AP Calculus AB Unit 1 Notes: Learning to Think in Limits

Defining Limits and Using Limit Notation

What a limit is (the core idea)

A limit describes what value a function’s outputs are approaching as the inputs get close to a particular number. The key word is “approaching,” because limits are about nearby behavior—not necessarily what happens exactly at the input value.

In AP Calculus, the limit is the bridge between algebra and calculus: it formalizes the idea of “getting arbitrarily close,” which is how we define derivatives (instantaneous rate of change) and integrals (accumulation). If you understand limits conceptually, many later ideas feel natural instead of mysterious.

When you write

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

you are saying: “As x gets close to a (from either side), the values of f(x) get close to L.”

Two important clarifications:

1. The limit can exist even if f(a) is not defined.
2. The limit can exist even if f(a) is defined but not equal to the limit.

That’s why limits are so useful for analyzing holes (removable discontinuities), jumps, and other “weird” points.

Limit notation and how to read it

Here are common forms of limit notation you’ll use:

A two-sided limit exists only when the left-hand and right-hand limits agree:

\lim_{x \to a} f(x) = L \text{ exists if and only if } \lim_{x \to a^-} f(x) = L \text{ and } \lim_{x \to a^+} f(x) = L

If the one-sided limits are different, the two-sided limit does not exist.

Why “approach” matters more than “equal”

Students often get stuck thinking a limit is about plugging in. Sometimes it is—but only when the function behaves nicely. Limits deliberately ignore the function’s value at the exact point and focus on the trend near that point.

A helpful analogy: if you’re walking toward a door, the approach describes where you’re headed even if you never touch the door handle. Limits describe the destination of the function’s behavior.

Worked examples

Example 1: A limit can exist even if the function value is “wrong”

Let

f(x) = \frac{x^2 - 1}{x - 1}

Find

\lim_{x \to 1} f(x)

Reasoning: Direct substitution gives 0/0, which is indeterminate (it means “simplify first,” not “the limit is 0”). Factor the numerator:

x^2 - 1 = (x - 1)(x + 1)

So for x \ne 1,

f(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1

Now the nearby behavior is just the line x + 1, so

\lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 1) = 2

Note: f(1) is not defined (division by zero), but the limit still exists.

Example 2: Two-sided limit fails when sides disagree

Suppose a piecewise function behaves like this near x = 0:

• From the left, outputs approach 1.
• From the right, outputs approach 3.

Then

\lim_{x \to 0^-} f(x) = 1

\lim_{x \to 0^+} f(x) = 3

Because these are not equal,

\lim_{x \to 0} f(x) \text{ does not exist}

It’s not “infinite” or “zero”—it’s simply not a single approaching value.

Exam Focus

• Typical question patterns:
  • Interpret a statement like \lim_{x \to 2} f(x) = 5 in words and connect it to a graph.
  • Use one-sided limits to decide whether a two-sided limit exists.
  • Distinguish between f(a) and \lim_{x \to a} f(x).
• Common mistakes:
  • Treating 0/0 as an answer instead of a signal to simplify.
  • Assuming the limit equals the function value automatically.
  • Forgetting that a two-sided limit requires both one-sided limits to match.

Estimating Limit Values from Graphs and Tables

Why estimation matters

In real applications (and many AP questions), you don’t always get a clean formula that simplifies nicely. Graphs and tables let you infer a function’s behavior near a point—even when the function is defined piecewise, comes from data, or has a complicated expression.

Estimation is also a conceptual test: it checks whether you truly understand “approach” rather than relying only on algebra.

Estimating limits from graphs

When estimating from a graph, your job is to track what y = f(x) is approaching as x moves toward a.

A reliable process:

1. Locate x = a on the horizontal axis.
2. Trace the graph from the left (values with x < a) to estimate \lim_{x \to a^-} f(x).
3. Trace from the right (values with x > a) to estimate \lim_{x \to a^+} f(x).
4. Compare them. If they agree, that shared value is the two-sided limit.

Graph features that often appear:

• A hole (open circle) at x = a: the limit may still exist and equals the height of the hole.
• A filled point at x = a at a different height: that’s f(a), which may differ from the limit.
• A jump: left-hand and right-hand limits disagree.
• A vertical asymptote: values grow without bound (you’ll describe this with infinite limits, like approaching \infty or -\infty).

Estimating limits from tables

Tables approximate the “getting close” idea by sampling inputs near a from both sides. The crucial idea is to choose values that get closer and closer to a.

