Squeeze theorem for sequences: Worked examples — Hinglish deep dive

4.3.2 · D3 · Maths › Calculus III — Sequences & Series › Squeeze theorem for sequences

Definition Ground rules jo is poori page pe maan ke chalte hain

Index $n$ natural numbers pe chalta hai: $n \in N$ , yaani $n = 1, 2, 3, \dots$ Koi $n = 0$ nahi hai aur koi fractional $n$ nahi; neeche har sequence values ki ek list hai, har aise $n$ ke liye ek value.
Arrow " $\to \infty$ " ka matlab hai =="har had se baar badhta jaata hai jaise $n$ badhta hai"== — formally, koi bhi diwar $M$ batao, woh quantity eventually $M$ se badi ho jaati hai. Jab hum " $n + 4 \to \infty$ " likhte hain toh matlab hai denominator $n$ badhne ke saath infinity ki taraf chala jaata hai.
Arrow " $x_{n} \to L$ " (jahan $L$ ek finite number hai) ka matlab hai " $x_{n}$ , $L$ pe aa ke settle ho jaata hai" $\varepsilon$–$N$ sense mein, jo prerequisite note mein hai.
Kyunki $n \geq 1$ hai, seedhe saadhe facts jaise $n + 4 > 0$ , $n > 0$ , $n^{2} + 1 > 0$ automatically hold hote hain — hum inpar constantly depend karte hain taaki positive numbers se divide kar sakein bina inequalities palte.

Scenario matrix

Har squeeze problem inhi cells mein se ek mein aata hai. Pehle, table mein jo do words loosely use kiye hain, unhe precise kar lete hain:

Neeche jo examples hain woh cell ke hisaab se label kiye gaye hain taaki tum dekh sako ki poora map cover ho raha hai.

Cell	Ise yeh cell kya banata hai	Wall strategy	Example
A. Bounded oscillator ÷ grower	ek bounded oscillator ( $sin$ , $cos$ , $(- 1)^{n}$ ) ko ek grower ( $x_{n} \to \infty$ ) se divide kiya	$\pm$ bound ko grower se divide karo	Ex 1
B. Sign-changing middle	middle khud negative ho sakta hai	two-sided wall use karni padegi, kabhi sirf ek nahi	Ex 2
C. Product / factorial vs power	terms multiply hue hain, ek doosre ko tame karta hai	crude upper bound $\leq \frac{c}{n}$ , lower $0$	Ex 3
D. Root of a polynomial	$n$ th roots, $a^{1/ n}$	jaane-maane $n$ th roots se sandwich karo ( $n n \to 1$ )	Ex 4
E. Sum of many small bits	ek sum jisme term count $n$ ke saath badhta hai	har term ko same walls se bound karo	Ex 5
F. Degenerate / equal walls	walls middle pe collapse ho jaati hain, ya middle khud ek wall hai	direct read-off, "walls to different limits" ka dhyan rakho	Ex 6
G. Word / real-world	error term, physical bound	noise ko bounded maano, squeeze karo	Ex 7
H. Exam twist	walls jo alag-alag limits pe jaati lagti hain par jaati nahi	pehle walls simplify karo, phir squeeze karo	Ex 8

Prerequisites jo chahiye: Limit of a sequence (epsilon-N definition), Bounded sequences, Standard limits ( n-th roots, n!/n^n, ln n / n ), aur kabhi kabhi Algebra of limits (sum, product, quotient).

Worked examples

Cell A — Bounded oscillator over a grower

Worked example Ex 1 —

$b_{n} = \frac{3 cos ( 5 n ) + 2}{n + 4}$ Forecast: upar wala wiggle karta hai par ek fixed band mein rehta hai; neeche wala badhta hai. Padhne se pehle limit guess karo.

Wiggle ko bound karo. Yeh step kyun? $cos (5 n)$ ek oscillator hai: $- 1 \leq cos (5 n) \leq 1$ . $3$ se multiply karo aur $2$ add karo: $- 1 \leq 3 cos (5 n) + 2 \leq 5.$
$n + 4 > 0$ se divide karo. Kyun? Kyunki $n \geq 1$ hai, $n + 4 \geq 5 > 0$ hai, isliye divide karne se har inequality usi direction mein rehti hai (flip nahi hoti): $\frac{- 1}{n + 4} \leq b_{n} \leq \frac{5}{n + 4} .$
Walls ke limits lo. Kyun? Dono walls constant-over-grower hain, isliye dono $\to 0$ (denominator $n + 4 \to \infty$ ): $\frac{- 1}{n + 4} \to 0, \frac{5}{n + 4} \to 0.$
Squeeze karo. Dono walls ka same limit $\Rightarrow b_{n} \to 0$ .

