AP Calculus AB Unit 1 Notes: Learning Limits (Foundations for Derivatives)

Defining Limits and Using Limit Notation

What a limit is (the big idea)

A limit describes the value that a function’s outputs approach as the inputs get close to some number. The key word is “approach,” not “equal.” You are allowed to study the behavior near an input even if the function is undefined there or does something strange at that exact point.

For example, saying the limit is 5 near an input of 2 means:

• when you plug in numbers close to 2 (like 1.9, 1.99, 2.01, 2.1),
• the function values get close to 5.

This “nearby behavior” focus is exactly why limits matter: derivatives (rates of change) are defined using limits, and continuity (whether graphs have breaks) is also defined using limits. Limits are the bridge between algebraic formulas and the idea of “instantaneous change.”

Limit notation and how to read it

The standard notation is:

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

Read it as: “The limit of $f(x)$ as $x$ approaches $a$ equals $L$.”

Interpretation: as $x$ gets close to $a$ (from either side), $f(x)$ gets close to $L$.

A crucial point: $\lim_{x \to a} f(x)$ is about values of $f(x)$ when $x$ is _near_ $a$, not necessarily at $x=a$. That means the following can all happen:

• The limit exists and equals some number, even if $f(a)$ is undefined.
• The limit exists but $f(a)$ is a different number.
• The limit does not exist, even if $f(a)$ is defined.

One-sided limits: approaching from the left or right

Sometimes the behavior differs depending on whether you come from smaller or larger inputs. Then you use one-sided limits.

Left-hand limit:

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

Right-hand limit:

$\lim_{x \to a^+} f(x) = R$

The two-sided limit exists precisely when both one-sided limits exist and are equal:

$\lim_{x \to a} f(x) \text{ exists } \Longleftrightarrow \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$

If the left and right limits are different, the two-sided limit does not exist (often abbreviated DNE).

A notation reference table

Common misconceptions to avoid

Students often confuse three separate ideas:

1. $\lim_{x \to a} f(x)$ (nearby behavior)
2. $f(a)$ (actual value at the point)
3. Whether the graph has a hole, jump, or asymptote

It helps to say it out loud: “The limit asks what the function is trying to be near $a$.”

Worked examples

Example 1: A limit can exist even if the function is undefined at the point

Suppose $f(x) = \frac{x^2-1}{x-1}$. Consider $\lim_{x \to 1} f(x)$.

Direct substitution fails because $x=1$ makes the denominator zero. But the limit cares about nearby values, so simplify:

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

So for $x \ne 1$,

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

Now take the limit:

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

Even though $f(1)$ is undefined in the original formula, the limit exists and equals 2.

Example 2: A limit fails if one-sided limits disagree

Imagine a function with

$\lim_{x \to 3^-} f(x) = 1$

and

$\lim_{x \to 3^+} f(x) = 4$

Then the two-sided limit does not exist:

$\lim_{x \to 3} f(x) \text{ DNE}$

Because the function approaches different values from different sides.

Exam Focus

• Typical question patterns:
  • Given a graph or formula, evaluate $\lim_{x \to a} f(x)$ and explain whether it depends on one-sided behavior.
  • Determine whether a two-sided limit exists by comparing $\lim_{x \to a^-} f(x)$ and $\lim_{x \to a^+} f(x)$.
  • Distinguish between $\lim_{x \to a} f(x)$ and $f(a)$ (especially when there is a hole or a piecewise definition).
• Common mistakes:
  • Assuming the limit equals $f(a)$ automatically; always check whether the function is even defined at $a$.
  • Declaring a limit DNE just because there’s a hole; holes often mean the limit exists.
  • Forgetting that a two-sided limit requires both one-sided limits to match.

Estimating Limit Values from Graphs and Tables

Why estimation matters

Before you learn algebraic tools, limits are often understood visually and numerically. Graphs and tables make the “approach” idea concrete: you can watch inputs move closer to a target and see outputs settle toward a value.

In AP Calculus AB, you are often asked to:

• read a limit from a graph (including one-sided limits),
• estimate from a table of values,
• recognize when a limit does not exist (jumps, oscillations, vertical asymptotes).

These skills matter because real data and calculator outputs are often approximate. Limits give language and structure to “the trend.”

Reading limits from graphs

When estimating from a graph, focus on what the function values approach as you move along the curve toward the input.

A good mental routine:

1. Locate the input value $a$ on the horizontal axis.
2. Trace the graph from the left toward $a$ to estimate $\lim_{x \to a^-} f(x)$.
3. Trace from the right toward $a$ to estimate $\lim_{x \to a^+} f(x)$.
4. Compare: if they match, that common value is $\lim_{x \to a} f(x)$.