A good table-based approach:

1. Use values approaching a from the left: for example, a - 0.1, a - 0.01, a - 0.001.
2. Use values approaching a from the right: a + 0.1, a + 0.01, a + 0.001.
3. Look for a stable trend (the outputs settling toward a number).

Be cautious: rounding can hide the trend, and some functions oscillate so the outputs never settle.

Worked examples

Example 1: Estimating from a table

Suppose you are given this table of values near x = 2:

From the left, outputs move toward about 4. From the right, outputs also move toward about 4. So it is reasonable to estimate:

\lim_{x \to 2} f(x) = 4

Even if f(2) is not given—or is different—the limit can still be 4.

Example 2: Estimating from a graph with a hole

Imagine a graph that follows a smooth curve and has an open circle at the point (1, 2) , but also has a filled dot at (1, 5) .

From both sides, the curve approaches height 2 as x approaches 1, so

\lim_{x \to 1} f(x) = 2

But the actual function value is

f(1) = 5

This is a classic “limit exists but function value differs” situation.

Exam Focus

• Typical question patterns:
  • Given a graph, estimate \lim_{x \to a} f(x) and compare it with f(a).
  • Use a table to approximate one-sided limits and determine whether the two-sided limit exists.
  • Identify whether a limit is finite, infinite, or does not exist based on graphical behavior.
• Common mistakes:
  • Reading f(a) off the filled point and calling it the limit without checking the approach.
  • Using only one side of the table (forgetting to approach from both left and right).
  • Confusing “gets large” with “does not exist” instead of using the correct infinite-limit description.

Determining Limits Using Algebraic Properties and Manipulation

Why algebra still matters in calculus

Many AP limits are designed to look impossible by substitution because you get indeterminate forms like 0/0. The point of these problems is usually not “guess the answer,” but “recognize the structure and rewrite it.”

If you can transform an expression into a simpler equivalent expression (for x near a), you can often evaluate the limit by substitution after simplification.

Limit laws (algebraic properties)

When limits exist, they behave nicely with basic operations. If

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

and

\lim_{x \to a} g(x) = M

then common limit laws include:

\lim_{x \to a} (f(x) + g(x)) = L + M

\lim_{x \to a} (f(x) - g(x)) = L - M

\lim_{x \to a} (c f(x)) = cL

\lim_{x \to a} (f(x) g(x)) = LM

\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M} \text{ provided } M \ne 0

A major consequence: for polynomials and rational functions (where the denominator is not zero at the point), you can often evaluate limits by direct substitution.

Indeterminate forms and why 0/0 is special

If direct substitution gives:

• A nonzero number, you’re usually done.
• Something like \frac{3}{0}, that suggests the function may blow up (vertical asymptote behavior).
• \frac{0}{0}, you have an indeterminate form. It means “your algebraic form hides the real behavior.” You must simplify.

Common algebraic techniques

1) Factoring and canceling

If a factor causing the zero cancels, you remove the “hole” behavior and reveal the nearby function.

Example:

\lim_{x \to 3} \frac{x^2 - 9}{x - 3}

Substitute gives 0/0, so factor:

x^2 - 9 = (x - 3)(x + 3)

For x \ne 3,

\frac{x^2 - 9}{x - 3} = x + 3

Now substitute:

\lim_{x \to 3} (x + 3) = 6

A common misconception is thinking you “plug in after canceling” is a trick. It’s valid because the simplified expression matches the original for all x near 3 except at 3 itself—and limits only care about “near.”

2) Rationalizing with conjugates

Square roots often produce 0/0. Multiplying by a conjugate can eliminate the radical.

Example:

\lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x}

Direct substitution gives 0/0. Multiply numerator and denominator by the conjugate:

\frac{\sqrt{x + 9} - 3}{x} \cdot \frac{\sqrt{x + 9} + 3}{\sqrt{x + 9} + 3} = \frac{(x + 9) - 9}{x(\sqrt{x + 9} + 3)}

Simplify:

\frac{x}{x(\sqrt{x + 9} + 3)} = \frac{1}{\sqrt{x + 9} + 3}

Now substitute x = 0:

\lim_{x \to 0} \frac{1}{\sqrt{x + 9} + 3} = \frac{1}{6}

A frequent error is conjugating but not applying it to both numerator and denominator (you must multiply by 1).

3) Simplifying complex rational expressions

When you see fractions within fractions, clear denominators by multiplying numerator and denominator by a common denominator.