Verify: $n = 100$ pe, upar wala $[- 1, 5]$ mein range karta hai, neeche $= 104$ , toh $b_{n} \in [- \frac{1}{104}, \frac{5}{104}] \approx [- 0.0096, 0.048]$ — pehle se $0$ ke paas. Limit $= 0$ .

Cell B — Sign-changing middle (ek wall KAAFI nahi hai)

Worked example Ex 2 —

$b_{n} = \frac{( - 1 ) ^{n} n}{n + 1}$ Forecast: sign har step pe flip hota hai. Kya yeh settle hoga, aur kahan?

Pehle absolute value. Yeh step kyun? Jab middle sign change karta hai, toh sirf upper bound use karke usse trap nahi kar sakte (parent ki mistake #2). Size ko bound karo ( $n + 1 > 0$ use karke kyunki $n \geq 1$ ): $∣ b_{n} ∣ = \frac{n}{n + 1} .$
Size ko sandwich karo. Kyun? $∣ b_{n} ∣ \geq 0$ hamesha, aur $\frac{n}{n + 1} \leq \frac{n}{n} = \frac{1}{n}$ : $0 \leq ∣ b_{n} ∣ \leq \frac{1}{n} .$
Absolute value ko squeeze karo. Kyun? $\frac{1}{n} \to 0$ , toh $∣ b_{n} ∣ \to 0$ . Aur $∣ b_{n} ∣ \to 0 ⟺ b_{n} \to 0$ — ek sequence jiska $0$ se distance khatam ho jaata hai woh $0$ pe converge karti hai chahe sign kuch bhi ho.

Verify: $n = 100$ : $\frac{10}{101} \approx 0.099$ , sign alternating; $n = 1 0^{4}$ : $\frac{100}{10001} \approx 0.01$ . $0$ ki taraf shrink ho raha hai.

Cell C — Factorial / product vs power

Worked example Ex 3 —

$b_{n} = \frac{2 ^{n}}{n !}$ Forecast: exponential upar, factorial neeche. Kaun jeetega?

Product likh do. Yeh step kyun? Factor by factor compare karo ( $n \geq 1$ ke liye valid): $\frac{2 ^{n}}{n !} = \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \dots \frac{2}{n} .$
Pehle do aur last factor rakho, beech wale block ko $1$ se bound karo. Kyun? Har $k$ ke liye jahan $3 \leq k \leq n - 1$ hai, $\frac{2}{k} \leq \frac{2}{3} < 1$ hai, toh unhe sab multiply karne se cheez $\leq 1$ rehti hai: $\prod_{k = 3}^{n - 1} \frac{2}{k} \leq 1.$ Ab product ko split karo $= 2 \frac{2}{1} \cdot \frac{2}{2} \cdot \leq 1 k = 3 \prod n - 1 \frac{2}{k} \cdot last \frac{2}{n}$ ke roop mein, jisse milta hai $0 < \frac{2 ^{n}}{n !} \leq 2 \cdot 1 \cdot \frac{2}{n} = \frac{4}{n} (n \geq 3) .$
Squeeze karo. Kyun? Lower wall $0$ , upper wall $\frac{4}{n} \to 0$ , same limit.

Verify: $n = 10$ : $\frac{2 ^{10}}{10 !} = \frac{1024}{3628800} \approx 2.8 \times 1 0^{- 4}$ , aur $\frac{4}{10} = 0.4$ — bound hold karta hai, value bahut choti. Limit $= 0$ .

Cell D — Root of a polynomial (jaane-maane $n$ th roots se sandwich)

Worked example Ex 4 —

$b_{n} = n 5 n^{2} + 3$ Forecast: polynomially growing cheez ka $n$ th root. Kya root isse $1$ tak daba deta hai?