Important: the graph might show a filled dot at $x=a$ that is not on the “main curve.” That filled dot tells you $f(a)$, but the limit comes from the trend of the curve.

Recognizing common “limit shapes” on graphs

• Hole (removable discontinuity): The curve approaches one y-value from both sides, but there’s an open circle at $x=a$. Limit exists.
• Jump discontinuity: Left-hand and right-hand approach different y-values. Limit DNE.
• Vertical asymptote: Function values grow without bound. You may write an infinite limit like $\lim_{x \to a^+} f(x) = \infty$.
• Oscillation near a point: If the graph wiggles infinitely without settling to one value, the limit DNE.

Estimating limits from tables

A table gives discrete samples, so you approximate by choosing x-values that get closer to $a$ from both sides.

A strong approach:

• Look at values slightly less than $a$ (left side) and slightly greater than $a$ (right side).
• Check whether the outputs are stabilizing toward the same number.
• If the outputs blow up (very large positive or negative), that suggests an infinite limit.

Be careful: rounding can hide behavior. If the table is too coarse, you might need values closer to $a$ to see the trend.

Worked examples

Example 1: Graph shows a hole

Suppose a graph looks like a straight line approaching the point $(2, 5)$ but with an open circle at $(2, 5)$ and a filled dot at $(2, 1)$.

Then:

$\lim_{x \to 2} f(x) = 5$

but

$f(2) = 1$

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

Example 2: Estimating from a table

You are given values near $x=1$:

From the left, outputs approach 2; from the right, outputs also approach 2. So you estimate:

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

Even if the table does not include $x=1$, that is normal for limits.

Example 3: Infinite limit from a table

If near $x=3$ you see outputs like 100, 500, 2000 on one side, that suggests the function grows without bound. A careful statement might be:

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

But you should still check the other side; it could go to $-\infty$ or also to $\infty$.

Exam Focus

• Typical question patterns:
  • “Use the graph to estimate $\lim_{x \to a} f(x)$, $\lim_{x \to a^-} f(x)$, $\lim_{x \to a^+} f(x)$, and $f(a)$.”
  • “Use the table to estimate the limit; explain whether it exists.”
  • Identify whether a discontinuity appears removable (hole) or non-removable (jump/asymptote) based on one-sided behavior.
• Common mistakes:
  • Reading $f(a)$ (the filled dot) instead of the approaching value.
  • Using only one side of the table and assuming the other side matches.
  • Confusing very large values with “no limit” without recognizing infinite limits.

Determining Limits Using Algebraic Properties and Manipulation

Why algebra helps

Graphs and tables are great for intuition, but AP Calculus expects you to compute many limits exactly. Algebraic techniques let you turn “indeterminate forms” into expressions where substitution works.

A major theme: if a function is “nice” (like a polynomial), you can often substitute directly. The hard cases usually come from operations that create undefined expressions such as dividing by zero.

Limit laws (algebraic properties)

Limit laws let you break complicated expressions into simpler pieces, provided the relevant limits exist.

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then:

Sum law:

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

Difference law:

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

Constant multiple:

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

Product:

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

Quotient (as long as $M \ne 0$):

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

Power:

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

These laws justify why direct substitution works for polynomials and rational functions (except where the denominator becomes zero).

Direct substitution (when it works)

For a polynomial like $f(x) = x^3 - 2x + 7$, the limit at any number $a$ is found by substitution:

$\lim_{x \to a} (x^3 - 2x + 7) = a^3 - 2a + 7$

For a rational function, substitution works whenever the denominator is not zero at $a$.

Indeterminate form $\frac{0}{0}$ and what it really means

When substitution gives:

$\frac{0}{0}$

that does not mean the limit is 0. It means “the expression is undefined at the point, and you must simplify to see the nearby behavior.” Two common algebra tools are:

• factoring and canceling a common factor,
• rationalizing (especially when radicals are involved).

Factoring and canceling

You can cancel only factors, not terms, and only after factoring completely.

Example 1: Factoring

Compute:

$\lim_{x \to 2} \frac{x^2-4}{x-2}$

Substitution gives $\frac{0}{0}$. Factor:

$x^2 - 4 = (x-2)(x+2)$

Then for $x \ne 2$:

$\frac{x^2-4}{x-2} = x+2$

So:

$\lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2} (x+2) = 4$

Notice the limit exists because the “hole” comes from a removable factor.

