Visual walkthrough — Squeeze theorem (sandwich theorem)

We build every symbol from nothing. If a piece of notation shows up, we first say what it means in plain words and show you where it lives on a picture.

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Step 0 — The three characters and the one symbol we lean on

Before any proof, meet the cast.

• $f(x)$ — the filling. This is the function whose limit we secretly want but can't compute directly.
• $g(x)$ — the lower bread. It sits below $f$.
• $h(x)$ — the upper bread. It sits above $f$.
• $a$ — the input value we are sneaking up on (we look near $a$, never at it).
• $L$ — the height (a number on the vertical axis) both breads are heading for.

The one symbol we must build carefully is $|x-a|$. Read it out loud: "how far $x$ is from $a$". The two vertical bars mean "distance", so negatives are thrown away — $|3-5|=2$, same as $|5-3|=2$. Whenever you see $|x-a|<\delta$ (delta, a small positive number), translate it as: "$x$ is within a distance $\delta$ of $a$." Picture a little band of width $2\delta$ centred on $a$ on the horizontal axis.

We also write $0<|x-a|$ — that little $0<$ means "$x$ is not exactly $a$". We stand next to $a$, never on it.

Here $\underset{x\to a}{\lim }g(x)=L$ is shorthand for "as $x$ slides toward $a$, the height $g(x)$ slides toward $L$." Look at figure s01: the green curve ($g$) and orange curve ($h$) both pinch toward the same dot at height $L$, and the magenta curve ($f$) is squished between them.

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Step 1 — Turn "the limit is $L$" into a box we can trap things in

WHAT. We restate what ${\lim }_{x\to a}g(x)=L$ actually promises, using the Limit definition (epsilon-delta).

WHY. "Heads toward $L$" is a feeling. To prove anything we need a testable statement. The epsilon–delta definition converts the feeling into a challenge-and-response game we can win.

The definition says: pick any tolerance $\epsilon >0$ (epsilon — a tiny target height, how close to $L$ you demand). Then there exists a window half-width ${\delta }_1>0$ so that

$0<|x-a|<{\delta }_1⟹\underset{\text{floor}}{\underbrace{L-\epsilon }}<\underset{\text{lower bread}}{\underbrace{g(x)}}<\underset{\text{ceiling}}{\underbrace{L+\epsilon }}.$

Term by term: $L-\epsilon$ and $L+\epsilon$ are a floor and ceiling a distance $\epsilon$ on either side of $L$ — think of a horizontal strip of height $2\epsilon$. The arrow $⟹$ means "forces". So: whenever $x$ is inside the window of half-width ${\delta }_1$, the bread $g(x)$ is trapped inside the strip.

PICTURE. In s02 the horizontal band $(L-\epsilon ,L+\epsilon )$ is the pink strip. The vertical band of half-width ${\delta }_1$ is where the green curve $g$ is guaranteed to live inside the pink strip.

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Step 2 — Do the exact same thing to the upper bread

WHAT. Repeat Step 1 for $h$.

WHY. Both breads must be pinned inside the strip; we've only pinned one so far.

Using ${\lim }_{x\to a}h(x)=L$ with the same $\epsilon$: there is a window half-width ${\delta }_2>0$ with

$0<|x-a|<{\delta }_2⟹L-\epsilon <\underset{\text{upper bread}}{\underbrace{h(x)}}<L+\epsilon .$

Same floor $L-\epsilon$, same ceiling $L+\epsilon$, same pink strip — only the actor changed to $h$, and its window half-width is a possibly-different ${\delta }_2$.

PICTURE. s03 shows the orange curve $h$ dipping into the pink strip once $x$ is inside its own window ${\delta }_2$. Notice ${\delta }_1$ and ${\delta }_2$ are generally different sizes — that's the whole point of the next step.

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Step 3 — Choose the smaller window so BOTH promises hold at once

WHAT. Define $\delta =\min ({\delta }_1,{\delta }_2)$ — "$\min$" means pick the smaller of the two.

WHY. $g$ behaves inside ${\delta }_1$; $h$ behaves inside ${\delta }_2$. Inside the smaller of the two windows, we are inside both, so both breads sit in the strip simultaneously. If we picked the larger one, we might step outside the other bread's guarantee.

$\delta =\min ({\delta }_1,{\delta }_2)⟹0<|x-a|<\delta \text{ makes both Step 1 and Step 2 true.}$

PICTURE. s04 overlays both vertical windows. The overlap — the narrower band $\delta$ — is shaded solid. Inside that solid band, green and orange are both caught in the pink strip.

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Step 4 — Chain the inequalities: the trap snaps shut

WHAT. Combine the sandwich hypothesis with the two strip-facts.

WHY. We now have three simultaneous truths inside the solid window: $g$ above the floor, $f$ between the breads, $h$ below the ceiling. Line them up left to right.

