A Hole from a Factor That Cancels

Classify the discontinuity of $f(x) = \frac{x^2 - x - 2}{x - 2}$ at $x = 2$ and remove it.

Check where $f$ is undefined: The denominator $x - 2$ is $0$ at $x = 2$ = $f(2)$ is undefined

Factor the numerator: $x^2 - x - 2 = (x - 2)(x + 1)$ = $f(x) = \frac{(x - 2)(x + 1)}{x - 2}$

Cancel the common factor: $f(x) = x + 1$ for every $x \neq 2$ = The factor $(x - 2)$ cancels completely

Find the limit: $\lim_{x \to 2} f(x) = 2 + 1$ = $= 3$, a finite limit

Classify and fix: Limit exists, $f(2)$ does not: a hole at $(2, 3)$. Define $f(2) = 3$. = Removable discontinuity

Answer: Removable discontinuity (hole) at $(2, 3)$. Setting $f(2) = 3$ makes $f$ continuous.

A Hole with the Wrong Value

Let $g(x) = x^2$ for $x \neq 1$ and $g(1) = 4$. What type of discontinuity is at $x = 1$?

Find the limit: Near $x = 1$ the rule is $x^2$, so $\lim_{x \to 1} g(x) = 1^2$ = $= 1$

Find the function value: The definition says $g(1) = 4$ = $g(1) = 4$

Compare: $\lim_{x \to 1} g(x) = 1 \neq 4 = g(1)$ = The limit exists but differs from $g(1)$

Classify and fix: The point $(1, 4)$ sits away from the curve; redefine $g(1) = 1$ = Removable discontinuity

Answer: Removable discontinuity at $x = 1$: there is a hole at $(1, 1)$ and a separate dot at $(1, 4)$. Redefining $g(1) = 1$ removes it.

Measuring a Jump

Let $h(x) = x + 2$ for $x < 1$ and $h(x) = x^2 + 5$ for $x \geq 1$. Classify the discontinuity at $x = 1$ and find the jump size.

Left-hand limit (use the rule for $x < 1$): $\lim_{x \to 1^-} (x + 2) = 1 + 2$ = $= 3$

Right-hand limit (use the rule for $x \geq 1$): $\lim_{x \to 1^+} (x^2 + 5) = 1 + 5$ = $= 6$

Compare the one-sided limits: Both are finite, but $3 \neq 6$ = Jump discontinuity

Jump size: $|6 - 3|$ = $= 3$

Answer: Jump discontinuity at $x = 1$ with jump size $3$. The value $h(1) = 6$ does not matter: no choice of $h(1)$ can close the gap.

An Infinite Discontinuity

Classify the discontinuity of $k(x) = \frac{x + 3}{x - 4}$ at $x = 4$.

Substitute $x = 4$: Numerator $4 + 3 = 7$, denominator $4 - 4 = 0$ = $\frac{7}{0}$: a nonzero number over zero

Approach from the right: $k(4.01) = \frac{7.01}{0.01} = 701$ = $\lim_{x \to 4^+} k(x) = \infty$

Approach from the left: $k(3.99) = \frac{6.99}{-0.01} = -699$ = $\lim_{x \to 4^-} k(x) = -\infty$

Classify: The one-sided limits are infinite = Infinite discontinuity, vertical asymptote $x = 4$

Answer: Infinite discontinuity at $x = 4$: the graph has the vertical asymptote $x = 4$.

Holes and Asymptotes in One Function

Find and classify every discontinuity of $r(x) = \frac{x^2 - 9}{x^2 - x - 6}$.

