Squeeze theorem (sandwich theorem): Question bank — Hinglish deep dive

4.1.5 · D5 · Maths › Calculus I — Limits & Derivatives › Squeeze theorem (sandwich theorem)

Theorem ki shape yaad karo shuru karne se pehle: humein teen cheezein ek saath chahiye — ek lower wall $g$ , ek upper wall $h$ , aur unke beech phansa filling $f$ jisme $g (x) \leq f (x) \leq h (x)$ point $a$ ke paas ho, PLUS dono walls ka ek hi limit $L$ hona zaroori hai. Ek bhi miss hua toh conclusion gir jaata hai. Neeche ke traps inhi ingredients mein se ek par attack karte hain.

True ya false — justify karo

Dono walls ko same limit ki taraf jaana chahiye tab hi koi conclusion nikle.

True — yahi poora engine hai. Agar

g \to 2

aur

h \to 4

hai, toh filling beech mein kahin bhi ghoom sakti hai, isliye kuch bhi force nahi hota.

Agar

g (x) \leq f (x) \leq h (x)

sirf ek hi point

x = a

par hold karta hai, toh theorem phir bhi apply hota hai.

False — limits

a

par ki value ko ignore karti hain aur poore neighbourhood par depend karti hain, isliye inequality

a

ke aaspaas ek interval par hold karni chahiye (sirf

a

par nahi bhi ho sakti).

Squeeze theorem prove kar sakta hai ki limit

L

equal hai, lekin kabhi nahi ki limit exist nahi karti.

True — yeh ek one-directional tool hai: matching walls existence force karti hain. Non-existence dikhane ke liye alag argument chahiye (jaise do paths alag values dein).

Agar dono walls

+ \infty

ki taraf jaati hain, toh squeeze theorem

f

ko bhi

+ \infty

ki taraf jaane par force karta hai.

True — "same limit"

+ \infty

(ya

- \infty

) bhi ho sakta hai; agar

g \to + \infty

aur

g \leq f

hai, toh

f \to + \infty

bhi hoga (yeh one-sided "push" version hai).

Squeeze theorem apply karne ke liye

f

ka ek explicit formula pata hona zaroori hai.

False — isi mein to khoobsurati hai ki

f

ek mystery ho sakta hai. Humein sirf do inequalities aur do wall-limits chahiye;

f

ki andar ki workings kabhi saamne aati hi nahi.

Agar

f

0

par squeeze kiya gaya hai, toh

f

a

ke paas kahin na kahin

0

ki value actually reach karni chahiye.

False — squeezing limit ko control karta hai, values ko nahi. Jaise

x^{2} sin (1/ x)

ka limit

0

hai lekin

0

ke arbitrarily close positive aur negative dono values leta hai.

Theorem sequences ke liye bhi same kaam karta hai jab

x \to a

ki jagah

n \to \infty

hota hai.

True — Limits of sequences version mein bas "sab

n

ke liye jo kaafi bade hain" likha hota hai "sab

x

ke liye jo

a

ke paas hain" ki jagah; trapping logic bilkul same rehti hai.

Agar

g \leq f \leq h

aur sirf lower wall

g

converge karti hai

L

par, toh theorem

f

ka limit deta hai.

False — dono walls

L

par chahiye. Sirf lower wall se ceiling free rehti hai, isliye

f

L

se door jaa sakta hai.

Error dhundho

lim_{x \to 0} x sin (1/ x) = 0

kyunki

- x \leq x sin (1/ x) \leq x

." — kya galat hai?

Jab

x < 0

ho toh claimed bounds ulte ho jaate hain, kyunki

- 1 \leq sin (1/ x) \leq 1

ko ek negative

x

se multiply karne par order flip ho jaata hai. Sahi bounds hain

- ∣ x ∣ \leq x sin (1/ x) \leq ∣ x ∣

"Kyunki

sin (1/ x)

0

par koi limit nahi hai, isliye product

x^{2} sin (1/ x)

ka bhi koi limit nahi hai." — error?

Limit exist kar sakti hai tab bhi jab ek factor ki nahi karti, Bounded times vanishing ke through: bounded (

sin

) times vanishing (

x^{2}

)

0

deta hai. Wild factor tame ho jaata hai shrinking wale se.

g \to 3

aur

h \to 3

lekin

g (x) \leq f (x) \leq h (x)

x = a

par fail karta hai, isliye main conclude nahi kar sakta." — error?

a

par ki exact value limit ke liye irrelevant hai; jab tak inequality

a

ke paas sab doosre

x

ke liye hold kare, squeeze phir bhi

lim f = 3

deta hai.

"Maine walls ko average kiya:

g \to 1

h \to 5

, toh

f \to 3

." — error?

Averaging koi theorem nahi hai. Alag wall-limits ke saath squeeze kuch nahi conclude karta;

f

[1, 5]

mein kisi bhi value par approach kar sakta hai ya bina limit ke oscillate kar sakta hai.

x^{2} sin (1/ x)

ko bound karne ke liye maine

- 1 \leq sin (1/ x) \leq 1

x^{2}

se multiply kiya aur 'safe rehne ke liye' signs flip kar diye." — error?