Rationalizing (using conjugates)

Rationalizing is useful when the expression contains radicals and substitution leads to $\frac{0}{0}$.

Example 2: Rationalizing

Compute:

$\lim_{x \to 9} \frac{\sqrt{x}-3}{x-9}$

Substitution gives $\frac{0}{0}$. Multiply by the conjugate:

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

Cancel $x-9$ (for $x \ne 9$):

$\frac{1}{\sqrt{x}+3}$

Now substitute:

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

Piecewise functions and algebraic limits

For piecewise functions, you usually compute one-sided limits using the formula that applies on that side.

If

• $f(x) = x+1$ for $x<2$,
• $f(x) = x^2$ for $x>2$,

then:

$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x+1) = 3$

and

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

Since 3 and 4 do not match, the two-sided limit does not exist.

Exam Focus

• Typical question patterns:
  • Evaluate a limit that gives $\frac{0}{0}$ by factoring or rationalizing.
  • Use limit laws to break apart expressions and justify direct substitution.
  • Compute one-sided limits of a piecewise function to determine if the two-sided limit exists.
• Common mistakes:
  • Canceling terms that are not factors (for example, canceling an $x$ across a sum).
  • Concluding “limit is 0” from seeing $\frac{0}{0}$.
  • Forgetting to check the denominator condition in the quotient law (the limiting denominator cannot be 0).

Selecting Procedures for Determining Limits

Why procedure choice is part of the skill

In AP Calculus, you’re not only expected to compute limits—you’re expected to choose an efficient, mathematically justified method. Two problems may look similar but require different tools. The goal is to recognize the “type” of limit and pick a strategy that avoids wasted time and algebra traps.

A good procedure choice starts with a diagnostic question:

What happens if I substitute the approach value right away?

There are three common outcomes:

1. You get a real number (like 7). Then you’re usually done.
2. You get an undefined form like $\frac{0}{0}$. Then you need algebraic manipulation.
3. You get something that signals divergence, like dividing by zero without cancellation, or unbounded growth. Then you analyze one-sided behavior or asymptotes.

Procedure map (how to decide)

Case 1: Direct substitution works

This typically applies to polynomials, and to rational functions where the denominator is not 0 at the point.

Example decision:

$\lim_{x \to 3} (x^2 - 5x)$

Substitution gives a number, so use direct substitution.

Case 2: Indeterminate form $\frac{0}{0}$

This is the classic signal to simplify.

Common tools:

• factoring and canceling,
• rationalizing,
• combining fractions into one fraction (then factor),
• rewriting complex fractions.

Example decision:

$\lim_{x \to 1} \frac{x^3-1}{x-1}$

Substitution gives $\frac{0}{0}$, so factor using difference of cubes.

Case 3: One-sided limits are necessary

You should switch to one-sided limits if:

• the function is piecewise,
• the graph has a jump,
• the expression involves absolute value near the point,
• the denominator approaches 0 without cancellation (possible vertical asymptote).

Example decision:

$\lim_{x \to 0} \frac{|x|}{x}$

You should not try to “simplify” by canceling $x$, because $|x|$ behaves differently for positive and negative x. Instead:

For $x>0$, $|x|=x$ so $\frac{|x|}{x}=1$, giving:

$\lim_{x \to 0^+} \frac{|x|}{x} = 1$

For $x<0$, $|x|=-x$ so $\frac{|x|}{x}=-1$, giving:

$\lim_{x \to 0^-} \frac{|x|}{x} = -1$

Because one-sided limits disagree, the two-sided limit DNE.

Case 4: Infinite limits and asymptotic behavior

If substituting gives a nonzero number over 0 (like $\frac{5}{0}$), the function may blow up to $\infty$ or $-\infty$. You often need one-sided analysis.

Example:

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

but

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

So:

$\lim_{x \to 2} \frac{1}{x-2} \text{ DNE}$

because the sides do not match.

A realistic “choose the method” worked problem

Compute:

$\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4}$

Step 1: Try substitution.

$\frac{\sqrt{4}-2}{4-4} = \frac{0}{0}$

Indeterminate, so you need manipulation. Because there’s a square root, rationalize.

Step 2: Multiply by the conjugate.

$\frac{\sqrt{x}-2}{x-4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2} = \frac{x-4}{(x-4)(\sqrt{x}+2)}$

Step 3: Cancel and substitute.

$\lim_{x \to 4} \frac{1}{\sqrt{x}+2} = \frac{1}{4}$

The main skill here was choosing rationalization quickly once you saw a radical and $\frac{0}{0}$.

