How to Use Squeeze Theorem (and when!) (AP)
1) What You Need to Know
What it is (and why you care)
The Squeeze Theorem (aka Sandwich Theorem) is your go-to tool for limits when:
- the expression oscillates (usually trig like \sin(1/x), \cos(1/x)),
- the expression is messy but you can trap it between two simpler functions,
- or you need to prove a standard limit (especially trig limits) from basic inequalities.
On AP Calc BC, it shows up most often in:
- limits as x \to 0 involving trig + powers of x,
- limits as x \to \infty (or sequences n \to \infty) where something is bounded,
- justification questions (“explain using Squeeze Theorem”).
The theorem (precise statement)
If for all x near a (possibly excluding a itself),
g(x) \le f(x) \le h(x)
and
\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L,
then
\lim_{x \to a} f(x) = L.
Same idea for sequences: if a_n \le b_n \le c_n for all sufficiently large n and \lim a_n = \lim c_n = L, then \lim b_n = L.
When you should use it
Use Squeeze Theorem when direct substitution fails and:
- you can find two functions that are easy to limit, and
- those bounds meet at the same limit.
Most common “Squeeze signals”:
- bounded trig: -1 \le \sin(\cdot) \le 1 and -1 \le \cos(\cdot) \le 1
- absolute value: -|u| \le u \le |u| and if |f(x)| \le g(x) with g(x) \to 0 then f(x) \to 0
- products with oscillation: something like x^k \sin(1/x) or \frac{\sin n}{n}
Critical reminder: Squeeze only works if you show the inequality holds in a punctured neighborhood of a (or for all sufficiently large n), and both bounds have the same limit.
2) Step-by-Step Breakdown
The “Sandwich” method (limits)
1) Identify the hard piece
- Usually the oscillating/unknown part: \sin(1/x), \cos(1/x), or something trapped by absolute value.
2) Write a true inequality that traps it
- For trig: -1 \le \sin(\theta) \le 1 and -1 \le \cos(\theta) \le 1 for any \theta.
- For absolute value: -|u| \le u \le |u|.
3) Multiply (or otherwise transform) the inequality carefully
- If you multiply by something that could be negative, the inequality direction can flip.
- A safe move: multiply by a nonnegative expression like |x|^k or x^2.
4) Compute the limits of the two bounds
- If both go to the same L, you’ve squeezed.
5) Conclude the middle limit equals that same L
- State explicitly: “Since g(x) \le f(x) \le h(x) and \lim g = \lim h = L, then \lim f = L.”
Mini worked walkthrough (classic oscillation)
Evaluate \lim_{x \to 0} x^2 \sin(1/x).
1) Use trig bound: -1 \le \sin(1/x) \le 1.
2) Multiply by x^2 (note x^2 \ge 0 so inequality direction stays):
-x^2 \le x^2\sin(1/x) \le x^2.
3) Take limits as x \to 0:
\lim_{x \to 0} (-x^2) = 0, \qquad \lim_{x \to 0} x^2 = 0.
4) Bounds match, so
\lim_{x \to 0} x^2\sin(1/x) = 0.
The “absolute value squeeze” (super common)
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.
Reason: -|f(x)| \le f(x) \le |f(x)| and 0 \le |f(x)| \le g(x).
Step-by-step for sequences
To evaluate \lim_{n \to \infty} b_n by squeeze:
1) Find a_n and c_n with a_n \le b_n \le c_n for all sufficiently large n.
2) Compute \lim a_n and \lim c_n.
3) If they match, conclude \lim b_n.
