Squeeze theorem (sandwich theorem)

WHAT is it?

The "bread" of the sandwich is $g$ (lower) and $h$ (upper). The "filling" $f$ is squeezed in the middle.

Why we need it: Some functions are too wiggly to compute their limit directly — like $x^2\sin(1/x)$ near $0$ , where $\sin(1/x)$ oscillates infinitely fast. We can't plug in, we can't simplify nicely. But we can bound it.

WHY is it true? (Derivation from first principles)

We use the $\varepsilon$ – $\delta$ definition of a limit. Recall: $\lim_{x\to a}f(x)=L$ means for every $\varepsilon>0$ there's a $\delta>0$ such that $0<|x-a|<\delta \implies |f(x)-L|<\varepsilon$ .

Step 1 — Translate the hypotheses. Since $\lim_{x\to a}g(x)=L$ : given $\varepsilon>0$ , there is $\delta_1>0$ with $0<|x-a|<\delta_1 \implies L-\varepsilon < g(x) < L+\varepsilon.$ Why this step? The definition of limit lets us force $g$ within $\varepsilon$ of $L$ .

Step 2 — Same for the top. Since $\lim_{x\to a}h(x)=L$ : there is $\delta_2>0$ with $0<|x-a|<\delta_2 \implies L-\varepsilon < h(x) < L+\varepsilon.$

Step 3 — Take the smaller window. Let $\delta=\min(\delta_1,\delta_2)$ . Then both bounds hold simultaneously when $0<|x-a|<\delta$ . Why this step? We need a single $\delta$ where everything is true at once.

Step 4 — Chain the inequalities. For $0<|x-a|<\delta$ : $L-\varepsilon < g(x) \le f(x) \le h(x) < L+\varepsilon.$ Why this step? The lower bound $L-\varepsilon$ comes from $g$ ; the upper bound $L+\varepsilon$ comes from $h$ . The middle is $f$ by hypothesis. So $f$ is trapped.

Step 5 — Read off the conclusion. $L-\varepsilon < f(x) < L+\varepsilon \iff |f(x)-L|<\varepsilon.$ That is exactly the definition of $\lim_{x\to a}f(x)=L$ . $\blacksquare$

HOW to use it (worked examples)

Worked example Example 1 — The classic:

$\displaystyle\lim_{x\to 0} x^2\sin\!\frac{1}{x}$ Goal: evaluate even though $\sin(1/x)$ oscillates wildly near $0$ .

Step 1. Bound the bounded part: for all $x\ne 0$ , $-1 \le \sin\frac1x \le 1.$ Why? Sine is always between $-1$ and $1$ , no matter how fast its input changes.

Step 2. Multiply by $x^2\ge 0$ (inequalities keep direction): $-x^2 \le x^2\sin\frac1x \le x^2.$ Why this step? Multiplying by a non-negative quantity preserves the inequality. (If $x^2$ could be negative we'd have to flip — but $x^2\ge0$ always.)

Step 3. Squeeze: $\lim_{x\to0}(-x^2)=0$ and $\lim_{x\to0}x^2=0$ . Both walls go to $0$ , so $\lim_{x\to 0} x^2\sin\frac1x = ==0==.$

Worked example Example 2 — Trig limit

$\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1$ Step 1. From a unit-circle area argument, for $0<x<\tfrac{\pi}{2}$ : $\cos x \le \frac{\sin x}{x} \le 1.$ Why? Area of triangle $\le$ area of sector $\le$ area of larger triangle gives $\sin x \le x \le \tan x$ ; divide by $\sin x$ and reciprocate.

Step 2. As $x\to 0^+$ , $\cos x \to 1$ and the constant $1\to 1$ . Why this step? Both walls converge to $1$ .

Step 3. Squeeze ⟹ $\lim_{x\to0^+}\frac{\sin x}{x}=1$ . By evenness of $\frac{\sin x}{x}$ the two-sided limit is $==1==$ .

