AP Calculus AB · Lesson Review · August 28, 2026

The Squeeze Theorem, Continuity, and the Intermediate Value Theorem

Trap an oscillating expression between simpler functions, formalize continuity, and use continuity to guarantee that an intermediate value exists.

43-minute edited lesson8 chaptersCaptions + corrected transcript

01 · Watch

Lesson video

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0:0001The Squeeze Theorem, Continuity, and IVT 0:0502The Squeeze Theorem 9:1003Continuity at a Point and on an Interval 12:2904One-Sided and Endpoint Continuity 20:2105Properties of Continuous Functions 25:0506Piecewise Continuity 33:2907The Intermediate Value Theorem 38:5408IVT Example: Prove a Zero Exists

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02 · Understand

Learning targets

Use known bounds to evaluate $\lim_{x\to0}x^2\sin\!\left(\frac1x\right)$ .
Test continuity at a point by comparing the limit and function value.
Apply one-sided continuity at the endpoints of a closed interval.
Find where a piecewise function is continuous by checking its junctions.
Use the Intermediate Value Theorem to prove that a zero exists.

03 · Vocabulary

Key definitions

Each theorem begins with hypotheses that must be verified before using its conclusion.

01Squeeze Theorem: If $g(x)\le f(x)\le h(x)$ near $a$ and both outer functions approach the same value $L$ , then $\lim_{x\to a}f(x)=L$ .
02Continuous at a point: The function is defined at $c$ , the limit exists, and $\lim_{x\to c}f(x)=f(c)$ .
03Continuous on a closed interval: The function is continuous on $(a,b)$ , continuous from the right at $a$ , and continuous from the left at $b$ .
04Piecewise continuity: Continuity of every formula on its own interval, plus agreement of the one-sided limits and function value at each junction.
05Existence theorem: A theorem that guarantees an object or value exists without necessarily providing a formula for finding it.
06Intermediate Value Theorem: A function continuous on $[a,b]$ takes every output value between $f(a)$ and $f(b)$ .

04 · Apply

Representative examples

State the hypotheses, show that they hold, and only then use the theorem.

Squeeze

Trap an oscillating factor

$-1\le\sin\!\left(\frac1x\right)\le1$
Since $x^2\ge0$ , multiply all three parts without reversing the inequalities.
$-x^2\le x^2\sin\!\left(\frac1x\right)\le x^2$
$\lim_{x\to0}(-x^2)=0=\lim_{x\to0}x^2$
Therefore $\lim_{x\to0}x^2\sin\!\left(\frac1x\right)=0$ .

Continuity

Check all three conditions at a point

$f(c)$ must be defined.
$\lim_{x\to c}f(x)$ must exist.
The two values must agree: $\lim_{x\to c}f(x)=f(c)$ .

Endpoints

Use the direction available inside the interval

At the left endpoint, require $\lim_{x\to a^+}f(x)=f(a)$ .
At the right endpoint, require $\lim_{x\to b^-}f(x)=f(b)$ .
That is why $\sqrt{x}$ is continuous on its domain $[0,\infty)$ even though there are no domain values to the left of zero.

Piecewise

Check the junctions at zero and pi

$f(x)=\begin{cases}\sin x,&x\ge\pi\\x^2+1,&0\le x<\pi\\e^x,&x<0\end{cases}$
At $x=0$ , both one-sided limits and $f(0)$ equal $1$ , so the function is continuous.
At $x=\pi$ , the left-hand limit is $\pi^2+1$ while the right-hand limit is $\sin\pi=0$ .
The function is continuous on $(-\infty,\pi)\cup(\pi,\infty)$ .

Intermediate value

Prove that a zero exists

Let $f(x)=x^3-4$ . As a polynomial, it is continuous on $[0,2]$ .
$f(0)=-4$ and $f(2)=4$ .
Because $0$ lies between $-4$ and $4$ , IVT guarantees $\exists c\in(0,2):f(c)=0$ .
The theorem guarantees existence; it does not calculate the exact value of $c$ .

05 · Avoid

Common mistakes

Treating $0\cdot\mathrm{DNE}$ as a number.

It is an indeterminate situation; use a theorem or another method.

Reversing an inequality unnecessarily.

Multiplying by $x^2$ never reverses an inequality because $x^2\ge0$ .

Checking only the formula pieces.

Piecewise continuity can fail where adjacent formulas meet.

Using a two-sided endpoint limit.

At a domain endpoint, use the one-sided limit from within the domain.

Skipping IVT hypotheses.

First state the closed interval and explain why the function is continuous there.

Claiming IVT finds the root.

It proves at least one root exists in the interval.

06 · Check yourself

Quick check

Try each question before opening its answer.

1Why can

\sin(1/x)

be bounded between

-1

and

1

Sine always has outputs in $[-1,1]$ , regardless of its input.

2What equation defines continuity at

x=c

$\lim_{x\to c}f(x)=f(c)$ , together with the existence of both quantities.

3Why is the piecewise function discontinuous at

x=\pi

The one-sided limits disagree: $\pi^2+1\ne0$ .

4What does IVT conclude from

f(0)=-4

and

f(2)=4

when

f

is continuous?

There is at least one $c\in(0,2)$ for which $f(c)=0$ .

07 · Revisit

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