For $0<|x-a|<\delta$:

$\underset{\text{from }g\text{ (Step 1)}}{\underbrace{L-\epsilon }}<\underset{\text{lower bread}}{\underbrace{g(x)}}\le \underset{\text{filling}}{\underbrace{f(x)}}\le \underset{\text{upper bread}}{\underbrace{h(x)}}<\underset{\text{from }h\text{ (Step 2)}}{\underbrace{L+\epsilon }}.$

Read the chain: the floor is below $g$; $g$ is below $f$; $f$ is below $h$; $h$ is below the ceiling. Transitivity (if $a<b$ and $b<c$ then $a<c$) lets us skip the middle men and keep just the outer walls squeezing $f$.

PICTURE. s05 zooms into the solid window: every one of the three curves is inside the pink strip, and the magenta $f$ is visibly pinned between green and orange, all three inside floor and ceiling.

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Step 5 — Read off the conclusion

WHAT. Drop the inner curves and keep only what surrounds $f$.

WHY. The definition of ${\lim }_{x\to a}f(x)=L$ is literally the statement "$f$ is inside the pink strip whenever $x$ is inside a window." We just proved exactly that.

$L-\epsilon <f(x)<L+\epsilon ⟺\underset{\text{"}f\text{ is within }\epsilon \text{ of }L\text{"}}{\underbrace{|f(x)-L|<\epsilon }}.$

The double arrow $⟺$ means "these say the same thing". Subtracting $L$ from all three parts of $L-\epsilon <f(x)<L+\epsilon$ gives $-\epsilon <f(x)-L<\epsilon$, which is precisely "distance from $f(x)$ to $L$ is under $\epsilon$".

Since our $\epsilon$ was arbitrary (we handled any tolerance you could name, by producing a matching $\delta$), we have satisfied the definition:

${\lim }_{x\to a}f(x)=L.■$

PICTURE. s06 shows the payoff: shrink $\epsilon$ (thinner pink strip) and the whole argument produces a new, thinner $\delta$ window — $f$ follows the breads into an ever-tighter pinch at the dot $(a,L)$.

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Step 6 — The degenerate cases (never leave the reader stranded)

Real functions misbehave. Here is every corner case, each with its own picture in s07.

(a) $f$ equals a wall. The inequalities are $\le$, not strict $<$. So $f$ may touch $g$ or $h$, even coincide with one on a whole stretch. Nothing breaks — a filling glued to the bread still arrives at $L$. (Left panel of s07.)

(b) $f$ is undefined at $a$. Classic case: $f(x)=x^2\sin (1/x)$ has no value at $x=0$. Irrelevant! The $0<|x-a|$ in every step means we never evaluate at $a$. The limit exists even where the function has a hole. (Middle panel.)

(c) The breads only pinch on one side. If $g$ and $h$ meet at $L$ only from the right, you get a one-sided limit $\underset{x\to a^{+}}{\lim }$. The identical argument runs with a half-window. This is how the parent's $\frac{\sin x}{x}$ proof works before evenness glues both sides.

(d) $g\to 2$ but $h\to 4$ — different heights. Now there is no common $L$, no single pink strip both breads enter. The trap has a gap; $f$ can wander anywhere between $2$ and $4$. The theorem gives nothing. This is the number-one misuse — see the parent's "same bread, same sandwich" mnemonic. (Right panel, showing the gap in grey.)

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The one-picture summary

Everything above, compressed: pick any $\epsilon$, get a $\delta$, watch $f$ get dragged to $L$.

Recall Feynman retelling — the walkthrough in plain words

Someone dares you: "Keep the middle rope within one centimetre of the ceiling hook." You reply, "Easy — I'll tell you how close to stand to the wall." You know two ropes, one above the middle rope and one below it, and both of those ropes you can already pin near the hook. So you find how close to the wall each of those two ropes needs its holder to stand, take the stricter of the two distances, and stand there. Now both outer ropes are within one centimetre of the hook — and the middle rope, tied between them, has no slack to escape. It's within one centimetre too. Whatever tolerance the dare demanded (one centimetre, one millimetre, anything), you answer with a matching "stand this close." Since you can always answer, the middle rope truly reaches the hook. You never once had to touch or measure the middle rope itself — the two outer ropes did all the work. That is the squeeze theorem.

Recall Quick self-check

Why do we take $\delta =\min ({\delta }_1,{\delta }_2)$ and not the max? ::: We need BOTH bread-guarantees active at once; only the smaller window sits inside both. Where in the proof do we use that $f$ is between the breads? ::: Step 4, the middle links $g(x)\le f(x)\le h(x)$ of the chain. Why is a hole in $f$ at $a$ harmless? ::: Every step assumes $0<|x-a|$, so $a$ itself is never tested. What fails if $g\to 2$ and $h\to 4$? ::: No common $L$, so no single strip traps $f$ — the theorem concludes nothing.

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Connections

• Limit definition (epsilon-delta) — the challenge–response game every step is built on.
• Limits of trigonometric functions — the pinch delivers ${\lim }_{x\to 0}\frac{\sin x}{x}=1$ (one-sided-then-glued, Step 6c).
• Bounded times vanishing — the "$f$ undefined but bounded" case (Step 6b) in disguise.
• Oscillating functions — the reason we bound instead of substitute.
• Continuity — once the pinch gives a limit, compare it to the function's value.
• Limits of sequences — swap $x\to a$ for $n\to \infty$; every figure still applies.