← Module 1 Visual Lab Calculus Student Support
Mathematical Architects · Calculus · Foundations

Get Ready: Module 1 — Limits & Continuity

Everyone starts somewhere. Before we open the door to calculus, we'll firm up the precalculus and algebra it stands on. This page rebuilds the handful of skills Module 1 leans on — clearly, one step at a time, with nothing assumed. Go at your own pace. There's no rush and no wrong place to begin.

4 Foundation Skills Worked Examples Self-Paced

Limits ask where a function is headed. Almost every limit you'll work either factors away a \(\tfrac{0}{0}\), or watches inputs run off to infinity along an asymptote. If factoring, \(f(x)\) notation, end behavior, or domain feel fuzzy right now, that's completely okay — we'll firm them up together below, then you'll be ready for the lab.

🧱

Skills to build first

Four short skills. Read the idea, follow the worked example, then try one yourself — the answer is one click away whenever you want it.

1 Skill

Factoring & rational expressions

What it is: rewriting a polynomial as a product of factors — pulling out a common factor, recognizing a difference of squares \(a^2 - b^2 = (a-b)(a+b)\), or factoring a trinomial. A rational expression is a fraction of polynomials; factoring lets you cancel matching factors. Why you need it: the most common limit move in Module 1 is escaping the \(\tfrac{0}{0}\) trap. You factor the top and bottom, cancel the troublemaker, and then substitute.

Worked example — simplify \(\dfrac{x^2 - 9}{x - 3}\)
  1. Factor the numerator. \(x^2 - 9\) is a difference of squares: \((x-3)(x+3)\).
  2. Write the fraction factored. \(\dfrac{(x-3)(x+3)}{x-3}\).
  3. Cancel the common factor \((x-3)\) (valid for \(x \ne 3\)): you're left with \(x + 3\).
  4. So \(\dfrac{x^2-9}{x-3} = x + 3\) for all \(x \ne 3\). (The point \(x = 3\) becomes a hole — exactly what makes a limit interesting.)

In a limit, after cancelling you'd substitute \(x = 3\) into \(x+3\) to get \(6\).

Try it 1 — Factor and simplify \(\dfrac{x^2 - 25}{x + 5}\).
Show answer

\(x^2 - 25 = (x-5)(x+5)\). Cancel \((x+5)\): \(\dfrac{(x-5)(x+5)}{x+5} = x - 5\) (for \(x \ne -5\)).

Try it 2 — Simplify \(\dfrac{x^2 + 5x + 6}{x + 2}\).
Show answer

Factor the trinomial: \(x^2 + 5x + 6 = (x+2)(x+3)\). Cancel \((x+2)\): the result is \(x + 3\) (for \(x \ne -2\)).

Practice more (free) Khan Academy → IXL →
2 Skill

Function notation \(f(x)\)

What it is: \(f(x)\) is a name for "the function \(f\), fed an input \(x\)." So \(f(3)\) means "put \(3\) in for \(x\) and report the output." You can also feed it an expression, like \(f(x+h)\). Why you need it: limits are written in this language — \(\lim_{x\to c} f(x)\) asks about the outputs of \(f\). And the very next thing you'll meet, the derivative, is built from \(f(x+h)\), so substituting a whole expression into a function needs to feel automatic.

Worked example — if \(f(x) = x^2 + 1\), find \(f(3)\) and \(f(x+h)\)
  1. \(f(3)\): the \(3\) replaces every \(x\). \(f(3) = 3^2 + 1 = 9 + 1 = 10\).
  2. \(f(x+h)\): the whole \((x+h)\) replaces every \(x\). \(f(x+h) = (x+h)^2 + 1\).
  3. Expand if needed. \((x+h)^2 = x^2 + 2xh + h^2\), so \(f(x+h) = x^2 + 2xh + h^2 + 1\).

The point \((3, 10)\) sits on the graph: input \(3\), output \(10\).

Try it 1 — If \(f(x) = 2x - 5\), find \(f(4)\).
Show answer

\(f(4) = 2(4) - 5 = 8 - 5 = 3\). So \(f(4) = 3\), i.e. the point \((4, 3)\).

Try it 2 — If \(g(x) = x^2\), write and simplify \(g(x+h) - g(x)\). (Hint: this is the top of the derivative formula.)
Show answer

\(g(x+h) = (x+h)^2 = x^2 + 2xh + h^2\). Subtract \(g(x) = x^2\): \(g(x+h) - g(x) = 2xh + h^2\).

Practice more (free) Khan Academy → IXL →
3 Skill

Asymptotes & end behavior (from Pre-Calculus)

What it is: an asymptote is a line the graph hugs but never crosses. A vertical asymptote occurs where a denominator hits zero (the graph rockets to \(\pm\infty\)); a horizontal asymptote describes end behavior — the height the graph levels off toward as \(x \to \pm\infty\). Why you need it: "limits at infinity" are literally end behavior, and "infinite discontinuities" are vertical asymptotes. Seeing them in advance makes those limits read like a picture you already know.

