinuities on (-2, 2), all jumps of size 1.

Example 7

Day 2: Two jumps that cancel

With Nizhoni’s p from Example 2, let q(x) = p(x) - 2\lfloor x \rfloor and r(x) = (x - 1)\,p(x). Classify every discontinuity of q and decide whether r is continuous at 1.

Solution

q at 1. Near 1, \lfloor x \rfloor is 0 on the left and 1 on the right. So

\begin{aligned} \lim_{x \to 1^-} q(x) &= 3 - 2(0) = 3 \\ \lim_{x \to 1^+} q(x) &= 5 - 2(1) = 3 \end{aligned}

and q(1) = 5 - 2 = 3. The jump of p (up 2) and the jump of -2\lfloor x \rfloor (down 2) cancel, and q is continuous at 1.

q at any other integer n. Here p is continuous, while -2\lfloor x \rfloor has one-sided limits -2(n - 1) and -2n, a jump of size 2. By the sum result, q has a jump of size 2 at every integer except 1. For example, at 2 the one-sided limits are 7 - 2 = 5 and 7 - 4 = 3. Between integers both terms are continuous, so these are all the discontinuities.

r at 1. The factor x - 1 is continuous with value 0 at 1, so by the product result r is continuous there. Directly: both one-sided limits are 0 \cdot 3 = 0 and 0 \cdot 5 = 0, and r(1) = 0 \cdot 5 = 0. Choosing a constant so that two pieces meet is the subject of Making a piecewise function continuous.

Boundaries

A discontinuity needs the function defined on both sides. At an end of a domain, such as 0 for \displaystyle \sqrt{x}, there is nothing to classify; Continuity on an interval handles ends with one-sided continuity. A number with a gap on both sides, like 0 for \displaystyle \frac{1}{x}, does count, even though \displaystyle \frac{1}{x} is continuous at every number of its domain.
A jump needs both one-sided limits to be numbers. If one side is infinite the discontinuity is infinite, even when the other side settles, as for e^{1/x} at 0.
Removing changes the function only at one point. The extension equals f everywhere else, and the simplified formula equals f only where f is defined.
The sum and product results need g continuous at a. Two broken functions can combine into anything: a continuous function (Example 7), a bigger jump, or a different kind.
The composition results need the inner limit and the outer behavior near it. A jump in the outer function matters only on the side from which the inner values arrive (Example 6).

Where this goes next

The next lesson, Making a piecewise function continuous, chooses constants in a function built from pieces so that the jump at a boundary has size 0. In differential calculus a function can have no derivative at a point for two reasons: a discontinuity of any of these kinds, or a continuous graph with a corner, cusp or vertical tangent, studied in Differentiability: corners, cusps and piecewise functions. Removable discontinuities return in every derivative computed from its definition: the difference quotient is undefined at h = 0, and when the derivative exists it is the value of the continuous extension there.

Exercises

Day 1: classifying from formulas

Find and classify every discontinuity of \dfrac{x^2 + x - 12}{x^2 - 9}.
Find and classify every discontinuity of \dfrac{x + 5}{x^2 + 5x}.
Let b(x) = x^2 for x \le 2 and b(x) = 10 - x for x > 2. Classify the discontinuity at 2, give its size, and say from which side b is continuous.
Classify the discontinuity of \dfrac{3\lvert x - 6 \rvert}{x - 6} at 6 and give its size.
Classify the discontinuity of \dfrac{1}{(x + 1)^2} at -1.
Classify the discontinuity of \cos\dfrac{1}{x} at 0.
Classify the discontinuity of (x - 2)\cos\dfrac{1}{x - 2} at 2.
Classify the discontinuity of \ln\lvert x - 3 \rvert at 3.

Day 1: one-sided facts and harder formulas

A function has \lim_{x \to 4^-} f(x) = 7, \lim_{x \to 4^+} f(x) = 7 and f(4) = 2. Classify the discontinuity at 4.
Classify the discontinuity of \dfrac{\sin x}{\lvert x \rvert} at 0 and give its size.
Find and classify every discontinuity of \dfrac{x^2 - x}{x^3 - x}.

