AP Calculus AB Unit 1 Notes: Infinity in Limits, Asymptotes, and Existence Theorems

Infinite Limits and Vertical Asymptotes

What an infinite limit means (concept first)

An infinite limit happens when, as you plug $x$ values closer and closer to some number $a$, the function values grow without bound (either upward or downward). The key idea is that the function is not approaching a finite output—it is “blowing up.”

For example, if as $x$ gets close to $a$ the values of $f(x)$ become arbitrarily large positive numbers, you write:

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

If the values become arbitrarily large negative numbers, you write:

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

This does not mean the limit equals a number called infinity. It means the function grows beyond any finite bound.

Why infinite limits matter

Infinite limits are how calculus makes the idea of a “vertical blow-up” precise. They matter because:

• They are tightly connected to vertical asymptotes, which describe the end behavior of graphs near certain vertical lines.
• They show up constantly with rational functions, logarithms, and some trigonometric functions.
• They help you reason about continuity: if a function has an infinite limit at $x=a$, it cannot be continuous there (and often isn’t even defined there).

One-sided infinite limits and “which way” the graph blows up

A common AP Calculus situation is that the left-hand and right-hand behaviors near $x=a$ are different. Then you use one-sided limits.

• Left-hand limit:

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

• Right-hand limit:

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

You might see:

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

and

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

In that case, the two-sided limit $\lim_{x \to a} f(x)$ does not exist as a single infinite direction, because the function does not head the same way from both sides.

Vertical asymptotes: what they are and how limits describe them

A vertical asymptote is a vertical line $x=a$ that the graph approaches as $x$ gets close to $a$ (from one side or both), while the function values become unbounded.

The limit language is:

• If

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

or

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

then $x=a$ is a vertical asymptote.

Important nuance: many textbooks and AP-style explanations treat “vertical asymptote at $x=a$” as meaning at least one one-sided limit is infinite. It does not require both sides to go to infinity, and it does not require the function to be undefined at $x=a$ (though that’s the most common situation).

How to find infinite limits for rational functions (the mechanism)

For a rational function

$f(x) = \frac{p(x)}{q(x)}$

infinite behavior near $x=a$ typically happens when:

• $q(a)=0$ (denominator is zero), and
• $p(a)\neq 0$ (numerator is not zero).

Then the fraction’s magnitude tends to blow up because you’re dividing a nonzero number by something approaching $0$.

To determine whether the function goes to $\infty$ or $-\infty$ from each side, you do a **sign analysis** near $x=a$:

1. Factor the denominator (and numerator if needed).
2. Determine the sign of each factor for $x$ just less than $a$ and just greater than $a$.
3. Combine signs to see whether the fraction is positive or negative while its magnitude grows.

A crucial idea: as $x \to a$, a factor like $x-a$ changes sign depending on the side:

• If $x \to a^-$, then $x-a$ is negative.
• If $x \to a^+$, then $x-a$ is positive.

Removable discontinuities vs vertical asymptotes (what can go wrong)

If both numerator and denominator are zero at $x=a$, you might have a removable discontinuity (a “hole”) rather than a vertical asymptote. Example structure:

$f(x)=\frac{(x-a)g(x)}{(x-a)h(x)}$

for $x\neq a$. Cancelling the common factor can remove the blow-up, producing a finite limit instead. Students often assume “denominator zero means vertical asymptote,” but cancellation changes everything.

Worked Example 1: Identify infinite limits and a vertical asymptote

Consider:

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

As $x$ approaches $2$, the denominator approaches $0$.

• For $x \to 2^-$, $x-2$ is a small negative number, so $\frac{1}{x-2}$ is a large negative number:

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

• For $x \to 2^+$, $x-2$ is a small positive number, so $\frac{1}{x-2}$ is a large positive number:

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

Therefore, $x=2$ is a vertical asymptote. The two-sided limit does not exist as a single infinity direction because the one-sided limits go to opposite infinities.

