A Hole and a Dot at the Same x

The graph of $f$ is the line $y = x + 1$ with an open circle at $(2, 3)$ and a filled dot at $(2, 5)$ (the first graph in this lesson). Find $\lim_{x \to 2} f(x)$ and $f(2)$.

Follow the curve from the left: As $x$ goes $1.9, 1.99, 1.999$, the line rises toward the open circle = Height approaches $3$

Follow the curve from the right: As $x$ goes $2.1, 2.01, 2.001$, the line falls toward the same open circle = Height approaches $3$

Compare the two sides: Both sides head for the same height = $\lim_{x \to 2} f(x) = 3$

Find the filled dot at $x = 2$: The filled dot sits at $(2, 5)$, away from the line = $f(2) = 5$

Answer: $\lim_{x \to 2} f(x) = 3$ but $f(2) = 5$. The limit follows the curve, the function value is the filled dot.

A Hole with No Dot

On the second graph in this lesson, $g$ follows $y = (x - 1)^2$ up to $x = 2$ and $y = 3 - x$ after $x = 2$, with an open circle at $(2, 1)$ and no filled dot at $x = 2$. Find $\lim_{x \to 2} g(x)$ and $g(2)$.

Left side: The parabola approaches the height $(2 - 1)^2 = 1$ = Approaches $1$

Right side: The line approaches the height $3 - 2 = 1$ = Approaches $1$

Compare: Both sides head for $1$ = $\lim_{x \to 2} g(x) = 1$

Look for a filled dot at $x = 2$: There is only the open circle, no filled dot = $g(2)$ is undefined

Answer: $\lim_{x \to 2} g(x) = 1$ even though $g(2)$ is undefined. A missing point does not stop the limit from existing.

A Jump: the Limit Does Not Exist

On the same graph, $g$ follows $y = x + 3$ up to a filled dot at $(-1, 2)$, and just to the right of $x = -1$ it continues as $y = (x - 1)^2$, starting from an open circle at $(-1, 4)$. Find $\lim_{x \to -1} g(x)$ and $g(-1)$.

Left side: The line $y = x + 3$ approaches the height $-1 + 3 = 2$ = Approaches $2$

Right side: The parabola approaches the height $(-1 - 1)^2 = 4$ = Approaches $4$

Compare: $2 \neq 4$: the two sides disagree = $\lim_{x \to -1} g(x)$ does not exist

Read the filled dot: The filled dot is at $(-1, 2)$ = $g(-1) = 2$

Answer: $\lim_{x \to -1} g(x)$ does not exist (the sides approach $2$ and $4$), yet $g(-1) = 2$. A function can have a value where it has no limit.

A Parabola with a Moved Point

The graph of $h$ is the parabola $y = 4 - x^2$, except that at $x = 1$ there is an open circle at $(1, 3)$ and a filled dot at $(1, 0)$. Find $\lim_{x \to 1} h(x)$, $h(1)$ and $\lim_{x \to 0} h(x)$.

Height of the hole: Both sides of the parabola approach $4 - 1^2 = 3$ = $\lim_{x \to 1} h(x) = 3$

Value at $x = 1$: The filled dot is at $(1, 0)$ = $h(1) = 0$

Limit at $x = 0$: Nothing special happens at $x = 0$: the curve passes smoothly through $(0, 4)$ = $\lim_{x \to 0} h(x) = 4$

Compare with $h(0)$: The point $(0, 4)$ is on the curve = $h(0) = 4$, same as the limit

Answer: $\lim_{x \to 1} h(x) = 3$, $h(1) = 0$, and $\lim_{x \to 0} h(x) = h(0) = 4$. Where the graph has no break, the limit equals the value.

Three Limits on One Graph

The graph of $p$: for $x < 1$ it is the horizontal line $y = 2$ ending in an open circle at $(1, 2)$; there is a filled dot at $(1, 4)$; for $1 < x \le 3$ it is the line $y = x + 1$, starting at an open circle at $(1, 2)$ and ending at a filled dot at $(3, 4)$; for $x > 3$ it is the line $y = 1$, starting at an open circle at $(3, 1)$. Find $\lim_{x \to 1} p(x)$, $\lim_{x \to 2} p(x)$ and $\lim_{x \to 3} p(x)$.