Day 2: removing a removable discontinuity

What value at 0 makes \displaystyle \dfrac{\sqrt{x + 16} - 4}{x} continuous there?
Write the continuous extension of \dfrac{x^2 - x}{x^3 - x} at both of its removable discontinuities as one formula, and give its domain.
Can k(x) = \dfrac{x^2 - 1}{\lvert x - 1 \rvert} be made continuous at 1 by choosing k(1)?

Day 2: compositions and functions built from pieces

Find and classify every discontinuity of \lfloor \cos x \rfloor on (0, 2\pi).
Classify the discontinuity of \lfloor -x^2 \rfloor at 0, and say how to remove it.
Classify (x - 3)\lfloor x \rfloor at 2 and at 3.
The function h has a removable discontinuity at 0 with \lim_{x \to 0} h(x) = 2 and h(0) = 5; the function j has \lim_{x \to 0^-} j(x) = 1 and \lim_{x \to 0^+} j(x) = 4. Classify h + j and h j at 0.
Classify the discontinuity of e^{-1/x^2} at 0.

Reasoning

Explain why a removable discontinuity can be removed by exactly one value.
Give two functions, each with a jump at 0, whose sum is continuous at 0.
Nizhoni says: “If \lvert f \rvert is continuous at a, then f is continuous at a.” Give a counterexample.
Doli changes the value of f at a and nowhere else. Explain why a jump, infinite or oscillating discontinuity at a stays the same kind.
Prove: if f has a jump of size J at a and g is continuous at a with g(a) \ne 0, then f g has a jump of size \lvert g(a) \rvert J at a.
Ahiga says every discontinuity of a rational function is removable or infinite. Explain why he is right.

