Interactive Learning

Welcome to Calculus I

Calculus is how we describe change: how a curve bends, how fast a quantity moves, and how small contributions add up to a total. Here you will move from limits and continuity into derivatives and their uses (including optimization), then into integration and the Fundamental Theorem—the bridge between instantaneous rate of change and accumulated growth.

Each chapter is built to be used, not skimmed: interact with live graphs, open worked examples line by line, and try practice problems with stepped solutions. Follow the numbered lessons in the sidebar at your own pace, take the Practice Exam when you are ready, and use Exam Remediation to focus on what still needs work.

⬡ Limits & Continuity ⬡ Differentiation ⬡ Optimization ⬡ Integration ⬡ Fundamental Theorem

Limits & Continuity

Start here: see what value a function is heading toward as $x$ gets close to a number—even when the function is not defined at that spot.

Topics in this chapter (review order)
1. Continuity — $f(a)$ defined, limit exists, limit $=f(a)$
2. Evaluate Basic Limits — direct substitution when $f$ is continuous at $a$
3. Solve Indeterminate Forms using Factoring — factor and cancel to resolve $\tfrac{0}{0}$
4. Rationalizing, and Simplifying Fractions — conjugate multiplication for radicals
5. Evaluating Limits at Infinity — divide by highest power, dominant terms
6. Special Limits & Theorems — Squeeze Theorem, trig limits ($\tfrac{\sin x}{x}$, $\tfrac{1-\cos x}{x^2}$), L'Hôpital
Big idea
What is a limit?

A limit describes the intended value of $f(x)$ as $x$ approaches $a$ — without $x$ ever needing to equal $a$. The value $f(a)$ may be undefined.

One-sided limits

The limit exists only if both $\lim_{x\to a^-}f(x)$ and $\lim_{x\to a^+}f(x)$ exist and agree. For piecewise jumps, the left and right limits can differ.

Indeterminate forms

When $\frac{0}{0}$ or $\frac{\infty}{\infty}$ arises, try: (1) factoring and canceling, (2) conjugate for radicals, (3) L'Hôpital's Rule: $\lim\frac{f}{g}=\lim\frac{f'}{g'}$.

The limit of $f(x)$ as $x$ approaches $a$ is the value $L$ that $f(x)$ gets arbitrarily close to — without $x$ ever needing to equal $a$.

$$\lim_{x \to a} f(x) = L$$

The value of $f(a)$ is irrelevant — $f$ need not even be defined at $a$.

Formal ε-δ Definition
$\displaystyle\lim_{x\to a}f(x)=L$ means: for every $\varepsilon>0$ there exists $\delta>0$ such that $$0<|x-a|<\delta \;\implies\; |f(x)-L|<\varepsilon$$
Limit Laws
$\lim[f\pm g]=\lim f\pm\lim g$
$\lim[f\cdot g]=\lim f\cdot\lim g$
$\lim[f/g]=\lim f/\lim g\;(g\neq0)$
$\lim[f]^n=(\lim f)^n$

A limit exists if and only if both one-sided limits exist and agree:

$$\lim_{x\to a}f(x)=L \;\iff\; \lim_{x\to a^-}f(x)=L \;\text{ and }\; \lim_{x\to a^+}f(x)=L$$

Limit Visualizer — $f(x)=\frac{x^2-4}{x-2}$ as $x\to 2$
x = 0.125
$x$ =
$f(x)$ =
Limit (as $x\to2$) = 4.0000

1. Continuity

Why learn this?

Trust your graphs. Many formulas only make sense when a quantity changes smoothly—no sudden jumps where “just plug in” would give a misleading answer.

Helpful hints

Three-part checklist. Continuity at $a$ needs $f(a)$ defined, $\lim_{x\to a}f(x)$ to exist, and that limit to equal $f(a)$—if any link fails, the function is discontinuous there.

Remember

Removable gap when the limit exists but $f(a)$ is missing or wrong · One-sided mismatch means the two-sided limit does not exist

$f$ is continuous at $a$ when $f(a)$ is defined, $\displaystyle\lim_{x\to a}f(x)$ exists, and $\displaystyle\lim_{x\to a}f(x)=f(a)$.