Andar ki cheez ko crudely trap karo. Yeh step kyun? $n \geq 1$ ke liye, $5 n^{2} + 3$ ka value $n$ aur $8 n^{2}$ ke beech mein hai ( $n \leq 5 n^{2} + 3$ aur $5 n^{2} + 3 \leq 8 n^{2}$ kyunki $n \geq 1$ ). Hum chahte hain walls jo aasaan $n$ th roots hon — yaani aisi quantities jinki limit hum standard limits se pehle se jaante hain: $n n \leq n 5 n^{2} + 3 \leq n 8 n^{2} = 8^{1/ n} \cdot (n n)^{2} .$
Standard limits yaad karo. Kyun? $n n \to 1$ aur $8^{1/ n} \to 1$ (koi bhi fixed $a > 0$ deta hai $a^{1/ n} \to 1$ ).
Squeeze karo. Kyun? Lower wall $n n \to 1$ ; upper wall $8^{1/ n} (n n)^{2} \to 1 \cdot 1^{2} = 1$ . Dono $1$ pe milte hain.

Neeche wali figure dekho: green lower wall $n n$ aur orange upper wall $n 8 n^{2}$ pehle bahut door hain (chhote $n$ ke liye bade number ka root bada hota hai) par dono $1$ ki red line par aa jaate hain. Blue middle $n 5 n^{2} + 3$ ke paas koi escape nahi — use bhi unke saath $1$ ki taraf khainch liya jaata hai. Yahi funnel squeeze hai.

Verify: $n = 10$ : $10503 \approx 1.86$ ; $n = 1000$ : $1000 5 \cdot 1 0^{6} + 3 \approx 1.0155$ . Dheere dheere $1$ ki taraf.

Cell E — Sum jisme term count badhta hai

Cell F — Degenerate walls (equal-limit check tumhe bachata hai)

Worked example Ex 6 — ek warning:

$b_{n} = \frac{n + ( - 1 ) ^{n} n}{2 n}$ Forecast: squeezable lagta hai — par dhyan rakho kaun si walls choose karte ho.

Naive walls try karo. Yeh step kyun? $- 1 \leq (- 1)^{n} \leq 1$ deta hai $n - n \leq n + (- 1)^{n} n \leq n + n$ , yaani ( $2 n > 0$ se divide karke kyunki $n \geq 1$ ) $\frac{0}{2 n} = 0 \leq b_{n} \leq \frac{2 n}{2 n} = 1.$ Walls hain $0$ aur $1$ — alag limits. Squeeze theorem yahan kuch nahi deta (parent mistake #1).
Actual values dekho. Kyun? Direct compute karo: agar $n$ even hai, $b_{n} = \frac{2 n}{2 n} = 1$ ; agar $n$ odd hai, $b_{n} = \frac{0}{2 n} = 0$ . Toh $b_{n}$ jump karta hai $0, 1, 0, 1, \dots$
Conclusion. Kyun? Sequence ka koi limit nahi hai — aur sahi baat hai, mismatched walls ne hamesha jhooth nahi bola.

Verify: terms hain $b_{1} = 0, b_{2} = 1, b_{3} = 0, b_{4} = 1$ . Do subsequential limits $0$ aur $1$ $\Rightarrow$ diverges. Sabak: koi bhi conclusion nikalne se pehle hamesha confirm karo ki dono walls ka ek hi limit ho.

Cell G — Real-world word problem

Worked example Ex 7 — sensor noise ka average nikalna

Ek thermometer reading $n$ par temperature $T_{n} = 20 + \frac{r _{n}}{n}$ °C report karta hai ( $n \in N$ , $n \geq 1$ ke saath), jahan noise $r_{n}$ ki koi bhi value $[- 3, 3]$ mein ho sakti hai (magnitude kabhi badi nahi). Kya reported temperature settle hoga, aur kahan?

Forecast: noise bounded hai, badhti hui readings ki count se divide ho rahi hai. Settle hone ki value guess karo.

Noise ko bounded maano. Yeh step kyun? Bataya gaya hai ki $∣ r_{n} ∣ \leq 3$ , yaani $- 3 \leq r_{n} \leq 3$ — ek bounded middle jise hum wall kar sakte hain.
$n > 0$ se divide karo aur $20$ add karo. Kyun? Kyunki $n \geq 1$ hai, positive $n$ se divide karne se direction preserve hoti hai: $20 - \frac{3}{n} \leq T_{n} \leq 20 + \frac{3}{n} .$
Squeeze karo. Kyun? Dono walls $\to 20$ (constant $\pm$ constant-over-grower). Same limit $\Rightarrow T_{n} \to 20$ .

Verify (units & sanity): har term °C mein hai; $n = 60$ readings par band $20 \pm 0.05$ °C hai. Physically noise average ho jaati hai — sahi reading $20 °C$ hai.