Koi flip ki zaroorat nahi.

x^{2} \geq 0

hamesha hota hai, aur non-negative quantity se multiply karne par inequality ka direction preserve hota hai; flip karna ek galat bound deta.

cos x \leq \frac{s i n x}{x} \leq 1

sab

x

ke liye hold karta hai, isliye limit

1

hai." — range claim mein error?

Unit-circle derivation assume karta hai

0 < x < \frac{π}{2}

; two-sided limit tab hi milta hai jab yeh note karo ki

\frac{s i n x}{x}

even hai, bound ko negative

x

tak extend karo. Dekho Limits of trigonometric functions.

Why questions

Proof mein sirf ek

δ

ki jagah

δ = min (δ_{1}, δ_{2})

kyun use hota hai?

Har wall apni window deti hai (

δ_{1}

g

ke liye,

δ_{2}

h

ke liye); chhotha wala lena guarantee karta hai ki dono

ε

-bounds us single interval par ek saath hold karein.

Squeeze theorem ki zaroorat hi kyun hai,

x = a

seedha substitute karne ki jagah?

Oscillating functions jaise

sin (1/ x)

ke liye substitute karne ko kuch hai hi nahi — function

a

par undefined ya bahut wildly vary kar raha hota hai, isliye direct evaluation possible nahi; bounding is problem ko sidestep karta hai.

ε

–

δ

argument close hone ke liye dono sides par same

L

kyun hona chahiye?

Chain

L - ε < g \leq f \leq h < L + ε

sirf

f

ko ek number

L

ε

ke andar pin karti hai; agar walls alag centres use karein, toh do

ε

-bands ek single

∣ f - L ∣ < ε

mein overlap nahi karenge.

x

ki jagah

x^{2}

se bound karna

0

ke paas bina case-splitting ke kyun kaam karta hai?

x^{2}

har real

x

ke liye non-negative hai, isliye multiply karne par order ek hi baar mein preserve hota hai;

x

ka sign

0

par badalta hai aur

x > 0

aur

x < 0

ke alag cases force karta.

"Squeeze" naturally Continuity se kyun connect hota hai?

Jab squeeze

lim_{x \to a} f = L

establish kar deta hai, tab

L

f (a)

se compare karna exactly continuity test hai; theorem aksar woh limit supply karta hai jise continuity phir function value se check karti hai.

Squeeze theorem ko "reverse" mein kyun nahi chala sakte taaki filling se walls ko bound karein?

Logic jaati hai known wall-limits se unknown filling-limit ki taraf.

f \to L

jaanana tumhe

g

h

ke baare mein kuch bhi force nahi karta — kai alag walls same

f

ko trap kar sakti hain.

Edge cases

Agar

g (x) = f (x) = h (x)

a

ke paas sab

x

ke liye (walls filling ko har jagah touch karti hain), toh kya theorem phir bhi apply hota hai?

Haan, trivially — equality

\leq

ka ek valid case hai, aur identical functions ka shared limit

L

hai; "squeeze" sirf ek single line hai jo ek point par collapse ho jaati hai.

Kya ho agar walls

L

par converge karti hain lekin

f

a

ke paas infinitely many points par undefined hai?

Theorem phir bhi limit deta hai agar

f

defined aur trapped hai ek punctured neighbourhood par; us set se bahar scattered undefined points theek hain, lekin

f

wahan exist karna chahiye jahan inequality claim ki gayi hai.

Dono walls

L = 0

ki taraf strictly positive values se jaati hain, yaani

0 < g \leq f \leq h

lim f

ka sign kya hoga?

Limit

0

hai, "strictly positive" nahi — functions ke beech strict inequalities limit mein non-strict ho jaati hain, isliye

lim f = 0

hai chahe

f > 0

throughout ho.

Ek sequence ke liye, bound

- \frac{1}{n} \leq \frac{c o s n}{n} \leq \frac{1}{n}

sirf

n \geq 1

ke liye hold karta hai. Kya missing

n = 0

matter karta hai?

Nahi — sequence limits sirf tail ki parwah karti hain (kisi point ke baad sab

n

), isliye finitely many early terms, ya ek undefined

n = 0

lim_{n \to \infty}

ko kabhi affect nahi karte.

Agar ek wall constant

L

hai aur doosri genuinely

L

ki taraf vary karti hai, toh kya yeh allowed hai?

Haan — ek constant function

g (x) = L

ek perfectly valid wall hai jiska

lim g = L

hai; sirf shared limit matter karta hai, yeh nahi ki dono walls move karein.

Recall Jaane se pehle ek one-line self-test

Agar tum bina dekhe bata sako: (1) dono walls, same limit, (2) ek point par nahi balki neighbourhood par inequality, (3) strict inequalities limit mein relax ho jaati hain, (4) sign-flip trap jab kisi aisi cheez se multiply karo jo negative ho sakti hai — tum is page ke har trap ko cover kar chuke ho.

Connections

Squeeze theorem (sandwich theorem) — parent theorem jinhe ye traps probe karte hain.
Limit definition (epsilon-delta) — kyun $δ = min$ aur neighbourhoods matter karti hain.
Bounded times vanishing — "tamed oscillation" traps ke peeche ka pattern.
Limits of trigonometric functions — $\frac{s i n x}{x}$ range subtlety.
Limits of sequences — tail-only edge cases.
Oscillating functions — kyun direct substitution fail hoti hai.
Continuity — jahan ek squeezed limit aage use hoti hai.