🧱 Shaky on the basics? Start with Foundations →
Limits & Continuity — Visual Lab
Module 1. Calculus begins with a single, slippery question: what value is a function heading toward — even at a point it never reaches? Slide a point in from each side and watch the answer emerge on the coordinate plane and in the numbers at once.
The limit \(\displaystyle\lim_{x \to c} f(x)\) asks where the outputs of \(f\) are headed as the input \(x\) creeps toward \(c\) — not what happens to land at \(c\) itself. That gap is the whole point: a function can aim squarely at a value through a hole it never touches. The limit only exists when the approach from the left and the approach from the right agree. Switch functions below and keep the graph, the two-sided table, and the verdict in perfect agreement.
Limit Explorer
Pick a function, then drag the approach distance slider to send a point in from both sides of \(x = c\). Use Zoom for a closer look, and the ×0.1 closer button to dive in fast. Read the two-sided table and the verdict: does the limit exist, and is the function continuous?
What you're seeing
- The dashed vertical line is \(x = c\). That's the target input. The limit is about the height the curve approaches as you slide toward this line — from either side.
- Two colored markers are your approaching points. One comes in from the left (\(x \to c^-\)), one from the right (\(x \to c^+\)). The distance slider controls how close they get.
- An open circle is a hole. On the removable function, the curve aims dead-on at a value the function never actually takes — \(f(c)\) is undefined, yet the limit exists.
- A gap between pieces is a jump. On the jump function the left and right pieces land at different heights, so the two-sided limit does not exist.
- The table is the limit, numerically. Each row halves the distance to \(c\); when both columns home in on the same number, that shared number is the limit.
Try this
- Start on "Removable hole." Drag the distance toward 0 (or tap ×0.1 a few times). Watch both table columns close in on \(4\) even though the graph has a hole at \(x = 2\). The limit is \(4\); the function value isn't.
- Switch to "Jump." Drive the distance down again. Notice the left column heads for \(1\) while the right heads for \(4\) — they never meet, so the verdict flips to does not exist.
- Compare to "Continuous." Here both sides and \(f(2)\) all equal \(4\). This is the definition of continuity: \(\lim_{x\to c} f(x) = f(c)\).
- Use Zoom. Tighten the window to \(\pm 0.5\) on the removable case. Up close, the hole and the surrounding line are obvious — this is exactly how a grapher hides (or reveals) a removable discontinuity.
Worked Examples
Two limits done the way you'll do them on paper: when direct substitution stalls at \(\tfrac{0}{0}\), and when the inputs run off to infinity.
Evaluate \(\displaystyle\lim_{x \to 2} \frac{x^{2}-4}{x-2}\)
This is exactly the "Removable hole" function in the lab above.
- Try direct substitution first. Plugging \(x = 2\) gives \(\dfrac{2^2 - 4}{2 - 2} = \dfrac{0}{0}\) — indeterminate. That's a signal to simplify, not a final answer.
- Factor the numerator. \(x^2 - 4 = (x-2)(x+2)\) (difference of squares).
- Cancel the common factor. For \(x \ne 2\), \(\dfrac{(x-2)(x+2)}{x-2} = x + 2\). The cancelled \((x-2)\) is the hole.
- Substitute into the simplified form. \(\displaystyle\lim_{x\to 2}(x+2) = 2 + 2 = 4\).
Evaluate \(\displaystyle\lim_{x \to \infty} \frac{3x^{2}-5}{x^{2}+2x}\)
"What height does the curve flatten toward as \(x\) runs to the right forever?"
- Compare the degrees. Top and bottom are both degree \(2\) — a tie. For a tie, the limit is the ratio of leading coefficients.
- Divide every term by the highest power, \(x^2\). \(\dfrac{3 - \frac{5}{x^2}}{1 + \frac{2}{x}}\).
- Send each \(1/x\) term to \(0\). As \(x \to \infty\), \(\dfrac{5}{x^2}\to 0\) and \(\dfrac{2}{x}\to 0\), leaving \(\dfrac{3 - 0}{1 + 0}\).
- Read the ratio. \(\dfrac{3}{1} = 3\) — the leading-coefficient ratio, as predicted.
Why it matters
Limits aren't a warm-up you leave behind — they're the engine the rest of calculus runs on.
Key Vocabulary
The precise words a mathematician uses to describe what the sliders are doing.
The single value \(L\) that \(f(x)\) approaches as \(x\) approaches \(c\): \(\lim_{x\to c} f(x) = L\). It describes the destination, not the arrival.
The value approached from a single direction: \(\lim_{x\to c^-}\) (from the left) or \(\lim_{x\to c^+}\) (from the right). The two-sided limit exists only if these agree.
\(f\) is continuous at \(c\) when \(\lim_{x\to c} f(x) = f(c)\) — the limit exists, the value exists, and they match (no holes, jumps, or breaks).
A hole: the two-sided limit exists, but \(f(c)\) is missing or misplaced. Re-defining the one point would "remove" it.
A jump: the one-sided limits disagree (a step). An infinite: \(f\) shoots to \(\pm\infty\) at a vertical asymptote.
The value \(f(x)\) approaches as \(x\to\pm\infty\); a finite answer is a horizontal asymptote \(y = L\) describing end behavior.
AP Calculus AB/BC — Unit 1 (Limits & Continuity)
This lab lives in Unit 1 of the AP Calculus AB/BC framework (equivalently, the opening unit of a collegiate Calculus I course), where students estimate limits numerically and graphically, evaluate them analytically, classify discontinuities, and apply the foundational theorems.
Note: Calculus is not a Texas TEKS/STAAR course — there is no STAAR EOC. Topic codes above reference the AP Calculus AB/BC Course and Exam Description (College Board) Unit 1, the standard collegiate Calculus I sequence.
Ready for the full course map? Head back to Calculus, read the Student Support, or check the Pacing Guide to see where Module 1 sits in the year. Shaky on factoring, asymptotes, or function notation? The Foundations page rebuilds them first.
Topic structure follows the College Board AP Calculus AB/BC Course and Exam Description, Unit 1 (Limits & Continuity), aligned to a collegiate Calculus I sequence. AP® is a trademark of the College Board, which does not endorse this site. Classroom use is non-commercial.