Continuity at a Point
  1. $f(a)$ is defined
  2. $\displaystyle\lim_{x\to a}f(x)$ exists
  3. $\displaystyle\lim_{x\to a}f(x)=f(a)$

In the Limit Visualizer above, the limit at $x=2$ is 4, but $f(2)$ is undefined — a removable discontinuity.

2. Evaluate Basic Limits

Why learn this?

Fast evaluation. A lot of limits come from “nice” functions built from polynomials, trig, exponentials, and logs—when things are continuous at the spot you care about, substitution is the first move.

Helpful hints

Check before you plug. Confirm $f$ is continuous at $a$ (polynomials and rationals where the denominator is nonzero are safe). If you get $\tfrac{0}{0}$ or other indeterminate forms, stop and simplify.

Remember

Continuous at $a$ $\Rightarrow$ $\lim_{x\to a}f(x)=f(a)$ · Composition of continuous functions is continuous on its domain

When $f$ is continuous at $a$, the limit equals $f(a)$. For polynomials, rationals (where defined), and compositions of continuous functions — just plug in.

Step 1 / 3
1
Check continuity. $f(x)=x^2+3x-1$ is a polynomial — continuous everywhere. Direct substitution applies.
2
Substitute. Plug in $x=2$: $2^2+3(2)-1=4+6-1=9$
Answer: $\displaystyle\lim_{x\to 2}(x^2+3x-1)=\boxed{9}$

3. Solve Indeterminate Forms using Factoring

Why learn this?

Rational expressions everywhere. Ratios of polynomials show up all over calculus; the limit can look like $\tfrac{0}{0}$ until you factor and cancel the piece that made both top and bottom zero.

Helpful hints

Factor first. Difference of squares, sum/difference of cubes, and grouping usually expose the $(x-a)$ factor that makes the numerator and denominator both vanish at $x=a$.

Remember

$\tfrac{0}{0}$ means “try algebra,” not “the limit is 0” · Cancel common factors valid for $x\to a$ because $x\neq a$ in the limit

When substitution yields $\tfrac{0}{0}$, factor and cancel the common factor. The limit of the simplified form is the same as the original.

Step 1 / 5
1
Check by substitution. Plugging $x=3$ gives $\tfrac{0}{0}$ — an indeterminate form. Direct substitution fails; we must simplify first.
2
Factor the numerator. $x^2-9=(x-3)(x+3)$
3
Cancel the common factor. Since $x\to3$ (not $x=3$), we may cancel $(x-3)$: $\dfrac{(x-3)(x+3)}{x-3}=x+3$
4
Substitute. $\displaystyle\lim_{x\to3}(x+3)=3+3=6$
Answer: $\displaystyle\lim_{x\to3}\frac{x^2-9}{x-3}=\boxed{6}$

4. Rationalizing, and Simplifying Fractions

Why learn this?

Radicals in measurements. Distances, norms, and wave speeds often involve square roots; rationalizing turns awkward $\sqrt{\cdot}$ differences into polynomials you can factor.

Helpful hints

Multiply by the conjugate. For $\sqrt{x}-c$, use $\sqrt{x}+c$ on top and bottom so $(\sqrt{x}-c)(\sqrt{x}+c)=x-c^2$ removes the radical from the numerator.

Remember

Conjugate flips the sign on the radical term · Simplify before substituting again

For limits with radicals (e.g. $\sqrt{x}-c$), multiply by the conjugate to remove the radical and reveal a cancelable factor.

Step 1 / 4
1
Identify the form. At $x=4$: numerator $=\sqrt{4}-2=0$, denominator $=0$. We have $\tfrac{0}{0}$ — rationalize.
2
Multiply by conjugate. $\dfrac{\sqrt{x}-2}{x-4}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+2}=\dfrac{x-4}{(x-4)(\sqrt{x}+2)}=\dfrac{1}{\sqrt{x}+2}$ for $x\neq4$
3
Substitute. $\displaystyle\lim_{x\to4}\frac{1}{\sqrt{x}+2}=\frac{1}{\sqrt{4}+2}=\frac{1}{4}$
Answer: $\displaystyle\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}=\boxed{\dfrac{1}{4}}$

5. Evaluating Limits at Infinity

Why learn this?

Long-run behavior. “What happens way out there?” is the same question as a limit at $\infty$: end behavior of a graph, long-term trends, horizontal asymptotes—not just one finite $x$.