Worked Example 2: Distinguish a hole from an asymptote

Consider:

$f(x)=\frac{x^2-9}{x-3}$

Factor the numerator:

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

So for $x\neq 3$:

$f(x)=\frac{(x-3)(x+3)}{x-3}=x+3$

Now the limit is finite:

$\lim_{x \to 3} \frac{x^2-9}{x-3} = \lim_{x \to 3} (x+3) = 6$

So there is not a vertical asymptote at $x=3$. Instead, the original expression is undefined at $x=3$ but approaches $6$—a removable discontinuity.

Notation reference (common on AP)

Exam Focus

• Typical question patterns:
  • “Evaluate” an infinite limit such as $\lim_{x \to a^-} \frac{p(x)}{q(x)}$ and state whether it is $\infty$ or $-\infty$.
  • Given a graph or table, identify where vertical asymptotes occur and match them to infinite one-sided limits.
  • Determine whether a discontinuity is removable (finite limit) or infinite (vertical asymptote).
• Common mistakes:
  • Assuming any denominator zero automatically gives a vertical asymptote (forgetting to check for cancellation).
  • Ignoring one-sided behavior and claiming a two-sided infinite limit exists when the sides go to opposite infinities.
  • Dropping sign analysis—getting $\infty$ vs $-\infty$ wrong.

Limits at Infinity and Horizontal Asymptotes

What “limit at infinity” is really saying

A limit at infinity describes what happens to $f(x)$ when $x$ becomes very large in the positive or negative direction. You are not “plugging in infinity.” Instead, you are looking at end behavior:

$\lim_{x \to \infty} f(x)$

and

$\lim_{x \to -\infty} f(x)$

If the function approaches a finite number $L$ as $x$ grows without bound, you write:

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

This means you can make $f(x)$ as close to $L$ as you want by choosing $x$ sufficiently large.

Why limits at infinity matter

Limits at infinity help you understand the “big picture” of a function’s graph—especially for rational functions and models. In applications, you often care about long-run behavior:

• In population or economics models, you might ask what value a quantity stabilizes around.
• In physics, a response might approach a steady-state value.

In graphing terms, these limits lead directly to horizontal asymptotes, which describe how the graph behaves far to the left or right.

Horizontal asymptotes: the limit connection

A horizontal asymptote is a horizontal line $y=L$ such that the graph approaches $L$ as $x$ goes to infinity (positive, negative, or both).

Formally:

• If

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

then $y=L$ is a horizontal asymptote (to the right).

• If

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

then $y=L$ is a horizontal asymptote (to the left).

A function can have:

• Two different horizontal asymptotes (one as $x \to \infty$ and another as $x \to -\infty$).
• No horizontal asymptote.

Also, having a horizontal asymptote does not mean the graph can’t cross it. It often can.

How to compute limits at infinity for rational functions

For rational functions,

$f(x)=\frac{p(x)}{q(x)}$

the end behavior is determined mainly by the degrees of the polynomials $p(x)$ and $q(x)$.

Let $n$ be the degree of $p(x)$ and $m$ be the degree of $q(x)$.

Case 1: Denominator degree larger (approaches 0)

If $n<m$, then

$\lim_{x \to \infty} \frac{p(x)}{q(x)} = 0$

and similarly for $x \to -\infty$ (for polynomial ratios, it will also be $0$). Intuition: the denominator grows faster, so the fraction shrinks toward zero.

Case 2: Same degree (ratio of leading coefficients)

If $n=m$, then the limit is the ratio of leading coefficients.

If

$p(x)=a_n x^n+\text{lower terms}$

and

$q(x)=b_n x^n+\text{lower terms}$

then

$\lim_{x \to \infty} \frac{p(x)}{q(x)} = \frac{a_n}{b_n}$

The same result holds for $x \to -\infty$.

