AP Calculus AB Unit 1 Study Guide: Limits and Continuity (Merged Notes)

The Big Idea of a Limit

A limit is the mathematical way to describe what value a function is approaching as the input gets close to some number. The key idea is that a limit is about nearby inputs, not necessarily what happens at the input itself.

For example, if you ask what happens to a function as the input gets close to 2, you are allowed to use inputs like 1.9, 1.99, 2.01, and 2.1, but you don’t actually need to use the input 2. That’s why limits are so powerful: they can describe behavior even when a function is undefined at a point, has a hole, or has a jump.

Why limits matter

Limits are the foundation for two major ideas you’ll soon study.

• Derivatives: the derivative is defined using a limit of an average rate of change as the interval shrinks.
• Integrals: definite integrals are defined using limits of Riemann sums as rectangles get thinner.

So Unit 1 is not “just” pre-derivative content. It’s building the language and logic calculus uses for everything else.

Limit notation and meaning

The notation

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

means: as the input gets close to %%LATEX1%% (from either side), the output gets close to %%LATEX2%%.

Important clarifications:

• %%LATEX3%% does not mean %%LATEX4%%.
• The function value $f(a)$ might not exist.
• Even if $f(a)$ exists, it might not equal the limit.

A common misconception is to treat a limit as “plug in %%LATEX7%%.” Sometimes that works, but only when the function is continuous at %%LATEX8%% (a concept you’ll formalize later).

A first example: when the limit exists but the function value is different

Suppose

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

If you plug in $x = 1$, you get an undefined form. But you can simplify:

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

For $x \ne 1$, this simplifies to

$f(x) = x + 1$

Now it’s clear that as $x \to 1$, the function approaches 2. So

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

even though $f(1)$ does not exist.

Exam Focus

• Typical question patterns:
  • Evaluate a limit from an algebraic expression that gives $\frac{0}{0}$ at first.
  • Decide whether a limit exists from a graph or description.
  • Distinguish clearly between %%LATEX18%% and %%LATEX19%%.
• Common mistakes:
  • Assuming “undefined at $a$” implies “limit does not exist.” A hole can still have a limit.
  • Saying a limit exists because you found a value by plugging in, even when the function has different behavior from left and right.

Reading Limits from Graphs and Tables

A big part of learning limits is training yourself to focus on behavior near the point, not the point itself. In practice, the AP exam expects you to be comfortable finding limits four common ways: from a graph, from a table, by algebraic properties (limit laws), and by algebraic manipulation (like factoring and canceling).

Estimating a limit from a graph

On a graph, to estimate

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

you look at what %%LATEX22%%-value the curve approaches as %%LATEX23%% gets close to $a$ from both sides.

Key idea: an open circle at $x = a$ might indicate the function value isn’t defined there, but you still can often see the height the graph approaches.

If the left side approaches one value and the right side approaches a different value, the two-sided limit does not exist.

One-sided limits visually

The left-hand limit is

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

and it describes what happens as %%LATEX27%% approaches %%LATEX28%% from values less than $a$.

The right-hand limit is

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

and it describes what happens as %%LATEX31%% approaches %%LATEX32%% from values greater than $a$.

A two-sided limit exists exactly when both one-sided limits exist and are equal.

Estimating a limit from a table

A table gives sample values of %%LATEX34%% close to %%LATEX35%% and the corresponding %%LATEX36%% values. Your job is to notice what %%LATEX37%% is approaching.

To do this well:

• Use values approaching from both sides (less than and greater than $a$).
• Ignore the value at $a$ if it’s present; the limit depends on nearby values.
• Watch for rapid growth (which might suggest an infinite limit).

Notation reference (common forms you must recognize)

Worked example (graph/table reasoning style)

Suppose a table near $x = 3$ shows:

• For %%LATEX49%%, %%LATEX50%%
• For %%LATEX51%%, %%LATEX52%%
• For %%LATEX53%%, %%LATEX54%%
• For %%LATEX55%%, %%LATEX56%%

These values suggest %%LATEX57%% is approaching %%LATEX58%% as $x \to 3$, so you would estimate

$\lim_{x \to 3} f(x) = 6$

Even if the table also says $f(3) = 10$, the limit is still about nearby values, not the function value at the point.

Exam Focus

• Typical question patterns:
  • Use a graph to state a two-sided limit and one-sided limits.
  • Use a table to estimate a limit and justify the estimate.
  • Determine whether a limit exists and explain why (often using left vs right behavior).
• Common mistakes:
  • Reading $f(a)$ instead of the approached value.
  • Declaring “DNE” based on one side only—always check both directions for two-sided limits.