Factor both parts: $r(x) = \frac{(x - 3)(x + 3)}{(x - 3)(x + 2)}$ = Undefined at $x = 3$ and $x = -2$

Cancel the common factor: $r(x) = \frac{x + 3}{x + 2}$ for $x \neq 3$ = $(x - 3)$ cancels, $(x + 2)$ stays

Handle $x = 3$: $\lim_{x \to 3} r(x) = \frac{3 + 3}{3 + 2}$ = $= \frac{6}{5}$: hole at $\left(3, \frac{6}{5}\right)$

Handle $x = -2$: Simplified numerator $-2 + 3 = 1 \neq 0$, denominator $0$ = Vertical asymptote $x = -2$

Answer: Removable discontinuity (hole) at $\left(3, \frac{6}{5}\right)$ and infinite discontinuity at $x = -2$.

Mistake: Drawing a vertical asymptote at every zero of the denominator

Why: A zero of the denominator can also be a hole. In $\frac{x^2 - x - 2}{x - 2}$ the factor $(x - 2)$ cancels, so there is no asymptote at $x = 2$.

Correct: Factor and cancel first. Only factors left in the denominator give vertical asymptotes; cancelled factors give holes.

Mistake: Calling a jump removable because both one-sided limits exist

Why: Removable needs the two-sided limit, which exists only when both sides agree. At a jump they are different numbers.

Correct: Equal one-sided limits: removable (or continuous). Different finite one-sided limits: jump.

Mistake: Using $f(a)$ as the height of the hole

Why: At a hole $f(a)$ is missing or wrong. The hole sits at the height of the limit.

Correct: Find the $y$-coordinate of a hole by substituting $x = a$ into the simplified function: the hole of $\frac{x^2 - 9}{x^2 - x - 6}$ is at $\left(3, \frac{6}{5}\right)$.

Mistake: Thinking a cancelled factor always means a hole, even when it appears twice in the denominator

Why: In $\frac{x - 1}{(x - 1)^2}$ only one factor cancels; $\frac{1}{x - 1}$ is left, which still blows up at $x = 1$.

Correct: Cancel as far as possible, then check: if $(x - a)$ is still in the denominator, it is an infinite discontinuity.

Parking Fees (Jumps)

Parking garages charge by the started hour, so the price function is constant on each hour and jumps at every full hour.

A garage charges 4 dollars for up to 1 hour and 7 dollars for more than 1 hour up to 2 hours. At $t = 1$ the price jumps by 3 dollars.

Lenses and Images at Infinity

For a lens with focal length $10$ cm, an object at distance $u$ produces an image at distance $v(u) = \frac{10u}{u - 10}$. When the object reaches the focal point, the image moves off to infinity.

At $u = 15$ cm the image is at $v = \frac{150}{5} = 30$ cm. At $u = 10.1$ cm it is already at $v = \frac{101}{0.1} = 1010$ cm.

Fixing a Division by Zero in Software

Programs often compute an average rate of change $\frac{f(t) - f(a)}{t - a}$. At $t = a$ the formula gives $\frac{0}{0}$ and the program crashes, although the values nearby are perfectly smooth: a removable discontinuity.

For $f(t) = t^2$ and $a = 3$, the rate $\frac{t^2 - 9}{t - 3}$ simplifies to $t + 3$, so the programmer stores the value $6$ for $t = 3$.

Q: Why is a hole called removable?

A: Because one point is enough to repair it. The limit $L$ already exists, so defining (or redefining) $f(a) = L$ makes $\lim_{x \to a} f(x) = f(a)$ true, and the function becomes continuous at $a$.

Q: Can a jump discontinuity ever be removed?

A: No. The left side heads to one height and the right side to another. Whatever value you give $f(a)$, it can match at most one side, so the two-sided limit still does not exist.

Q: Does a function value at $x = a$ change the type?

A: No. The type depends only on the limits near $a$. A jump stays a jump whether $f(a)$ is the left value, the right value or undefined, and a hole stays removable whether $f(a)$ is missing or wrong.

Q: Are there other types of discontinuity?

A: Yes, but they are rarer. For example $\sin\left(\frac{1}{x}\right)$ oscillates faster and faster near $x = 0$, so it has no one-sided limits at all (an oscillating discontinuity). Jumps, infinite and oscillating discontinuities are all non-removable; many books call only the infinite and oscillating ones essential.