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Reading Time: 5 min
Last Updated: January 26, 2026
Main Ideas: 4
Reading Time: 5 min
Last Updated: January 26, 2026
Main Ideas: 4

Topic 1.2 Notes – Defining Limits and Using Limit Notation

Verified for 2027 AP® Calculus AB Exam
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You’ve probably talked about “approaching” a value before. Now we make that precise using notation and connect the idea to graphs, tables, and formulas. This is the language everything else in calculus builds on.

What a Limit Is

Here’s the core statement:

lim⁡x→cf(x)=R \lim_{x \to c} f(x) = R

This means:

As xx gets arbitrarily close to cc (but not equal to cc), the values of f(x)f(x) get arbitrarily close to RR.

Two phrases matter:

  • Arbitrarily close → as close as we want.
  • Near the point, not at the point → the limit is about behavior around cc, not necessarily the value at cc.

So three things can be true:

  • The limit exists and equals RR.
  • The function value f(c)f(c) might equal RR.
  • Or f(c)f(c) might be different from RR, or even undefined.

Those are separate questions.

The AP does not test the epsilon-delta definition, but you should understand the idea: if you can make f(x)f(x) as close as you want to RR by choosing xx close enough to cc, the limit is RR.

Limit Notation and How to Read It

lim⁡x→cf(x)=R \lim_{x \to c} f(x) = R

Read it naturally:
“The limit of f(x)f(x) as xx approaches cc is RR.”

Break it apart:

  • x→cx \to c → what input is approaching
  • f(x)f(x) → the function
  • The whole expression → the output value being approached

You’ll also see one-sided limits.

One-Sided Limits

Left-hand limit:
lim⁡x→c−f(x) \lim_{x \to c^-} f(x)

Right-hand limit:
lim⁡x→c+f(x) \lim_{x \to c^+} f(x)

The full limit exists only if:

lim⁡x→c−f(x)=lim⁡x→c+f(x) \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)

If left and right disagree, the limit does not exist. On quizzes, this is one of the fastest ways to lose points.

Representing Limits Three Ways

AP expects you to move between symbolic, numerical, and graphical representations easily.

1. Analytical (Symbolic)

If the function is nice and continuous at cc, just substitute.

Example:

lim⁡x→3(2x2−5x+1) \lim_{x \to 3} (2x^2 - 5x + 1)

Polynomials are continuous. Plug in:

2(3)2−5(3)+1=18−15+1=4 2(3)^2 - 5(3) + 1 = 18 - 15 + 1 = 4

That’s it.

If substitution gives something weird like 00\frac{0}{0}, that signals you need to simplify first. It usually means there’s a hole.

2. Numerical (Tables)

You pick values approaching cc from both sides and look for a pattern.

Example structure:

x (left of c)f(x)x (right of c)f(x)
1.94.82.15.2
1.994.982.015.02
1.9994.9982.0015.002

If both sides are getting closer to 5, then:

lim⁡x→2f(x)=5 \lim_{x \to 2} f(x) = 5

If one side trends toward 3 and the other toward 6, the limit does not exist.

On calculator sections, tables are common. Always check both sides.

3. Graphical

You trace the graph as xx approaches cc from left and right.

Here’s a classic removable discontinuity example.

Removable discontinuity with a hole at (2, 4)

The graph follows the line y=x+2y = x + 2 but has an open circle at (2,4)(2,4). Even though there’s a hole at x=2x=2, the graph approaches 4 from both sides. So:

lim⁡x→2x2−4x−2=4 \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4

The limit exists even though the function is undefined at that point.

Now compare that with a jump discontinuity.

Jump discontinuity at x = 1

As x→1−x \to 1^-, the graph approaches 2. As x→1+x \to 1^+, it approaches 5. No agreement, so the limit does not exist.

What AP Really Cares About

You need to:

  • Write limit notation correctly.
  • Interpret what a limit statement is saying in words.
  • Move between graph, table, and formula.
  • Check left and right behavior when necessary.

If a question says “Explain why the limit exists,” they want reasoning like:
“The left-hand and right-hand limits both equal 4.”

That language earns points.

Key Takeaways

A limit describes what f(x)f(x) approaches as xx approaches cc, not necessarily the value f(c)f(c).
The limit exists only if lim⁡x→c−f(x)=lim⁡x→c+f(x)\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x).
Direct substitution works for functions that are continuous at that point.
Getting 00\frac{0}{0} means simplify first, not that the limit is zero.
A hole in the graph can still have a limit; a jump cannot.

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Notes

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