Helpful hints

Divide by the highest power. Factor $x^n$ from numerator and denominator so every remaining term is a constant or tends to 0; then read off the horizontal asymptote or blow-up.

Remember

Equal degrees → ratio of leading coefficients · Denominator wins → limit $0$ · Numerator wins → $\pm\infty$ (watch signs)

Divide every term by the highest power of $x$. Terms with $x$ in the denominator vanish. The limit equals the ratio of leading coefficients when degrees match.

When the degrees don't match, the ratio of leading coefficients no longer applies:

  • Denominator degree > numerator degree — After dividing by the highest power, the numerator vanishes while the denominator tends to a nonzero constant. The limit is 0. Example: $\displaystyle\lim_{x\to\infty}\frac{2x+1}{x^3-5}=\frac{0}{1}=0$.
  • Numerator degree > denominator degree — The fraction behaves like a polynomial (e.g. $x^2$, $-x^3$) and grows without bound. The limit is $\pm\infty$, with the sign determined by the leading coefficients. Example: $\displaystyle\lim_{x\to\infty}\frac{5x^4-3x}{2x^2+1}\to+\infty$ (positive leading term dominates).

Summary: compare degrees first. If equal → ratio of leading coefficients. If denominator wins → 0. If numerator wins → $\pm\infty$.

Step 1 / 4
1
Identify the dominant terms. Both numerator and denominator grow without bound. Divide every term by the highest power present: $x^3$.
2
Simplify. $\displaystyle\lim_{x\to\infty}\frac{4-2/x^2}{3+5/x-1/x^3}$
3
Take the limit. Every term with $x$ in the denominator vanishes: $\dfrac{4-0}{3+0-0}=\dfrac{4}{3}$
Answer: $\displaystyle\lim_{x\to\infty}\frac{4x^3-2x}{3x^3+5x^2-1}=\boxed{\dfrac{4}{3}}$ Rule: the limit equals the ratio of leading coefficients.

6. Special Limits & Theorems

Why learn this?

Oscillating and messy ratios. The Squeeze Theorem and standard trig limits are the standard toolkit when algebra alone does not resolve the limit.

Helpful hints

Check the setup. L'Hôpital only applies when you truly have $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ (and the pieces are differentiable nearby). The Squeeze Theorem needs a lower and upper “fence” with the same limit.

Remember

$\lim_{x\to 0}\frac{\sin x}{x}=1$ · L'Hôpital differentiates numerator and denominator separately

Squeeze Theorem
If $g(x)\le f(x)\le h(x)$ near $a$ (except possibly at $a$) and $\displaystyle\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$, then $$\lim_{x\to a}f(x)=L$$

Use when $f$ is trapped between two functions with the same limit — e.g. to prove $\lim_{x\to 0}\frac{\sin x}{x}=1$.

L'Hôpital's Rule
When $\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms arise: $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$

Use for $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ when factoring or algebraic tricks fail.

Special Trig Limits
$$\lim_{x\to 0}\frac{\sin x}{x}=1$$ $$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$

Fundamental limits used when trig appears in $\tfrac{0}{0}$ forms. Prove via Squeeze Theorem or L'Hôpital.

Step 1 / 5
1
Identify the form. At $x=0$: numerator $= e^0-1-0=0$, denominator $=0$. We have a $\tfrac{0}{0}$ form — L'Hôpital applies.
2
Apply L'Hôpital once. Differentiate top and bottom: $\displaystyle\lim_{x\to0}\frac{e^x-1}{2x}$. Still $\tfrac{0}{0}$.
3
Apply L'Hôpital again. $\displaystyle\lim_{x\to0}\frac{e^x}{2}$. Now substitute $x=0$ directly.
4
Evaluate. $\dfrac{e^0}{2}=\dfrac{1}{2}$
Answer: $\displaystyle\lim_{x\to0}\frac{e^x-1-x}{x^2}=\boxed{\dfrac{1}{2}}$
Practice Problems
3 per topic · 15 per refresh (97 in pool)
1. Direct Substitution & Basic Limits
2. Solve Indeterminate Forms using Factoring
3. Rationalizing, and Simplifying Fractions
4. Evaluating Limits at Infinity
5. Special Limits & Theorems (Squeeze, Trig Limits)