Common mistakes (steel-manned)

Recall Feynman: explain it to a 12-year-old

Imagine two elevators, a slow one above you and a fast one below you, and you're stuck on a platform between them. If both elevators are heading to the 3rd floor, you must end up on the 3rd floor too — you're squished between them and can't go anywhere else. The squeeze theorem says: if the function above and the function below both arrive at the same number, the function in the middle is forced to arrive there as well. We don't even need to watch the middle function — the two outside ones drag it along.

Active-recall flashcards

#flashcards/maths

State the squeeze theorem precisely.

g(x)\le f(x)\le h(x)

near

a

and

\lim_{x\to a}g=\lim_{x\to a}h=L

, then

\lim_{x\to a}f=L

What is the ONE condition people most often forget?

The two bounding functions must tend to the same limit

L

Why must the inequality hold on a neighborhood, not just at

a

Limits depend on behavior near

a

, not the value at

a

itself.

Evaluate

\lim_{x\to0}x^2\sin(1/x)

and name the bounds used.

=0

; bounds

-x^2\le x^2\sin(1/x)\le x^2

When you multiply an inequality near

0

x

, what trap appears?

x

may be negative, flipping the inequality; use

x^2

|x|

instead.

Which

\varepsilon

–

\delta

trick lets one

\delta

serve both bounds?

Take

\delta=\min(\delta_1,\delta_2)

so both bounds hold at once.

Does the squeeze theorem work for sequences?

Yes — replace

x\to a

with

n\to\infty

; e.g.

\cos n/n\to0

Why can't we just plug

x=0

into

x^2\sin(1/x)

\sin(1/x)

is undefined/oscillating at

0

; so we bound instead of substitute.

Connections

Limit definition (epsilon-delta) — the engine that powers the proof.
Limits of trigonometric functions — squeeze gives $\lim_{x\to0}\frac{\sin x}{x}=1$ .
Bounded times vanishing — general pattern: bounded $\times$ (something $\to 0$ ) $=0$ .
Continuity — once a limit is found, it tests continuity.
Limits of sequences — same theorem, $n\to\infty$ version.
Oscillating functions — squeeze tames infinite oscillation.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Squeeze theorem ka idea bilkul simple hai: agar koi function $f(x)$ do aur functions ke beech "phasa" (sandwich) hua hai — neeche $g(x)$ aur upar $h(x)$ — aur dono walls same value $L$ tak ja rahe hain, toh beech wala $f(x)$ bhi majboori mein $L$ tak hi jayega. Function ko bhaagne ki jagah hi nahi milti. Isiliye ise sandwich theorem bhi kehte hain — do bread slices same height pe milengi toh filling bhi wahi height pe hogi.

Yeh kab use karte hain? Jab function itna wiggly ya oscillating hai ki direct limit nikalna mushkil hai. Classic example: $x^2\sin(1/x)$ , jahan $\sin(1/x)$ zero ke paas infinitely fast oscillate karta hai. Hum direct plug nahi kar sakte. Par hum jaante hain $\sin$ hamesha $-1$ aur $1$ ke beech rehta hai, toh $-x^2 \le x^2\sin(1/x) \le x^2$ . Dono walls $0$ pe ja rahe hain, isliye answer bhi $0$ .

Ek important dhyaan rakhne ki baat: dono walls ka limit same hona chahiye. Agar ek $2$ pe jaa raha hai aur dusra $4$ pe, toh kuch conclude nahi hota — average mat lo, woh galat hai. Aur ek bada trap: jab inequality ko $x$ se multiply karte ho zero ke paas, $x$ negative bhi ho sakta hai jo inequality ko flip kar deta hai. Isliye $x^2$ ya $|x|$ se multiply karna safe hai kyunki woh hamesha non-negative hote hain. Yeh chhoti baatein exam mein marks bachati hain.

WHAT is it?

WHY is it true? (Derivation from first principles)

HOW to use it (worked examples)

Common mistakes (steel-manned)

Active-recall flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus I — Limits & Derivatives

Connections