How to Use Squeeze Theorem (and when!) (AP)

What You Need to Know

The big idea (and why it matters)

The Squeeze Theorem (aka Sandwich Theorem) is a limit tool you use when:

• the function you actually care about is hard/impossible to take the limit of directly, but
• you can trap it between two simpler functions with the same limit.

It shows up constantly on AP Calc AB for limits involving:

• trig expressions like $\sin(1/x)$, $\cos x$, $\frac{\sin x}{x}$
• expressions that oscillate but are multiplied by something going to $0$
• absolute value bounds
• sequences (limits as $n \to \infty$)

The theorem (state it cleanly)

If $g(x) \le f(x) \le h(x)$ for all $x$ in some open interval around $a$ (except possibly at $a$), and
$\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L,$
then
$\lim_{x\to a} f(x) = L.$

Sequence version: If $a_n \le b_n \le c_n$ for all sufficiently large $n$ and
$\lim_{n\to\infty} a_n = \lim_{n\to\infty} c_n = L,$
then
$\lim_{n\to\infty} b_n = L.$

Critical reminder: You must have a true inequality (a “squeeze”) holding near the limit point, and the two outer limits must be the same.

When you should think “Squeeze!”

Use it when you see:

• Oscillation: $\sin(\cdot)$ or $\cos(\cdot)$ with input blowing up (like $\sin(1/x)$ as $x\to 0$)
• A small factor times a bounded factor: something like $x^2\sin(1/x)$ as $x\to 0$
• Absolute values: you can often bound with $|f(x)|$
• Classic trig inequality situations: limits involving $\frac{\sin x}{x}$ or $\frac{1-\cos x}{x}$

Step-by-Step Breakdown

How to apply Squeeze Theorem (limit version)

1. Identify the “problem child” function $f(x)$ whose limit is not straightforward.
2. Find two simpler bounding functions $g(x)$ and $h(x)$ such that $g(x) \le f(x) \le h(x)$ near the point.
   • Most common: bound trig by $-1$ and $1$, then multiply by something small.
3. Take limits of the bounds.
4. Check the key condition: make sure
   $\lim g(x) = \lim h(x).$
5. Conclude: then $\lim f(x)$ equals that common value.

The fastest “AP-style” squeeze pattern

If you can show
$|f(x)| \le g(x)$
and $\lim_{x\to a} g(x)=0$,
then automatically
$\lim_{x\to a} f(x)=0.$

This is just Squeeze with $-g(x) \le f(x) \le g(x)$.

Mini worked walkthrough (annotated)

Example skeleton: $\lim_{x\to 0} x^2\sin(1/x)$

1. “Problem child”: $\sin(1/x)$ oscillates, no limit.
2. Use bound: $-1 \le \sin(1/x) \le 1$.
3. Multiply by $x^2$ (note: for all $x$, $x^2 \ge 0$ so inequalities keep direction):
   $-x^2 \le x^2\sin(1/x) \le x^2.$
4. Limits of bounds as $x\to 0$:
   $\lim_{x\to 0} (-x^2)=0, \quad \lim_{x\to 0} x^2=0.$
5. Squeeze conclusion:
   $\lim_{x\to 0} x^2\sin(1/x)=0.$

Decision point: If multiplying an inequality by something that might be negative, stop and handle sign carefully (or use absolute values).

Key Formulas, Rules & Facts

Core squeeze statements

Go-to bounding inequalities (high yield)

Canonical trig limits you often combine with squeeze

Important: On AP Calc AB, you’re generally allowed to use $\lim_{x\to 0} \frac{\sin x}{x}=1$ as a known limit, but you should still know how squeeze proves it in case the question explicitly asks.

Examples & Applications

Example 1 (oscillation times small): $\lim_{x\to 0} x\cos(1/x)$

• Key bound: $-1 \le \cos(1/x) \le 1$.
• Multiply by $|x|$ using absolute value:
  $|x\cos(1/x)| \le |x|.$
• Since $\lim_{x\to 0} |x| = 0$, squeeze gives
  $\lim_{x\to 0} x\cos(1/x) = 0.$

Exam twist: They may ask if $\lim_{x\to 0} \cos(1/x)$ exists (it does not), but $x\cos(1/x)$ does.

Example 2 (classic trig limit via squeeze): $\lim_{x\to 0} \frac{\sin x}{x}$

For $x>0$ near $0$, a standard inequality is
$\sin x \le x \le \tan x.$
Divide by $\sin x$ (positive for small positive $x$):
$1 \le \frac{x}{\sin x} \le \frac{1}{\cos x}.$
Now take reciprocals (all positive, so inequality flips correctly when reciprocating):
$\cos x \le \frac{\sin x}{x} \le 1.$
As $x\to 0$,
$\lim_{x\to 0} \cos x = 1, \quad \lim_{x\to 0} 1 = 1,$
so
$\lim_{x\to 0} \frac{\sin x}{x} = 1.$

Exam twist: You might need to do it two-sided. The same limit holds for $x<0$, so the two-sided limit is $1$.