When Can You Substitute? Limits and Continuity Intuition

A lot of limit problems become easy if you know when simple substitution is valid.

Direct substitution and its hidden assumption

If you try to compute

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

and %%LATEX64%% is a “nice” expression (like a polynomial) near %%LATEX65%%, you can often substitute $x = a$:

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

This works when %%LATEX68%% is continuous at %%LATEX69%% (or at least has no problematic feature at $a$). In particular, to find the limit of a simple polynomial, you can plug in the number that the variable is approaching.

Polynomials are continuous everywhere, so substitution always works for them.

Rational functions

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

are continuous wherever $q(x) \ne 0$, so substitution works as long as you are not plugging into a zero denominator.

Indeterminate forms: why $\frac{0}{0}$ is special

When substitution gives

$\frac{0}{0}$

that does not mean the limit is 0, and it does not mean the limit is undefined. It means “this algebraic form doesn’t tell you the answer yet.” You have to rewrite/simplify the function to reveal the approaching behavior.

Algebraic properties (limit laws)

The limit laws (often listed in algebraic-properties summaries) let you break complicated expressions into simpler pieces when the relevant limits exist.

If

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

and

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

then:

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

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

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

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

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

provided

$M \ne 0$

And for integer powers:

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

These laws explain why substitution works so often: if a function is built from continuous pieces using these operations, it inherits continuity.

Worked example (direct substitution)

Evaluate

$\lim_{x \to 2} (x^3 - 4x + 1)$

Because this is a polynomial, substitution is valid:

$\lim_{x \to 2} (x^3 - 4x + 1) = 2^3 - 4(2) + 1 = 1$

Exam Focus

• Typical question patterns:
  • Evaluate limits quickly using substitution (polynomials, rational functions with nonzero denominator).
  • Identify an indeterminate form and choose a technique (factor, rationalize, etc.).
  • Use limit laws to justify steps.
• Common mistakes:
  • Using the quotient law when the denominator limit is 0.
  • Treating $\frac{0}{0}$ as a numerical result instead of an indeterminate form.

Algebraic Techniques for Evaluating Limits

When substitution fails (often because of $\frac{0}{0}$), you need algebra to rewrite the function in a form that shows its behavior near the point.

Technique 1: Factoring and canceling (removable discontinuities)

If you have a rational expression where both numerator and denominator go to 0, factoring is often the fastest path. The key idea is that if both numerator and denominator share a factor that becomes 0 at $x = a$, canceling that factor removes the “hole behavior” (a removable discontinuity) and reveals the value the function approaches.

Example: classic factor-and-cancel

Evaluate

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

Factor the numerator:

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

So (for %%LATEX91%%) the expression simplifies to %%LATEX92%%, and then

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

Example (from common “cancel the removable discontinuity” practice)

Simplify the rational expression:

$\frac{(x+3)(x+2)}{(x+3)(x-3)}$

The factor %%LATEX95%% can be removed (for %%LATEX96%%), which indicates a removable discontinuity (a hole) at $x = -3$. After cancellation, the simplified expression is:

$\frac{x+2}{x-3}$

Technique 2: Rationalizing (conjugates)

Rationalizing is useful when a limit involves a square root and substitution gives $\frac{0}{0}$. The move is to multiply by the conjugate to eliminate the radical.

Example: rationalize a difference of square roots

Evaluate

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

Multiply by the conjugate:

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

The numerator becomes a difference of squares:

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

Cancel (for $x \ne 4$) and substitute:

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

Technique 3: Simplifying complex fractions

Complex fractions hide the main algebraic structure. A reliable approach is to combine the “small fractions” in the numerator (or multiply top and bottom by a common denominator).

Example: clear denominators

Evaluate

$\lim_{x \to 0} \frac{\frac{1}{x+1} - 1}{x}$

First simplify the numerator:

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

Then the whole expression becomes (for $x \ne 0$):

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

Now substitute:

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

Technique 4: Special trigonometric limits

These trig limits as the input approaches 0 appear frequently in AP Calculus AB (angles in radians):

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

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

A very useful generalization is:

$\lim_{x \to 0} \frac{\sin(ax)}{x} = a$

which follows from rewriting:

$\frac{\sin(ax)}{x} = a \cdot \frac{\sin(ax)}{ax}$

Another common pattern is:

$\lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} = \frac{a}{b}$

(assuming %%LATEX115%% and %%LATEX116%% are constants and $b \ne 0$).