At $x = 1$, both sides: Left: $y = 2$. Right: $y = x + 1 \to 1 + 1 = 2$. Same height = $\lim_{x \to 1} p(x) = 2$

Compare with $p(1)$: The filled dot is at $(1, 4)$ = $p(1) = 4 \neq 2$

At $x = 2$: The line $y = x + 1$ passes smoothly through $(2, 3)$ = $\lim_{x \to 2} p(x) = 3$

At $x = 3$, both sides: Left: $y = x + 1 \to 4$. Right: $y = 1$. Different heights = $\lim_{x \to 3} p(x)$ does not exist

Answer: $\lim_{x \to 1} p(x) = 2$ (although $p(1) = 4$), $\lim_{x \to 2} p(x) = 3$, and $\lim_{x \to 3} p(x)$ does not exist (the sides approach $4$ and $1$).

Mistake: Reading the filled dot as the limit: answering $\lim_{x \to 2} f(x) = 5$ because there is a dot at $(2, 5)$

Why: The limit describes where the curve is heading near $x = 2$. A single dot away from the curve does not change where the curve is heading.

Correct: Follow the curve from both sides: it heads for the open circle at height $3$, so $\lim_{x \to 2} f(x) = 3$, while $f(2) = 5$.

Mistake: Saying the limit does not exist because there is a hole

Why: A hole only means $f(a)$ is missing. The limit never looks at $x = a$ itself, only at the values around it.

Correct: If both sides approach the height of the hole, the limit exists and equals that height, even when $f(a)$ is undefined.

Mistake: At a jump, giving the height of the filled dot (or of one side) as the limit

Why: A two-sided limit needs both sides to agree. At a jump, the left side and the right side approach different heights.

Correct: Check both sides. If they approach different heights, write that the limit does not exist, whatever the filled dot shows.

Mistake: Answering with the $x$-value: writing $\lim_{x \to 2} f(x) = 2$

Why: The $2$ under the limit sign is where $x$ goes. The limit itself is a height, a $y$-value.

Correct: Read the answer on the $y$-axis: the height the curve approaches as $x$ gets close to $2$.

Shipping Costs with a Weight Limit

Couriers charge by weight class, so the graph of cost against weight is a step graph with jumps at the class boundaries.

A courier charges 5 dollars up to 2 kg and 8 dollars above 2 kg. Just below 2 kg the graph sits at 5, just above it sits at 8, so the limit of the cost as the weight approaches 2 kg does not exist.

A Glitch in Sensor Data

Weather stations and fitness trackers plot readings over time. When one reading is lost or wrong, the graph gets a hole or a stray dot, and the limit recovers the real trend.

A heart rate graph that approaches 120 beats per minute from both sides of one moment, with a single faulty reading of 0 at that moment, still has a limit of 120 there.

Holes on a Graphing Calculator

Graphing software draws $f(x) = \frac{x^2 - 9}{x - 3}$ as what looks like a straight line, but tracing at $x = 3$ gives an error. The graph has a hole there, and its height is the limit.

For $f(x) = \frac{x^2 - 4}{x - 2}$ the graph looks like $y = x + 2$ with a hole at $(2, 4)$, so $\lim_{x \to 2} f(x) = 4$ while $f(2)$ is undefined.

Q: Can the limit exist if $f(a)$ is undefined?

A: Yes. If the graph has a hole at $x = a$ and both sides approach the height of the hole, the limit equals that height. For example $\frac{x^2 - 4}{x - 2}$ is undefined at $x = 2$, but its limit there is $4$.

Q: If there is a filled dot at $x = a$, does that mean the limit exists?

A: No. The dot only gives $f(a)$. At a jump the graph has a filled dot, but the two sides approach different heights, so the limit does not exist.

Q: What do I write when a limit does not exist?

A: Write $\lim_{x \to a} f(x)$ does not exist (often shortened to DNE) and give the reason, for example: the left side approaches $2$ and the right side approaches $4$.

Q: How accurate is reading a limit from a graph?

A: Only as accurate as the graph. Use grid lines and labelled points, and when the height is not a clear grid value, confirm it with a table of values or with algebra (the next lessons).