Calculus I › Limits › full formula sheet

f(x) ≤ g(x) ≤ h(x),  lim f = lim h = L  ⇒  lim g = L

The squeeze theorem

When direct laws fail on oscillation, trap the function between two that agree.

On this page the sandwich f(x) ≤ g(x) ≤ h(x) only needs to hold near a (a itself may be excluded), and the outer limits must both equal the same L. That sameness is the entire theorem.

Notation in this lesson

f ≤ g ≤ h
the sandwich: g is the filling, f and h the bread
L
the common outer limit — both pieces of bread must tend to the same L
“the vise must close”
if the outer limits don't match, the theorem says nothing

Where it comes from

The problem: the limit laws are helpless against oscillation. lim(x→0) sin(1/x) doesn’t exist — the function hits 1 and −1 infinitely often near 0 — so no splitting law can touch anything containing it. The squeeze theorem is the rescue: if you can trap the wild function between two tame ones heading to the same place, the wild one has nowhere else to go.

Before reading on: sin(1/x) oscillates forever near 0 — so how can x·sin(1/x) still tend to 0? What squeezes it?
f(x) ≤ g(x) ≤ h(x) near a,   lim(x→a) f = lim(x→a) h = L  ⇒  lim(x→a) g = Lthe outer limits must match — that sameness is the squeezeSay it: if f of x is below g of x is below h of x near a, and both outer limits equal L, then the limit of g of x is L

The naive “squeeze” that isn’t:

−1 ≤ sin(1/x) ≤ 1
⇒
“lim(x→0) sin(1/x) = 0”?
Invalid. The outer “functions” tend to −1 and 1 — different limits. No squeeze happens; indeed the limit DNE. A sandwich with the bread going two different directions squeezes nothing.
−|x| ≤ x·sin(1/x) ≤ |x|
⇒
lim(x→0) x·sin(1/x) = 0
Valid. Both outer functions tend to 0 — the same L. The oscillation is crushed between them. This is the theorem working as intended.

The intuition is a vise: g is clamped between f and h, and the vise closes to a single point L. Whatever g does inside — wiggle, jump, oscillate — it gets carried to L anyway. The proof is one epsilon-delta vise-turn.

Derivation

Assume f(x) ≤ g(x) ≤ h(x) near a (possibly excluding a), with lim(x→a) f = lim(x→a) h = L. We prove lim(x→a) g = L. This proof is short enough to show in full.

|f − L| < ε
⇒
f > L − ε
Step 1 — keep the lower half. From lim f = L, some δ1 gives |f−L| < ε. We only need the lower bound: f > L − ε. (The upper half is useless — f is the floor of the sandwich.)
|h − L| < ε
⇒
h < L + ε
Step 2 — keep the upper half. From lim h = L, some δ2 gives |h−L| < ε. We only need h < L + ε — the ceiling of the sandwich.
δ = min(δ1, δ2, δ3)
⇒
L − ε < f ≤ g ≤ h < L + ε
Step 3 — close the vise. δ3 is where the sandwich f ≤ g ≤ h holds. Inside the min-delta neighborhood, the chain gives L−ε < g < L+ε, i.e. |g−L| < ε. ∎

Full proof shown. Notice the division of labor: f supplies the floor, h supplies the ceiling, and the sandwich inequality chains them through g. If the outer limits differed (L1 ≠ L2), the chain would read L1−ε < g < L2+ε — a gap, not a vise — which is exactly why the bad “squeeze” of sin(1/x) proves nothing.

How to use it

The procedure, every time:

  1. Spot the pattern: bounded oscillation (sin/cos of something wild) multiplied by something → 0 — or any function visibly trapped between two friendly ones.
  2. Build the sandwich. The workhorse: −1 ≤ sin(□) ≤ 1 and −1 ≤ cos(□) ≤ 1, multiplied through by the vanishing factor (use absolute values to keep inequalities facing the right way).
  3. Check the outer limits are EQUAL. This is the #1 exam error: a “sandwich” with bread heading to different places squeezes nothing.
  4. Conclude: the trapped limit equals the common outer limit.

Which side does the inequality face?

Multiplying −1 ≤ sin(1/x) ≤ 1 by x is dangerous when x < 0 (inequalities flip). The safe move: multiply by |x| instead, giving −|x| ≤ x·sin(1/x) ≤ |x|. Since |x| → 0, the squeeze still closes.

Common mistake: bounding g between f → 2 and h → 3 and concluding lim g = 2.5 (or anything). Unequal outer limits mean no conclusion — the theorem’s hypothesis fails.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the classic: lim(x→0) x·sin(1/x)

  1. Spot the pattern: bounded oscillation sin(1/x) × vanishing x. Product law is illegal (lim sin(1/x) DNE).
  2. Build the sandwich: −1 ≤ sin(1/x) ≤ 1; multiply by |x|: −|x| ≤ x·sin(1/x) ≤ |x|. (Why |x|? For x < 0 we have x·sin(1/x) ≥ x·1 = x = −|x| — the absolute value keeps both bounds valid.)
  3. Outer limits match: −|x| → 0 and |x| → 0.
  4. Conclude: the limit is 0.
Common mistake: multiplying by x instead of |x| and writing x ≤ x·sin(1/x) ≤ −x — backwards for x < 0. The |x| version is bulletproof.
Your turn: limx→0 x·cos(1/x) = ?