Conceptually, the limit

$\lim_{x \to 0} \frac{\sin x}{x} = 1$

can be justified with unit-circle geometry and an early use of the Squeeze Theorem.

Example: trig limit using the special form

Evaluate

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

Rewrite:

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

So

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

Exam Focus

• Typical question patterns:
  • Evaluate limits that initially produce $\frac{0}{0}$ using factoring or rationalizing.
  • Evaluate a limit requiring a special trig limit and an algebraic rewrite.
  • Match an algebraic simplification to a graph interpretation (hole vs asymptote).
• Common mistakes:
  • Canceling terms that are not factors (for example, “canceling” inside sums).
  • Using $\lim_{x \to 0} \frac{\sin x}{x} = 1$ when the angle is not approaching 0 (without rewriting).

One-Sided Limits and Piecewise-Defined Functions

Piecewise functions are where one-sided limits become unavoidable, because the rule for $f(x)$ can change depending on whether the input is less than or greater than a boundary value.

The formal relationship: two-sided vs one-sided

The two-sided limit

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

exists if and only if both one-sided limits exist and are equal:

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

If they are unequal, the two-sided limit does not exist, which often corresponds to a jump discontinuity.

How to evaluate limits of piecewise functions

When you see a piecewise definition, your strategy should be:

1. Identify which formula applies for values just less than $a$ (left side).
2. Identify which formula applies for values just greater than $a$ (right side).
3. Compute the left-hand and right-hand limits separately.
4. Compare them.

Example: one-sided limits from a piecewise rule

Let the function be defined by:

$f(x)=x+2 \text{ for } x<1$

$f(x)=x^2 \text{ for } x \ge 1$

Compute the one-sided limits at $x = 1$.

Left-hand limit (use the rule for $x < 1$):

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

Right-hand limit (use the rule for $x \ge 1$):

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

Because $3 \ne 1$, the two-sided limit does not exist:

$\lim_{x \to 1} f(x) \text{ does not exist}$

Also note the function value is determined by the rule that includes equality, so:

$f(1)=1$

This is a classic situation where a function value exists but the two-sided limit does not.

One-sided limits and domain restrictions

Sometimes one-sided limits arise because the function is only defined on one side. For example,

$f(x)=\sqrt{x}$

is only defined for %%LATEX140%% (in real numbers). So at %%LATEX141%% you can discuss

$\lim_{x \to 0^+} \sqrt{x}$

but a left-hand limit in the reals doesn’t apply.

Exam Focus

• Typical question patterns:
  • Compute one-sided limits for a piecewise function and decide if the two-sided limit exists.
  • Use one-sided limits to determine whether a function has a jump at a point.
  • Interpret limit notation at domain endpoints.
• Common mistakes:
  • Using the wrong piece of the function for the left-hand or right-hand limit.
  • Claiming a two-sided limit exists by computing only one side.

Infinite Limits and Vertical Asymptotes

Not all limits approach a finite number. Sometimes a function grows without bound as you approach a certain input. That behavior is captured by infinite limits.

What an infinite limit means

If

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

it means that as %%LATEX144%% gets close to %%LATEX145%%, $f(x)$ grows arbitrarily large (positive). Similarly,

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

means the function decreases without bound.

Important: infinity is not a number you “reach,” so you should not treat $\infty$ like a normal value you can plug into other algebra.

Vertical asymptotes and their connection to limits

A vertical asymptote occurs at

$x = a$

if the function goes to %%LATEX150%% or %%LATEX151%% as %%LATEX152%% approaches %%LATEX153%% from at least one side. In many algebraic situations (especially rational functions), the simplified function is undefined at the asymptote, and the vertical line is not something the graph passes through in the usual way. The key calculus idea is unbounded growth near the line.

Be careful with rational functions:

• If a factor cancels, you get a hole (removable discontinuity), not a vertical asymptote.
• If it does not cancel, a vertical asymptote is likely.

One-sided infinite limits

Often the left and right behaviors differ:

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

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

In that case, the two-sided limit does not exist, but you can still fully describe the behavior.

Worked example: detect a vertical asymptote

Evaluate the behavior of

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

as $x \to 2$.

From the left, the denominator is a small negative number, so the fraction becomes very negative:

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

From the right, the denominator is a small positive number, so the fraction becomes very positive:

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

So there is a vertical asymptote at $x = 2$, and the two-sided limit does not exist.

