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)
- Identify the “problem child” function f(x) whose limit is not straightforward.
- 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.
- Take limits of the bounds.
- Check the key condition: make sure
\lim g(x) = \lim h(x). - 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)
- “Problem child”: \sin(1/x) oscillates, no limit.
- Use bound: -1 \le \sin(1/x) \le 1.
- 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. - Limits of bounds as x\to 0:
\lim_{x\to 0} (-x^2)=0, \quad \lim_{x\to 0} x^2=0. - 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
| Rule / Fact | When to use | Notes |
|---|---|---|
| If g(x) \le f(x) \le h(x) and \lim g = \lim h = L, then \lim f = L | Direct squeeze setup | Inequality must hold in a punctured neighborhood of the limit point |
| If |f(x)| \le g(x) and \lim g = 0, then \lim f = 0 | Fastest squeeze to prove a limit is 0 | Rewrite as -g \le f \le g |
| Sequence squeeze: if a_n \le b_n \le c_n eventually and limits of outer sequences match | Limits as n\to\infty | “Eventually” means for all n \ge N |
Go-to bounding inequalities (high yield)
| Bound | When it helps | Notes |
|---|---|---|
| -1 \le \sin u \le 1 and -1 \le \cos u \le 1 | Anything with trig oscillation | Works for any real u |
| 0 \le \sin^2 u \le 1 and 0 \le \cos^2 u \le 1 | Squares of trig | Often makes nonnegative bounds |
| |\sin u| \le 1 and |\cos u| \le 1 | Absolute value squeeze | Leads to |\text{(something)}| \le \text{small} |
| If x>0 and x near 0: \sin x \le x \le \tan x | Classic trig-limit squeezes | Used to prove \lim_{x\to 0} \frac{\sin x}{x}=1 |
| From the classic result: for all x\ne 0 near 0, |\sin x| \le |x| | Controlling sine by its input | Very common shortcut once you know it |
Canonical trig limits you often combine with squeeze
| Limit | Why it matters | Typical use |
|---|---|---|
| \lim_{x\to 0} \frac{\sin x}{x}=1 | Foundation trig limit | Rewrite expressions to use it |
| \lim_{x\to 0} \frac{1-\cos x}{x}=0 | Often asked; can be squeezed | Use identity 1-\cos x = 2\sin^2(x/2) |
| \lim_{x\to 0} \frac{1-\cos x}{x^2}=\frac12 | Very common | Use 1-\cos x = 2\sin^2(x/2) and the sine-over-angle limit |
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