Cell H — Exam twist (walls jo sirf lagte hain alag hain)

Worked example Ex 8 —

$b_{n} = \frac{n sin n + n ^{2}}{n ^{2} + cos n}$ Forecast: do oscillators, upar aur neeche dono blow up karte hain. Darta hai — walls aapas mein ladti lagti hain. Phir bhi guess karo.

Upar aur neeche ke oscillators ko bound karo. Yeh step kyun? $- n \leq n sin n \leq n$ aur $- 1 \leq cos n \leq 1$ . Walls banane ke liye unhe worst case se replace karo. Fraction ki sabse chhoti value ke liye sabse chota upar, sabse bada neeche; sabse badi value ke liye sabse bada upar, sabse chota neeche: $\frac{n ^{2} - n}{n ^{2} + 1} \leq b_{n} \leq \frac{n ^{2} + n}{n ^{2} - 1} (n \geq 2, bottom > 0) .$ $n \geq 2$ kyun? Hume $n^{2} + cos n > 0$ chahiye; kyunki $cos n \geq - 1$ hai, $n^{2} - 1 > 0$ $n \geq 2$ ke liye (yaad rakho $n \in N$ ). Yeh "eventually" clause hai — finitely many early terms se koi fark nahi padta.
Har wall ko simplify karo. Kyun? Top aur bottom dono ko $n^{2}$ se divide karo: $\frac{1 - \frac{1}{n}}{1 + \frac{1}{n ^{2}}} \to 1, \frac{1 + \frac{1}{n}}{1 - \frac{1}{n ^{2}}} \to 1.$ Jo walls alag lagte the woh dono $1$ par aa jaate hain.
Squeeze karo. Dono walls $\to 1 \Rightarrow b_{n} \to 1$ .

Verify: $n = 1000$ : numerator $\approx 1 0^{6} \pm 1000$ , denominator $\approx 1 0^{6} \pm 1$ , ratio $\approx 1.00 \pm 0.001$ . Limit $= 1$ .

Active recall

Recall Cell ko strategy se match karo

Oscillator over a grower ( $\frac{3 c o s ( 5 n ) + 2}{n + 4}$ , Ex 1)? → Cell A: grower pe $\pm$ bound, limit $0$ .
Har step sign flip hota hai ( $\frac{( - 1 ) ^{n} n}{n + 1}$ )? → Cell B: $∣ b_{n} ∣$ ko squeeze karo, limit $0$ .
$\frac{2 ^{n}}{n !}$ ? → Cell C: upar se $\frac{4}{n}$ se bound karo, neeche se $0$ se, limit $0$ .
$n 5 n^{2} + 3$ ? → Cell D: jaane-maane $n$ th roots se sandwich karo, limit $1$ .
Walls $0$ aur $1$ tak jaati hain? → Cell F: theorem kuch nahi kehta; actual divergence check karo.

Recall One-line takeaways (prompt ::: answer)

Ex 6 diverge kyun hua? ::: Dono walls ke alag-alag limits the ( $0$ aur $1$ ), toh squeeze fail hua — aur terms sach mein $0, 1$ alternate karti hain. Ex 2 mein $b_{n}$ nahi $∣ b_{n} ∣$ kyun bound kiya? ::: Middle sign change karta hai; sirf ek two-sided (absolute-value) wall hi dono directions ko trap karti hai. Ex 8 mein $n \geq 2$ kyun chahiye? ::: Denominator positive rakhne ke liye; squeeze ko sirf eventually hold karna hota hai. Real-world Ex 7 ka limit? ::: $20$ °C — bounded noise jo badhte $n$ se divide hoti hai woh khatam ho jaati hai. Is page pe index $n$ ka domain kya hai? ::: $n \in N$ , yaani $n = 1, 2, 3, \dots$ — positive denominators guarantee karne ke liye use kiya.

Connections

Squeeze theorem for sequences — parent note: statement + $ε$ – $N$ proof.
Standard limits ( n-th roots, n!/n^n, ln n / n ) — Cells C & D mein walls yahan se aati hain.
Bounded sequences — har oscillator wall ( $∣ sin ∣, ∣ cos ∣, ∣ (- 1)^{n} ∣ \leq 1$ ).
Algebra of limits (sum, product, quotient) — walls simplify karne ke liye use hota hai (Cells E, H).
Squeeze theorem for functions — real line pe same idea.