Answer: 0.

Bounded × vanishing: −|x| ≤ x·cos(1/x) ≤ |x|, and both outer limits are 0.

Example 2 — stronger vanishing: lim(x→0) x²·cos(1/x²)

  1. Spot the pattern: bounded cos(1/x²) × x² → 0.
  2. Sandwich: −x² ≤ x²·cos(1/x²) ≤ x² (x² ≥ 0, so no flipping worries).
  3. Outer limits: both → 0. Conclude: 0.
Common mistake: thinking the wilder oscillation (1/x²) changes anything. It doesn’t — bounded is bounded, no matter how fast it wiggles.
Your turn: limx→0 x³·sin(1/x) = ?

Answer: 0.

−|x|³ ≤ x³·sin(1/x) ≤ |x|³ (the absolute value avoids sign flips for x < 0); outer limits → 0.

Example 3 — at infinity: lim(x→∞) (sin x)/x

  1. Spot the pattern: bounded sin x over growing x — the quotient law is illegal (denominator → infinity).
  2. Sandwich: −1 ≤ sin x ≤ 1; divide by x > 0: −1/x ≤ (sin x)/x ≤ 1/x.
  3. Outer limits: ±1/x → 0. Conclude: 0.
Common mistake: “infinity over infinity” hand-waving. The squeeze makes it rigorous in one line.
Your turn: limx→∞ (cos x)/x = ?

Answer: 0.

−1/x ≤ (cos x)/x ≤ 1/x (x > 0), and ±1/x → 0.

Before reading on: −1 ≤ sin(1/x) ≤ 1 is a valid sandwich — so does it prove the limit is 0? Check the bread before you answer.

Example 4 — the trap: “−1 ≤ sin(1/x) ≤ 1 proves lim(x→0) sin(1/x) = 0”

  1. Check the outer limits: the “bread” here is the constant functions −1 and 1, whose limits are −1 and 1 — not equal.
  2. Verdict: the hypothesis fails; the argument proves nothing. (And indeed the limit DNE: sin(1/x) hits ±1 arbitrarily close to 0.)
  3. The lesson: a valid squeeze needs the vise to close to a point. −1 and 1 leave a gap the function can dance in forever.
Common mistake: this exact fallacy, offered confidently. When you grade your own work, always ask: “do my outer limits match?”
Your turn: Does 0 ≤ |sin(1/x)| ≤ 1 prove limx→0 |sin(1/x)| = 0?

Answer: no.

The bread's limits are 0 and 1 — they don't match, so the hypothesis fails and the argument proves nothing. (Indeed the limit DNE.)

Memorization tips

  • Bounded × vanishing = 0: the pattern behind every squeeze-theorem limit. See oscillation times something → 0 — think squeeze.
  • The outer limits must match: chant it. A sandwich whose bread goes two directions squeezes nothing.
  • Multiply by |x|, not x: absolute values keep inequalities facing the right way when the vanishing factor changes sign.
  • The sandwich need only hold near a: the point a itself may be excluded — limits don’t care about the point.
  • −1 ≤ sin ≤ 1 is the workhorse: almost every squeeze starts here. cos works identically.
  • Squeeze also works at infinity: (sin x)/x → 0 uses the same vise, with 1/x as the closing bread.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and this law is yours.

← Back to the Calculus I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the squeeze theorem?

If f(x) ≤ g(x) ≤ h(x) near a (a itself may be excluded) and lim(x→a) f = lim(x→a) h = L — the SAME L — then lim(x→a) g = L. The wild middle function is trapped.

Why must the outer limits be equal?

The proof chains L−ε < f ≤ g ≤ h < L+ε. With different outer limits L₁ ≠ L₂ you'd get a gap L₁−ε < g < L₂+ε instead of a vise — no conclusion follows. That's why −1 ≤ sin(1/x) ≤ 1 proves nothing.

When should I reach for the squeeze theorem?

When a bounded oscillation (sin/cos of something wild) is multiplied by something → 0, or whenever a function is visibly trapped between two friendly ones with a common limit. Pattern: bounded × vanishing = 0.

Why multiply by |x| instead of x in the sandwich?

Multiplying an inequality by a negative x flips its direction. Using |x| keeps both bounds valid on both sides of 0, and |x| → 0 still closes the vise.

Does the sandwich need to hold AT a?

No — only near a. Limits ignore the point itself, so f, g, h may even be undefined at a.

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