Squeeze theorem for sequences — Hinglish notes

4.3.2 · Maths › Calculus III — Sequences & Series

Ye actually kya kehta hai

The bounds only need to hold eventually (for $n \geq N_{0}$ ). Limits start ke kisi bhi finite chunk ko ignore karte hain.
Do outer sequences ko same limit $L$ par converge karna zaroori hai. Agar wo alag limits par jaayein, toh theorem kuch nahi kehta.

YE KYUN sach hai — scratch se derivation

Setup. Koi bhi $ε > 0$ fix karo. Hume $N$ dhundhna hai jaise ki $∣ b_{n} - L ∣ < ε$ ho saare $n \geq N$ ke liye.

Step 1 — $a_{n} \to L$ use karo. Kyun? Ye lower wall hai. Limit ki definition se, ek $N_{1}$ exist karta hai jisme $∣ a_{n} - L ∣ < ε (n \geq N_{1}) ⟹ L - ε < a_{n} .$ Hum sirf left half rakhte hain — lower wall bas yehi de sakti hai.

Step 2 — $c_{n} \to L$ use karo. Kyun? Ye upper wall hai. Ek $N_{2}$ exist karta hai jisme $∣ c_{n} - L ∣ < ε (n \geq N_{2}) ⟹ c_{n} < L + ε .$

Step 3 — Squeeze ke saath combine karo. Kyun? Ab hum inequalities stack karte hain. Maano $N = max (N_{0}, N_{1}, N_{2})$ jisse teeno facts ek saath hold karein. $n \geq N$ ke liye: $L - ε < a_{n} \leq b_{n} \leq c_{n} < L + ε .$ Sabse bahari pieces padhne par: $L - ε < b_{n} < L + ε ⟺ ∣ b_{n} - L ∣ < ε .$

Step 4 — Conclude karo. Kyun? Humne ek arbitrary $ε$ ke liye required $N$ produce kar diya. Ye bilkul $lim_{n \to \infty} b_{n} = L$ ki definition hai. $■$

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Quick self-test (cover the answers)

$b_{n} \to L$ conclude karne ke liye squeeze mein teen conditions bolo. → $a_{n} \leq b_{n} \leq c_{n}$ eventually; $a_{n} \to L$ ; $c_{n} \to L$ (same $L$ ).
One-sided bounding kyun kaafi nahi hai? → Ye sequence ko doosri taraf escape karne se rok nahi sakta.
$\frac{c o s ( n ^{2} )}{n}$ ki kya limit hai, aur walls kya hain? → $0$ , walls $\pm \frac{1}{n}$ .
True/false: bounds $n = 1$ se hold karni chahiye. → False, sirf $n \geq N_{0}$ ke liye.

Squeeze theorem (sequences) — statement

Agar

a_{n} \leq b_{n} \leq c_{n}

saare

n \geq N_{0}

ke liye aur

a_{n} \to L

c_{n} \to L

, tab

b_{n} \to L

Do outer sequences par key requirement

Dono ko SAME limit

L

par converge karna zaroori hai.

Bound "eventually" hi kyun hold karni chahiye?

Limits initial terms ki finite number ko ignore karti hain.

lim_{n \to \infty} \frac{s i n n}{n}

aur use ki gayi walls

0

; walls

- \frac{1}{n} \leq \frac{s i n n}{n} \leq \frac{1}{n}

lim_{n \to \infty} \frac{n !}{n ^{n}}

0

(upar se

\frac{1}{n}

se bounded, neeche se

0

se).

lim_{n \to \infty} n n

1

(logs lo:

\frac{l n n}{n} \to 0

One-sided bounding limit kyun nahi deta?

Sequence unbounded side par unboundedly escape kar sakti hai.

Lower wall

a_{n}

ε

-step kya deta hai

L - ε < a_{n}

for

n \geq N_{1}

Upper wall

c_{n}

ε

-step kya deta hai

c_{n} < L + ε

for

n \geq N_{2}

Teeno facts combine karne ke liye use hone wala index

N = max (N_{0}, N_{1}, N_{2})

Recall Feynman: 12-saal ke bachche ko samjhao

Socho tum ek lift mein apni tall mummy aur tall daddy ke beech dabbe ho, aur dono ek hi room mein jaate hain. Tum alag room mein nahi ja sakte — tum unके beech ho, toh tum usi room mein pahunch jaate ho. Ek sequence jo do dusri sequences ke beech trap hai aur dono number $L$ par jaati hain, use bhi $L$ par hi land karna hoga. Hum ye tab use karte hain jab beech wali cheez (jaise $sin n$ ) itni zyada wiggle kare ki directly follow karna mushkil ho — hum bas neat walls lagaate hain jo dono same jagah calm ho jaayein.

Connections

Limit of a sequence (epsilon-N definition) — squeeze proof pure $ε$ – $N$ hai.
Bounded sequences — walls provide karta hai ( $∣ sin n ∣ \leq 1$ , etc.).
Algebra of limits (sum, product, quotient) — alternative tool jab sequence tame ho.
Monotone convergence theorem — limit ki existence ka ek aur tool.
Squeeze theorem for functions — continuous analogue; same $ε$ logic.
Standard limits ( n-th roots, n!/n^n, ln n / n ) — classic squeeze applications.