Exam Focus

• Typical question patterns:
  • “Evaluate the limit. Show your work.” The hidden task is selecting factoring vs rationalizing vs one-sided analysis.
  • Limits involving absolute value or piecewise definitions where one-sided limits must be computed explicitly.
  • Limits that are infinite or DNE, requiring correct notation like $\infty$, $-\infty$, or DNE with justification.
• Common mistakes:
  • Forcing algebraic cancellation when the issue is actually one-sided behavior (especially with $|x|$ or piecewise functions).
  • Treating infinite limits as “DNE” without recognizing you can state $\lim_{x \to a} f(x) = \infty$ when appropriate.
  • Stopping after manipulation without actually taking the limit (students simplify but forget the final substitution step).

Squeeze Theorem

What the Squeeze Theorem says (in plain language)

The Squeeze Theorem is a way to find a limit when a function is trapped between two other functions whose limits you already know.

If you can show that near $x=a$,

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

and also

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

then the “middle” function must approach the same value:

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

Why this is true: if the bottom and top functions both get arbitrarily close to $L$ as you approach $a$, there’s no room for the trapped function to go anywhere else.

Why it matters in calculus

Some limits are hard or impossible to compute by algebraic simplification, especially when functions oscillate. The Squeeze Theorem turns a messy limit into an inequality argument.

This theorem is especially famous for limits involving trig functions and absolute values. In a first AP Calculus limits unit, you commonly use it to handle expressions like “something times $\sin(1/x)$” near 0, where the sine oscillates but stays bounded.

How to apply it (a practical process)

1. Identify a function that is hard to evaluate directly (often because it oscillates).
2. Find simple bounds for it. Common bounding facts include:

$-1 \le \sin(\theta) \le 1$

and

$-1 \le \cos(\theta) \le 1$

3. Multiply the inequality by something that shrinks to 0 (often $|x|$ or $x^2$) to trap the expression between two functions with the same limit.
4. Take limits of the bounding functions.
5. Conclude the squeezed limit.

Worked examples

Example 1: Oscillation squeezed to 0

Evaluate:

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

Direct substitution does not work because $\sin(1/x)$ has no limit as $x \to 0$ (it oscillates). But sine is always between -1 and 1:

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

Now multiply by $x$. To avoid sign issues, a clean approach is to multiply by $|x|$ instead, because it is always nonnegative:

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

As $x \to 0$,

$\lim_{x \to 0} (-|x|) = 0$

and

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

So the middle limit is 0:

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

Finally, note that $x\sin(1/x)$ is also squeezed by similar bounds (or you can argue its absolute value is bounded by $|x|$), so:

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

The important lesson: you didn’t need to know how $\sin(1/x)$ behaves exactly—only that it stays bounded.

Example 2: A common bounding trick with absolute value

Evaluate:

$\lim_{x \to 0} x^2 \cos\left(\frac{5}{x}\right)$

Cosine is bounded:

$-1 \le \cos\left(\frac{5}{x}\right) \le 1$

Multiply by $x^2$ (always nonnegative):

$-x^2 \le x^2\cos\left(\frac{5}{x}\right) \le x^2$

Both bounds go to 0:

$\lim_{x \to 0} (-x^2) = 0$

and

$\lim_{x \to 0} x^2 = 0$

Therefore:

$\lim_{x \to 0} x^2 \cos\left(\frac{5}{x}\right) = 0$

What can go wrong (common pitfalls)

The Squeeze Theorem only works if:

• the inequality is true in a neighborhood around the point (not necessarily at the point), and
• the two outer limits both exist and are equal.

If the outer limits are different, you cannot conclude anything definite about the middle limit.

Also, be careful when multiplying inequalities by an expression that could be negative (like $x$). If you multiply by a negative number, the inequality signs reverse. That’s why using $|x|$ is a popular safe move.

Exam Focus

• Typical question patterns:
  • Evaluate limits involving bounded oscillations, especially products like $x\sin(1/x)$ or $x^2\cos(1/x)$ as $x \to 0$.
  • Provide a brief inequality-based justification (not just an answer) showing what squeezes what.
  • Identify appropriate bounds using facts like $-1 \le \sin(\theta) \le 1$.
• Common mistakes:
  • Writing bounds incorrectly (for example, forgetting sine and cosine are always between -1 and 1).
  • Multiplying inequalities by $x$ without handling sign changes; using $|x|$ often avoids this.
  • Assuming “bounded” automatically means the limit is 0; you still need the shrinking factor (like $x$ or $x^2$) to force the squeeze.