Case 3: Numerator degree larger (no horizontal asymptote)

If $n>m$, the rational function does not approach a finite constant; it tends to $\infty$, $-\infty$, or behaves like a slant (oblique) or higher-degree polynomial asymptote after division. In AP Calculus AB, you are expected to recognize that a horizontal asymptote does not exist when the numerator degree is bigger than the denominator degree.

A frequent “what goes wrong” moment: students try to force a horizontal asymptote even when the function grows without bound.

A helpful technique: dividing by the highest power

To evaluate

$\lim_{x \to \infty} \frac{p(x)}{q(x)}$

you can divide numerator and denominator by the highest power of $x$ appearing in the denominator (or just by the highest overall power). This makes the dominant terms visible because terms like

$\frac{1}{x}$

and

$\frac{1}{x^2}$

go to $0$ as $x \to \infty$.

Worked Example 1: Same degree rational function

Find:

$\lim_{x \to \infty} \frac{3x^2-5}{2x^2+7x+1}$

The degrees match (both are 2). The limit is the ratio of leading coefficients:

$\lim_{x \to \infty} \frac{3x^2-5}{2x^2+7x+1} = \frac{3}{2}$

So $y=\frac{3}{2}$ is a horizontal asymptote (as $x \to \infty$ and also as $x \to -\infty$).

Worked Example 2: Denominator grows faster

Find:

$\lim_{x \to \infty} \frac{5x-1}{x^2+4}$

Here the numerator degree is 1 and denominator degree is 2, so the denominator grows faster. The limit is:

$\lim_{x \to \infty} \frac{5x-1}{x^2+4} = 0$

Thus $y=0$ is a horizontal asymptote.

Worked Example 3: Different left and right behavior (non-rational example)

Consider:

$f(x)=\frac{|x|}{x}$

For large positive $x$, $|x|=x$ so $f(x)=1$, giving:

$\lim_{x \to \infty} \frac{|x|}{x} = 1$

For large negative $x$, $|x|=-x$ so $f(x)=-1$, giving:

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

This shows a function can approach different horizontal asymptotes on the left and right.

Connecting to graphs and real interpretation

When you see $\lim_{x \to \infty} f(x)=L$, picture walking to the far right of the graph. The curve may wiggle, but it trends closer and closer to the line $y=L$. In real-world terms, $L$ can represent a “steady-state” or “long-term expected” value.

Exam Focus

• Typical question patterns:
  • Compute $\lim_{x \to \infty} \frac{p(x)}{q(x)}$ using degree comparison or leading coefficients.
  • Identify horizontal asymptotes from an algebraic expression or from a graph.
  • Compare $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$ to decide whether there are one or two horizontal asymptotes.
• Common mistakes:
  • Thinking a horizontal asymptote is a “barrier” the graph cannot cross.
  • Mixing up vertical vs horizontal asymptotes (vertical relates to $x \to a$, horizontal relates to $x \to \pm\infty$).
  • Using the ratio of leading coefficients when degrees are not equal (degree comparison comes first).

Intermediate Value Theorem

What the theorem says (and what it doesn’t)

The Intermediate Value Theorem (IVT) is a guarantee about existence. It says that if a function is continuous on an interval, then it takes on every output value between its values at the ends.

A standard statement is:

If $f$ is continuous on the closed interval $[a,b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in $[a,b]$ such that:

$f(c)=N$

This theorem does not tell you what $c$ is, and it does not say there is only one such $c$. It only guarantees at least one exists.

Why IVT matters in calculus

IVT is one of the main tools in Unit 1 for proving that solutions to equations exist without solving them exactly. On the AP exam, it’s often used to justify statements like:

• “There is at least one root between $x=1$ and $x=2$.”
• “There exists a value $c$ where the function equals a given target.”

It also connects to the big idea of continuity: IVT is false if the function has a jump, hole, or vertical asymptote on the interval.