Example:

\lim_{x \to 1} \frac{\frac{1}{x} - 1}{x - 1}

Rewrite the numerator:

\frac{1}{x} - 1 = \frac{1 - x}{x}

So the expression becomes:

\frac{\frac{1 - x}{x}}{x - 1} = \frac{1 - x}{x(x - 1)}

Note that 1 - x = -(x - 1), so:

\frac{1 - x}{x(x - 1)} = \frac{-(x - 1)}{x(x - 1)} = -\frac{1}{x}

Now substitute:

\lim_{x \to 1} -\frac{1}{x} = -1

4) Trig limits you’ll use repeatedly

A central result (often justified using geometry and the Squeeze Theorem) is:

\lim_{x \to 0} \frac{\sin(x)}{x} = 1

From that, a very common variant follows:

\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1

If you need \lim_{x \to 0} \frac{\sin(kx)}{x}, rewrite:

\frac{\sin(kx)}{x} = k \cdot \frac{\sin(kx)}{kx}

So:

\lim_{x \to 0} \frac{\sin(kx)}{x} = k

Example:

\lim_{x \to 0} \frac{\sin(5x)}{x}

Rewrite:

\frac{\sin(5x)}{x} = 5 \cdot \frac{\sin(5x)}{5x}

Take the limit:

\lim_{x \to 0} 5 \cdot \frac{\sin(5x)}{5x} = 5 \cdot 1 = 5

A common mistake is trying to substitute x = 0 immediately and concluding 0/0 means “DNE.” In trig limits, 0/0 usually signals “use a known trig limit or an identity.”

Exam Focus

• Typical question patterns:
  • Evaluate limits that give 0/0 by factoring, canceling, or rationalizing.
  • Use limit laws to break complicated expressions into simpler limits.
  • Evaluate trig limits by rewriting to match \frac{\sin(x)}{x}.
• Common mistakes:
  • Canceling terms that are not factors (you can cancel factors, not addends).
  • Forgetting domain restrictions after simplification (the simplified expression may not define the original point, but that’s okay for limits).
  • Misusing trig identities or skipping the rewrite needed to create \frac{\sin(x)}{x}.

Selecting Procedures for Determining Limits

The skill: choosing a method, not just doing algebra

On many AP questions, the hardest part is recognizing what you should try first. Limits problems often look different on the surface, but they fall into patterns. A good strategy is to treat limit evaluation like diagnosing a problem: you try the simplest method, then adjust based on what you see.

A practical decision process

Step 1: Try direct substitution

If the function is continuous at x = a (polynomials always are; rational functions are if the denominator is nonzero), then:

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

So start by substituting. If you get a real number immediately, you’re done.

Step 2: If substitution gives 0/0, simplify

This is the most common AP signal to use algebraic manipulation.

• If you see polynomials: try factoring.
• If you see square roots: try rationalizing.
• If you see a complex fraction: rewrite using common denominators.
• If you see trig functions: try identities and known trig limits.

Step 3: If values blow up (division by a small number)

If substitution makes the denominator 0 but numerator is not 0, the function may have a vertical asymptote at that input. The limit might be an infinite limit (approaching \infty or -\infty) or might not exist if the sides go different directions.

For example, consider:

\lim_{x \to 2} \frac{1}{(x - 2)^2}

As x approaches 2, the denominator approaches 0 but is always positive (because of the square), so the expression grows without bound:

\lim_{x \to 2} \frac{1}{(x - 2)^2} = \infty

But for:

\lim_{x \to 2} \frac{1}{x - 2}

you get different one-sided behavior:

\lim_{x \to 2^-} \frac{1}{x - 2} = -\infty

\lim_{x \to 2^+} \frac{1}{x - 2} = \infty

So the two-sided limit does not exist.

Step 4: If the function oscillates or is trapped

If the expression involves something like \sin(1/x) near x = 0, substitution doesn’t help and simplification may not settle it. Then you consider whether:

• the limit does not exist due to oscillation, or
• the oscillation is multiplied by something that shrinks to 0, allowing the Squeeze Theorem.

Worked examples

Example 1: Choosing direct substitution

Evaluate:

\lim_{x \to -1} (x^3 + 4x)

Polynomials are continuous everywhere, so substitute:

(-1)^3 + 4(-1) = -1 - 4 = -5

Thus:

\lim_{x \to -1} (x^3 + 4x) = -5

Example 2: Recognizing a vertical asymptote situation

Evaluate:

\lim_{x \to 0^+} \frac{3}{x}

As x approaches 0 from the right, x is positive and very small, so 3/x becomes very large and positive. Therefore:

\lim_{x \to 0^+} \frac{3}{x} = \infty

If the problem asked for \lim_{x \to 0} \frac{3}{x} (two-sided), you would check both sides and find one side goes to -\infty, so the two-sided limit does not exist.