Answers

Removable at 3 with limit \displaystyle \frac{7}{6}; infinite at -3. The function is \displaystyle \frac{(x + 4)(x - 3)}{(x - 3)(x + 3)} = \frac{x + 4}{x + 3} for x \ne \pm 3. At -3 the top tends to 1, so the limit is \infty from the right and -\infty from the left.
Removable at -5 with limit \displaystyle -\frac{1}{5}; infinite at 0. For x \ne -5, 0 the function is \displaystyle \frac{x + 5}{x(x + 5)} = \frac{1}{x}, which tends to \infty from the right of 0 and -\infty from the left.
A jump of size 4, continuous from the left. The one-sided limits are 2^2 = 4 and 10 - 2 = 8, and b(2) = 4.
A jump of size 6. For x < 6 the function is -3; for x > 6 it is 3.
Infinite: both one-sided limits are \infty, since the top is 1 and the bottom is positive and tends to 0.
Oscillating. At \displaystyle x = \frac{1}{2\pi k} the value is 1 and at \displaystyle x = \frac{1}{(2k + 1)\pi} it is -1, with both lists approaching 0 from the right (and their negatives from the left); the values stay between -1 and 1.
Removable, limit 0: \displaystyle 0 \le \left\lvert (x - 2)\cos\frac{1}{x - 2} \right\rvert \le \lvert x - 2 \rvert, so the squeeze theorem gives 0, and the function is undefined at 2.
Infinite: \lvert x - 3 \rvert \to 0^+ from both sides, so both one-sided limits are -\infty.
Removable: the limit is 7 but the value is 2. Redefining f(4) = 7 removes it.
A jump of size 2. For x > 0 it is \displaystyle \frac{\sin x}{x} \to 1; for x < 0 it is \displaystyle -\frac{\sin x}{x} \to -1, using The squeeze theorem and the limit of sine x over x.
Removable at 0 (limit 1) and at 1 (limit \displaystyle \frac{1}{2}); infinite at -1. The function is \displaystyle \frac{x(x - 1)}{x(x - 1)(x + 1)} = \frac{1}{x + 1} for x \ne 0, \pm 1; at -1 it tends to \infty from the right and -\infty from the left.
\displaystyle \frac{1}{8}. Multiplying top and bottom by \displaystyle \sqrt{x + 16} + 4 gives \displaystyle \frac{1}{\sqrt{x + 16} + 4} for x \ne 0, which tends to \displaystyle \frac{1}{4 + 4} = \frac{1}{8}.
\displaystyle \frac{1}{x + 1}, for x \ne -1. It equals the function for x \ne 0, \pm 1, and its values 1 at 0 and \displaystyle \frac{1}{2} at 1 are the limits there.
No. For x < 1, \displaystyle k(x) = \frac{(x - 1)(x + 1)}{-(x - 1)} = -(x + 1) \to -2; for x > 1, k(x) = x + 1 \to 2. A jump of size 4, which no value removes.
Jumps of size 1 at \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2}. On \displaystyle \left(0, \frac{\pi}{2}\right), 0 < \cos x < 1, so the function is 0; on \displaystyle \left(\frac{\pi}{2}, \frac{3\pi}{2}\right), -1 \le \cos x < 0, so it is -1; on \displaystyle \left(\frac{3\pi}{2}, 2\pi\right) it is 0 again. At both jumps the value is \lfloor 0 \rfloor = 0: continuous from the left at \displaystyle \frac{\pi}{2}, from the right at \displaystyle \frac{3\pi}{2}. At \pi, where \cos x = -1, the function is -1 on both sides and at \pi: continuous.
Removable, limit -1. For 0 < \lvert x \rvert < 1, -x^2 lies in (-1, 0), so the function is -1, but at 0 it is \lfloor 0 \rfloor = 0. Redefining the value at 0 as -1 removes it.
At 2: a jump of size 1, one-sided limits (-1)(1) = -1 and (-1)(2) = -2, value -2, so continuous from the right. At 3: continuous, since x - 3 is 0 there; both one-sided limits are 0 \cdot 2 = 0 and 0 \cdot 3 = 0, and the value is 0.
Both jumps. h + j has one-sided limits 2 + 1 = 3 and 2 + 4 = 6, size 3; h j has 2 \cdot 1 = 2 and 2 \cdot 4 = 8, size 6. The value h(0) = 5 plays no part.
Removable, limit 0. From both sides \displaystyle -\frac{1}{x^2} \to -\infty, and e^u \to 0 as u \to -\infty. The function is undefined at 0, and the value 0 removes it.
A value c removes the discontinuity when the new function is continuous at a, which means c = \lim_{x \to a} f(x). Changing the value at a does not change the limit, and a limit has only one value, so c is forced.
One answer: f(x) = 1 for x \ge 0 and 0 for x < 0; g(x) = -1 for x \ge 0 and 0 for x < 0. Each jumps by 1 at 0, and f + g is 0 everywhere.
One answer: f(x) = 1 for x \ge 0 and f(x) = -1 for x < 0, at a = 0. Then \lvert f \rvert is the constant 1, continuous, but f has a jump of size 2.
One-sided limits at a use only values at x \ne a, which did not change. The kind is decided by the one-sided limits alone, so it stays the same. A removable discontinuity is the one exception: setting the value equal to the limit ends the discontinuity.
Let L^- and L^+ be the one-sided limits of f, with \lvert L^+ - L^- \rvert = J. By the one-sided product law, f g has one-sided limits L^- g(a) and L^+ g(a), both numbers. Their difference has size \lvert g(a) \rvert \lvert L^+ - L^- \rvert = \lvert g(a) \rvert J, which is not 0 because g(a) \ne 0 and J \ne 0. So f g has a jump of that size.
A rational function is defined on both sides of each zero a of its bottom, so these are its only discontinuities. Cancel every common factor x - a of top and bottom. If the bottom left over is not 0 at a, that form is continuous at a, so the limit exists: removable. If it is 0 at a, the top left over is not 0 at a, so near a a nonzero number is divided by a quantity tending to 0, and each one-sided limit is \infty or -\infty: infinite. Both one-sided limits exist as numbers and differ in neither case, so no jump occurs, and nothing oscillates.