Worked example — end behavior of \(\dfrac{2x + 1}{x - 3}\)
  1. Vertical asymptote: where the bottom is zero. \(x - 3 = 0 \Rightarrow x = 3\). Near \(x = 3\), outputs blow up.
  2. Horizontal asymptote: compare degrees. Top and bottom are both degree \(1\) (a tie), so the asymptote is the ratio of leading coefficients: \(\tfrac{2}{1} = 2\).
  3. Confirm with big \(x\). Divide each term by \(x\): \(\dfrac{2 + \frac{1}{x}}{1 - \frac{3}{x}} \to \dfrac{2}{1} = 2\) as \(x \to \infty\).
  4. So \(y = 2\) is the horizontal asymptote; the graph flattens toward height \(2\) at both far ends.

Quick rule: bottom-heavy \(\to\) horizontal asymptote \(y = 0\); equal degrees \(\to\) ratio of leading coefficients; top-heavy \(\to\) no horizontal asymptote (it runs off).

Try it 1 — Find the vertical and horizontal asymptotes of \(\dfrac{5}{x + 2}\).
Show answer

Vertical: \(x + 2 = 0 \Rightarrow x = -2\). Horizontal: the bottom's degree is larger (1 > 0), so \(y = 0\).

Try it 2 — What height does \(\dfrac{4x^2}{2x^2 + 7}\) approach as \(x \to \infty\)?
Show answer

Equal degrees (both \(2\)) → ratio of leading coefficients \(\tfrac{4}{2} = 2\). So it approaches \(2\) (horizontal asymptote \(y = 2\)).

Practice more (free) Khan Academy → IXL →
4 Skill

Domain of a function

What it is: the domain is the set of all inputs \(x\) a function is allowed to take. The two usual deal-breakers: you can't divide by zero, and you can't take an even root of a negative. Why you need it: the points excluded from the domain are precisely where something interesting happens to a limit — a hole, a jump, or a vertical asymptote. Knowing the domain tells you where to look.

Worked example — domain of \(\dfrac{x + 1}{x^2 - 4}\)
  1. Find what breaks it. It's a fraction, so the danger is the denominator equalling zero.
  2. Set the denominator to zero and solve. \(x^2 - 4 = 0 \Rightarrow (x-2)(x+2) = 0 \Rightarrow x = 2\) or \(x = -2\).
  3. Exclude those inputs. Every other real number is fine.
  4. State the domain. All real \(x\) except \(x = 2\) and \(x = -2\): \(\;\{x \mid x \ne \pm 2\}\).

Those two excluded points are exactly where you'd check for a vertical asymptote or a removable hole in a limit problem.

Try it 1 — What is the domain of \(\dfrac{3}{x - 5}\)?
Show answer

Denominator zero at \(x - 5 = 0 \Rightarrow x = 5\). Domain: all real numbers except \(x = 5\).

Try it 2 — What is the domain of \(\sqrt{x - 1}\)?
Show answer

An even root needs a non-negative inside: \(x - 1 \ge 0 \Rightarrow x \ge 1\). Domain: \([1, \infty)\).

Practice more (free) Khan Academy → IXL →
✅

Quick Readiness Check

Six short questions spanning all four skills. Try each one first, then reveal the answer to check yourself.

  1. Factor and simplify \(\dfrac{x^2 - 16}{x - 4}\).
    Show answer
    Show answer

    \(x^2 - 16 = (x-4)(x+4)\); cancel \((x-4)\) to get \(x + 4\) (for \(x \ne 4\)).

  2. If \(f(x) = x^2 - 2x\), find \(f(5)\).
    Show answer
    Show answer

    \(f(5) = 5^2 - 2(5) = 25 - 10 = 15\).

  3. Find the vertical asymptote of \(\dfrac{x}{x - 7}\).
    Show answer
    Show answer

    Denominator zero at \(x - 7 = 0 \Rightarrow x = 7\). Vertical asymptote: \(x = 7\).

  4. What height does \(\dfrac{6x}{2x + 1}\) approach as \(x \to \infty\)?
    Show answer
    Show answer

    Equal degrees (both \(1\)) → ratio of leading coefficients \(\tfrac{6}{2} = 3\). Horizontal asymptote \(y = 3\).

  5. What is the domain of \(\dfrac{x - 1}{x + 3}\)?
    Show answer
    Show answer

    Denominator zero at \(x = -3\), so the domain is all real numbers except \(x = -3\).

  6. Simplify \(\dfrac{x^2 + 7x + 12}{x + 3}\), then substitute \(x = -3\) into the result.
    Show answer
    Show answer

    \(x^2 + 7x + 12 = (x+3)(x+4)\); cancel \((x+3)\) to get \(x + 4\). At \(x = -3\): \(-3 + 4 = 1\). (That's the limit as \(x \to -3\)!)

If these feel comfortable, you're ready for the module. If one or two felt shaky, that's your cue — scroll back up to that skill, re-read the worked example, and try a couple from the free practice links. No score to hit, no clock running. Come back when it clicks.

Floor poured. When the skills above feel steady, step into the module — the Limit Explorer lets you see every one of these ideas move at once.

Module 1 Overview & Visual Lab Jump to the Limit Explorer Student Support