Worked example: hole vs vertical asymptote

Consider

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

For $x \ne 1$,

$g(x) = \frac{1}{x+2}$

So as $x \to 1$,

$\lim_{x \to 1} g(x) = \frac{1}{3}$

This indicates a removable discontinuity (a hole) at $x = 1$, not a vertical asymptote.

Exam Focus

• Typical question patterns:
  • Determine whether a rational function has a vertical asymptote or a removable discontinuity at a given $x$-value.
  • Evaluate one-sided infinite limits from algebra or a graph.
  • Interpret limit statements involving $\infty$.
• Common mistakes:
  • Treating $\infty$ as a number and performing arithmetic with it as if it were finite.
  • Forgetting to simplify first, leading to calling a hole a vertical asymptote.

Limits as $x$ Goes to Infinity: End Behavior and Horizontal Asymptotes

Limits can also describe what happens far to the left or right on a graph.

Meaning of a limit at infinity

The statement

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

means that as %%LATEX172%% becomes very large, %%LATEX173%% gets close to $L$.

Similarly,

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

describes end behavior as $x$ becomes very negative.

Horizontal asymptotes

A horizontal asymptote occurs at

$y = L$

if

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

or

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

A function can have one horizontal asymptote, two different horizontal asymptotes (one for %%LATEX180%%, one for %%LATEX181%%), or no horizontal asymptote. Also, a horizontal asymptote describes end behavior, not a “barrier,” so it can be crossed.

Rational functions and degree comparison (horizontal asymptote rules)

For rational functions

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

where %%LATEX183%% and %%LATEX184%% are polynomials, end behavior is determined by the degrees.

Let the degree of %%LATEX185%% be %%LATEX186%% and the degree of %%LATEX187%% be %%LATEX188%%.

• If $n < m$, then

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

so the horizontal asymptote is

$y = 0$

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

• If %%LATEX193%%, then the function does not approach a finite horizontal value. In particular, the limit as %%LATEX194%% grows without bound (often described informally as “the limit is infinity”), so there is no horizontal asymptote.

Example: degree numerator less than degree denominator

Evaluate

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

As $x$ grows, the denominator grows faster (quadratic vs linear), so

$\lim_{x \to \infty} \frac{3x+1}{x^2+5} = 0$

Example: equal degrees

Evaluate

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

The highest power in both numerator and denominator is $x^2$, so the limit equals the ratio of leading coefficients:

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

A reliable way to see this is to divide numerator and denominator by $x^2$:

$\frac{5 - \frac{1}{x^2}}{2 + \frac{7}{x}}$

As %%LATEX203%%, the fractions go to 0, leaving %%LATEX204%%.

Exam Focus

• Typical question patterns:
  • Compute limits at infinity for rational functions using degree/leading coefficient reasoning.
  • Identify horizontal asymptotes from limits (or vice versa).
  • Compare end behavior for %%LATEX205%% and %%LATEX206%%.
• Common mistakes:
  • Thinking a horizontal asymptote cannot be crossed.
  • Using substitution “$x = \infty$” instead of using degree comparison or dividing by the highest power.

Continuity: Definition, Meaning, and How to Check It

Limits set the stage for continuity, which formalizes the idea that a graph can be drawn without lifting your pencil (with an important caveat: calculus cares about specific conditions, not just a sketch).

What it means for a function to be continuous at a point

A function %%LATEX208%% is continuous at %%LATEX209%% if all three of these conditions hold:

1. $f(a)$ exists.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

Continuity on an interval

A function is continuous on an interval if it is continuous at every point in the interval. At endpoints, continuity is interpreted with one-sided limits:

• At a left endpoint $a$, check

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

• At a right endpoint $b$, check

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

Types of discontinuities you should recognize

1. Removable discontinuity (a hole): an otherwise continuous curve has a hole in it. It’s “removable” because one can remove the discontinuity by filling the hole.
2. Jump discontinuity: the curve “breaks” at a particular place and starts somewhere else; the limits from the left and the right both exist, but they do not match.
3. Infinite (essential) discontinuity: the curve has a vertical asymptote and the function grows without bound.

Removing discontinuities (fixing a removable discontinuity)

You can remove a removable discontinuity by redefining the function at the problematic input value so that the function value equals the limit (in other words, “fill the hole”). Algebraically, this is frequently done by factoring out a common root between the numerator and denominator to reveal the simplified expression that gives the correct limit.