3) Key Formulas, Rules & Facts
Core Squeeze Theorem rules
| Fact / Rule | When to use | Notes |
|---|---|---|
| g(x) \le f(x) \le h(x) and \lim g = \lim h = L implies \lim f = L | Any limit or sequence limit | Inequality must hold near a (not just at points). |
| If |f(x)| \le g(x) and \lim g = 0 then \lim f = 0 | Proving a limit is 0 | Often fastest approach. |
| Sequence version: a_n \le b_n \le c_n and \lim a_n = \lim c_n = L implies \lim b_n = L | Limits as n \to \infty | Inequality must hold for all large enough n. |
Standard bounds you should know cold
| Bound | Use it to squeeze | Notes |
|---|---|---|
| -1 \le \sin(\theta) \le 1 | Anything with \sin(\cdot) | Works for all real \theta. |
| -1 \le \cos(\theta) \le 1 | Anything with \cos(\cdot) | Also for all real \theta. |
| 0 \le \sin^2(\theta) \le 1 and 0 \le \cos^2(\theta) \le 1 | Nonnegative squeezes | Great when multiplying by positives. |
| |\sin(\theta)| \le 1 and |\cos(\theta)| \le 1 | Absolute value squeezes | Often cleaner than writing -1 \le \cdot \le 1. |
| -\,|u| \le u \le |u| | Handling sign issues | Lets you “trap” any expression. |
Useful “squeeze-ready” patterns
| Pattern | Typical conclusion | Why it works |
|---|---|---|
| x^k \sin(1/x) as x \to 0 with k>0 | Limit is 0 | Because |x^k\sin(1/x)| \le |x^k| \to 0. |
| \frac{\sin x}{x} as x \to 0 | Limit is 1 | Classic squeeze using geometry/inequalities (often given/assumed). |
| \frac{\sin(n)}{n} as n \to \infty | Limit is 0 | |\sin(n)| \le 1 so \left|\frac{\sin(n)}{n}\right| \le \frac{1}{n} \to 0. |
| \sin(x) bounded + denominator grows | Limit tends to 0 | Use bounded-over-growing squeeze. |
Warning: Squeeze is not “the limit of the middle is between the limits.” The bounds’ limits must be the same.
4) Examples & Applications
Example 1: Oscillation + power of x
Evaluate \lim_{x \to 0} x\cos(1/x).
- Bound: -1 \le \cos(1/x) \le 1.
- Multiply by |x| (safe, nonnegative):
-|x| \le x\cos(1/x) \le |x|.
- Limits: \lim_{x \to 0} (-|x|)=0 and \lim_{x \to 0} |x|=0.
So
\lim_{x \to 0} x\cos(1/x)=0.
Exam variation: They may give x^3\sin(5/x) or \sqrt{|x|}\cos(1/x). Same idea: trap trig between -1 and 1, then the outside factor goes to 0.
Example 2: Sequence squeeze
Evaluate \lim_{n \to \infty} \frac{\sin(n^2)}{n}.
- Since |\sin(n^2)| \le 1,
\left|\frac{\sin(n^2)}{n}\right| \le \frac{1}{n}.
- And \lim_{n \to \infty} \frac{1}{n} = 0.
Therefore
\lim_{n \to \infty} \frac{\sin(n^2)}{n} = 0.
Exam variation: could be \frac{\cos(\sqrt{n})}{n^{1/3}} or \frac{5\sin(n)}{n^2}.
Example 3: Squeezing a “difference that goes to 0”
Evaluate \lim_{x \to 0} \big(\cos(x) - 1\big)\sin(1/x).
- Use |\sin(1/x)| \le 1:
\left|\big(\cos(x) - 1\big)\sin(1/x)\right| \le |\cos(x)-1|.
- As x \to 0, \cos(x) \to 1, so |\cos(x)-1| \to 0.
So by absolute value squeeze,
\lim_{x \to 0} \big(\cos(x) - 1\big)\sin(1/x)=0.
Exam variation: Replace \cos(x)-1 with any factor that goes to 0 (like x^2, \ln(1+x), e^x-1).
Example 4: The classic trig limit (why squeeze is famous)
Evaluate \lim_{x \to 0} \frac{\sin x}{x}.
A standard squeeze proof uses inequalities (often derived geometrically) that imply for x near 0:
\cos x \le \frac{\sin x}{x} \le 1
(for x>0; a similar argument handles x