Example 3: Deciding between “DNE” and “Squeeze candidate”

Consider:

\lim_{x \to 0} \sin\left(\frac{1}{x}\right)

As x approaches 0, 1/x becomes huge in magnitude, and \sin(1/x) oscillates between -1 and 1 without settling to a single number. There is no “approaching value,” so:

\lim_{x \to 0} \sin\left(\frac{1}{x}\right) \text{ does not exist}

But if you instead had:

\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)

the factor x shrinks to 0 and can trap the oscillation—this becomes a Squeeze Theorem situation (shown in the next section).

Exam Focus

• Typical question patterns:
  • “Evaluate the limit or state that it does not exist” where the key is choosing simplification vs one-sided analysis.
  • Determine whether a limit is finite, infinite, or DNE based on algebraic form.
  • Use a known trig limit by rewriting an expression into a standard form.
• Common mistakes:
  • Assuming any vertical asymptote automatically means “DNE” instead of checking one-sided infinite limits.
  • For oscillatory expressions, guessing a value (like 0) instead of testing whether outputs settle.
  • Not trying direct substitution first and wasting time on unnecessary algebra.

Squeeze Theorem

What it is and why it works

The Squeeze Theorem is a way to find a limit when a function is trapped between two other functions that approach the same value. Conceptually, it’s like saying: if you’re walking in a narrow hallway whose walls meet at a point, you must also end up at that meeting point.

Formally, if for all x near a (except possibly at a itself) you have:

g(x) \le f(x) \le h(x)

and

\lim_{x \to a} g(x) = L

\lim_{x \to a} h(x) = L

then:

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

The power of the theorem is that you don’t need to understand the messy function directly—you just need good bounds.

Where squeezing shows up in AP Calculus

Squeeze Theorem is especially useful when:

• you have trig functions bounded by -1 and 1,
• the function oscillates but is multiplied by something small,
• geometry or inequalities naturally trap a function.

A key bounded fact you’ll often use is:

-1 \le \sin(x) \le 1

and similarly:

-1 \le \cos(x) \le 1

Worked examples

Example 1: Classic squeeze with oscillation

Evaluate:

\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)

Start with the bound on sine:

-1 \le \sin\left(\frac{1}{x}\right) \le 1

Multiply every part by x. To avoid sign issues when multiplying inequalities, it’s cleaner to use absolute value:

\left|\sin\left(\frac{1}{x}\right)\right| \le 1

Multiply by |x|:

\left|x \sin\left(\frac{1}{x}\right)\right| \le |x|

Also, absolute values are always nonnegative, so:

0 \le \left|x \sin\left(\frac{1}{x}\right)\right| \le |x|

Now take limits as x \to 0. The outer expressions both go to 0:

\lim_{x \to 0} 0 = 0

\lim_{x \to 0} |x| = 0

So by squeezing,

\lim_{x \to 0} \left|x \sin\left(\frac{1}{x}\right)\right| = 0

If the absolute value approaches 0, the original expression must also approach 0, so:

\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0

This example highlights a common big idea: wild oscillation doesn’t necessarily destroy a limit if the oscillation is forced to shrink in amplitude.

Example 2: Establishing a trig limit (idea-level)

One of the most important limits in calculus is:

\lim_{x \to 0} \frac{\sin(x)}{x} = 1

A standard justification uses geometry on the unit circle to create inequalities that trap \frac{\sin(x)}{x} between two expressions that both approach 1 as x \to 0. On the AP exam, you are usually allowed to use this limit as a known result, but it’s worth understanding that it comes from squeezing—meaning it’s not magic, it’s a consequence of bounding.

What can go wrong when using Squeeze

Squeeze Theorem is not “if something looks small, the limit is 0.” You must provide (or clearly identify) two bounding functions with the same limit. Also, the inequality must hold in some interval around a (not just at a few points).

Exam Focus

• Typical question patterns:
  • Evaluate limits of the form x^n \sin(1/x) or x^n \cos(1/x) as x \to 0.
  • Use bounding facts like -1 \le \sin(x) \le 1 to trap a complicated function.
  • Justify a limit by explicitly writing inequalities and taking limits of the bounds.
• Common mistakes:
  • Forgetting to handle sign changes when multiplying inequalities (using absolute value often avoids this).
  • Choosing bounds that do not approach the same limit, which means you cannot conclude anything about f(x).
  • Writing an inequality that is not true for all x sufficiently close to the target value.