Worked example: checking continuity at a point

Define the function by:

$f(x) = \frac{x^2 - 1}{x - 1} \text{ for } x \ne 1$

$f(1) = 5$

Is %%LATEX219%% continuous at %%LATEX220%%?

• %%LATEX221%% exists, and %%LATEX222%%.
• For $x \ne 1$,

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

So

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

• The limit exists, but it does not equal the function value:

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

So the function is not continuous at %%LATEX227%%. This is removable: continuity would be “fixed” by redefining %%LATEX228%% to be 2.

Real-world interpretation: why continuity matters

Continuity is often a modeling assumption. If a function represents the temperature of a room over time, it’s reasonable to assume temperature changes continuously, not with instantaneous jumps (unless something physically discontinuous happens, like a sensor glitch). In contrast, if a function represents the cost to ship a package based on weight brackets, jumps can be realistic.

Exam Focus

• Typical question patterns:
  • Check continuity at a point using the 3-condition definition.
  • Classify a discontinuity (removable, jump, infinite) from a graph or formula.
  • Determine values of constants in a piecewise function so that it is continuous.
• Common mistakes:
  • Checking only whether $f(a)$ exists, without checking whether the limit matches.
  • Confusing “limit exists” with “function is continuous.” Continuity needs equality with the function value.

Theorems That Connect Limits and Continuity (Including the Squeeze Theorem and IVT)

Calculus becomes much easier when you rely on theorems that let you reason about limits and continuity without re-deriving everything from scratch.

Continuity of common function types

These are standard facts you can safely use:

• Polynomials are continuous for all real inputs.
• Rational functions are continuous wherever the denominator is not 0.
• Root functions like $\sqrt{x}$ are continuous on their domains.
• Trigonometric functions like %%LATEX231%% and %%LATEX232%% are continuous for all real inputs.

Composition and continuity

If a function %%LATEX233%% is continuous at %%LATEX234%% and a function %%LATEX235%% is continuous at %%LATEX236%%, then the composition %%LATEX237%% is continuous at %%LATEX238%%.

In limit language: if

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

and %%LATEX240%% is continuous at %%LATEX241%%, then

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

Example: limit using continuity of a composition

Evaluate

$\lim_{x \to 0} \cos(3x)$

As %%LATEX244%%, %%LATEX245%%. Since cosine is continuous,

$\lim_{x \to 0} \cos(3x) = \cos(0) = 1$

The Squeeze Theorem

The Squeeze Theorem is used when a function is difficult to evaluate directly, but you can trap it between two simpler functions with the same limit.

If for all inputs in some interval around $a$,

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

and

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

and

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

then

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

Example: classic squeeze limit

Evaluate

$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$

Because

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

and since $x^2 \ge 0$,

$-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$

Now

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

and

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

So

$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$

A common misconception is thinking oscillation always forces “no limit.” Oscillation can prevent a limit, but if it’s multiplied by something that shrinks to 0 fast enough, the product can still have a limit.

The Intermediate Value Theorem (IVT)

The Intermediate Value Theorem says: if %%LATEX259%% is continuous on the closed interval from %%LATEX260%% to %%LATEX261%%, and %%LATEX262%% is any number between %%LATEX263%% and %%LATEX264%%, then there exists at least one number between %%LATEX265%% and %%LATEX266%% such that

$f(c) = C$

Intuition: a continuous graph can’t “jump over” intermediate heights.

IVT tells you a solution exists, but it does not tell you where it is exactly, and it does not tell you it is unique.

Example: using IVT to prove a root exists

Let

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

This is a polynomial, so it is continuous everywhere.

Compute:

$f(1) = -1$

$f(2) = 5$

Because 0 lies between %%LATEX271%% and %%LATEX272%%, IVT guarantees at least one number $c$ in the interval from 1 to 2 such that

$f(c)=0$

So the equation

$x^3 - x - 1 = 0$

has at least one solution in the open interval from 1 to 2.

Exam Focus

• Typical question patterns:
  • Use continuity of standard functions and compositions to evaluate limits by substitution.
  • Apply the Squeeze Theorem to evaluate a limit that involves oscillation or absolute value.
  • Use IVT to justify that an equation has a solution on an interval.
• Common mistakes:
  • Applying IVT when the function is not continuous on the entire interval.
  • Using Squeeze Theorem without establishing a valid inequality around the target function.
  • Treating IVT as a method to find the solution rather than prove it exists.