Example 3 (make it look like sine-over-angle): $\lim_{x\to 0} \frac{1-\cos x}{x}$

This is a great squeeze + identity situation.
Use
$1-\cos x = 2\sin^2(x/2).$
Then
$\frac{1-\cos x}{x} = \frac{2\sin^2(x/2)}{x}.$
Rewrite to expose $\frac{\sin(x/2)}{x/2}$:
$\frac{2\sin^2(x/2)}{x} = \sin(x/2) \cdot \frac{\sin(x/2)}{x/2}.$
Now use bounds:

• $|\sin(x/2)| \le |x/2| \to 0$
• $\frac{\sin(x/2)}{x/2} \to 1$
  So the product goes to $0$:
  $\lim_{x\to 0} \frac{1-\cos x}{x} = 0.$

Exam twist: Many students try L’Hôpital (not AB). This method is the intended AB path.

Example 4 (sequence squeeze): $\lim_{n\to\infty} n\sin\left(\frac{1}{n}\right)$

This looks like a product, but it’s really the sine-over-angle limit in disguise.
Rewrite:
$n\sin\left(\frac{1}{n}\right) = \frac{\sin(1/n)}{1/n}.$
As $n\to\infty$, we have $1/n \to 0$, so
$\lim_{n\to\infty} \frac{\sin(1/n)}{1/n} = 1.$

Alternative squeeze view: Use $|\sin u| \le |u|$ with $u=1/n$ to see it’s bounded and behaving nicely.

Common Mistakes & Traps

1. Forgetting the inequality must hold near the point

   • What goes wrong: You show $g(x) \le f(x) \le h(x)$ for “some” $x$ but not in a full neighborhood around $a$.
   • Fix: Explicitly state “for all $x$ sufficiently close to $a$ (and $x\ne a$)” and make sure your bound is actually always true there.

2. Outer limits aren’t equal (so you can’t conclude anything)

   • What goes wrong: You find bounds, but $\lim g(x) \ne \lim h(x)$.
   • Fix: Squeeze only works when the two outer limits match. If they don’t, you need different bounds or a different method.

3. Multiplying/dividing inequalities without checking sign

   • What goes wrong: You multiply by an expression that can be negative (like $x$ near $0$) and forget the inequality flips.
   • Fix: Use $|x|$ and absolute values whenever sign is unclear, or split into cases $x>0$ and $x<0$.

4. Trying to squeeze with one bound

   • What goes wrong: You only find $f(x) \le h(x)$ but no lower bound.
   • Fix: Often the missing bound is $-h(x)$ after taking absolute value: show $|f(x)| \le h(x)$.

5. Assuming $\sin(1/x) \to 0$ as $x\to 0$

   • What goes wrong: You confuse “bounded” with “approaches $0$.”
   • Fix: Remember $\sin(1/x)$ oscillates between $-1$ and $1$; it has **no limit**. Only products like $x\sin(1/x)$ can be squeezed.

6. Using the squeeze theorem when direct limit laws already work

   • What goes wrong: You waste time building inequalities unnecessarily.
   • Fix: First try algebraic simplification and known limits. Use squeeze when you hit oscillation, absolute values, or indeterminate behavior that’s hard to simplify.

7. Not matching the squeeze to the exact expression

   • What goes wrong: You bound $\sin(1/x)$ by $[-1,1]$ but forget the multiplying factor (like $x^2$) changes the bounds.
   • Fix: After bounding, multiply through carefully to produce bounds that look like something with an easy limit.

8. Confusing two-sided limits with one-sided inequalities

   • What goes wrong: You use an inequality valid only for $x>0$ (like $\sin x \le x \le \tan x$) but claim a two-sided conclusion without addressing $x<0$.
   • Fix: Either do both sides or use absolute value inequalities that work for all $x$ near the point.

Memory Aids & Quick Tricks

Quick Review Checklist

• You can use Squeeze when you can prove $g(x) \le f(x) \le h(x)$ near the point and $\lim g = \lim h$.
• The inequality must hold for all $x$ sufficiently close to $a$ (except possibly at $a$).
• Fast version: if $|f(x)| \le g(x)$ and $g(x)\to 0$, then $f(x)\to 0$.
• Default trig bounds: $-1 \le \sin u \le 1$ and $-1 \le \cos u \le 1$.
• Watch signs when multiplying/dividing inequalities; use $|\cdot|$ to stay safe.
• For trig limits, try rewriting until you see $\frac{\sin u}{u}$.
• If outer limits don’t match, you can’t conclude a limit from squeeze.

You’ve got this—most squeeze problems are just one good bound away from being easy.