The intuition: “no teleporting” for continuous graphs

If a function is continuous on $[a,b]$, you can draw its graph from $x=a$ to $x=b$ without lifting your pencil. If the graph starts at height $f(a)$ and ends at height $f(b)$, it must pass through every intermediate height in between—because there’s no break where it could “teleport” from one side to the other.

That mental picture is exactly what IVT formalizes.

The continuity condition is the whole game

IVT requires continuity on the entire interval $[a,b]$. That means you must check (or be told) that the function has no discontinuities between $a$ and $b$.

Common situations:

• Polynomials are continuous everywhere, so IVT always applies on any interval.
• Rational functions are continuous on intervals that do not include points where the denominator is zero.
• Piecewise functions require checking that the pieces connect without breaks on the interval.

A classic mistake is to check only $f(a)$ and $f(b)$, see a sign change, and conclude a root exists—without noticing a vertical asymptote or hole inside the interval.

A special but very common IVT use: proving a root exists

To show there is a solution to $f(x)=0$ on $[a,b]$, you can use IVT with $N=0$.

If:

• $f$ is continuous on $[a,b]$,
• $f(a)$ and $f(b)$ have opposite signs (one positive, one negative),

then there exists at least one $c$ in $[a,b]$ such that:

$f(c)=0$

Why? Because $0$ is between a positive number and a negative number.

Worked Example 1: Existence of a root (polynomial)

Show that the equation has a solution between $x=1$ and $x=2$:

$x^3-x-1=0$

Let:

$f(x)=x^3-x-1$

Because $f$ is a polynomial, it is continuous on $[1,2]$.

Compute endpoint values:

$f(1)=1^3-1-1=-1$

$f(2)=2^3-2-1=5$

Since $-1$ and $5$ have opposite signs, $0$ lies between them. By IVT, there exists at least one $c$ in $[1,2]$ such that:

$f(c)=0$

So the equation has at least one solution in that interval.

Worked Example 2: Showing IVT cannot be applied (discontinuity)

Consider:

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

Can you use IVT to claim there is a $c$ in $[0,2]$ such that $f(c)=0$?

First notice: $f$ is not continuous on $[0,2]$ because it is undefined at $x=1$ (vertical asymptote). So IVT does not apply.

Also, in this particular case, $\frac{1}{x-1}$ is never equal to $0$ anyway. The bigger lesson is methodological: discontinuities inside the interval break the guarantee.

Worked Example 3: Hitting a target value $N$

Suppose $f$ is continuous on $[2,5]$, and you know:

$f(2)=3$

and

$f(5)=11$

Explain why there must be a number $c$ in $[2,5]$ such that $f(c)=7$.

Because $7$ is between $3$ and $11$, IVT applies (continuity is given). Therefore there exists at least one $c$ in $[2,5]$ with:

$f(c)=7$

Notice you cannot find $c$ from IVT alone; you would need more information or a numerical method.

How IVT shows up with limits and continuity (connecting this section)

This unit ties IVT to your earlier work on limits and continuity:

• If you can show a function is continuous on $[a,b]$ (often using known continuity rules), IVT becomes available immediately.
• Infinite limits and vertical asymptotes are red flags: if there’s a vertical asymptote inside $[a,b]$, the function is not continuous there, and IVT may fail.
• Limits at infinity don’t directly affect IVT (since IVT is about finite intervals), but both ideas build your overall understanding of how functions behave on intervals and near boundaries.

Exam Focus

• Typical question patterns:
  • “Use the Intermediate Value Theorem to show that a solution exists” for an equation $f(x)=0$ on an interval.
  • “Explain why there exists $c$ such that $f(c)=N$” given continuity and endpoint values.
  • Identify whether IVT can be applied to a given function on a given interval (often involving rational/piecewise functions).
• Common mistakes:
  • Forgetting to verify continuity on the entire interval (especially missing discontinuities where a denominator is zero).
  • Claiming IVT finds the exact value of $c$ (it only guarantees existence).
  • Using IVT when $N$ is not actually between $f(a)$ and $f(b)$ (no “between,” no guarantee).