Squeeze theorem for sequences: Exercises — Hinglish deep dive

4.3.2 · D4 · Maths › Calculus III — Sequences & Series › Squeeze theorem for sequences

Definition Is poore page par use hone wale ground rules

Index $n$ positive whole numbers mein chalta hai $n = 1, 2, 3, \dots$ (natural numbers $N$ ). Isliye $n$ hamesha ek positive integer hai — khaaskar $n > 0$ , yahi wajah hai ki hum $n$ se divide kar sakte hain bina kisi inequality ko flip kiye. Isi tarah $n^{2}, n^{2} + 1, n^{2} + n$ sab positive hain.
Ek wall do outer sequences mein se ek hoti hai jo hamare target ko trap karti hai; "lower wall" $a_{n}$ neeche rehti hai, "upper wall" $c_{n}$ upar rehti hai.
$N_{0}$ ek threshold index hai: ek specific whole number jiske baad se hamare sandwich $a_{n} \leq b_{n} \leq c_{n}$ ka guarantee hona pakka hai. $N_{0}$ se pehle koi promise nahi. Yeh "eventually" ka idea hai: ek limit sirf sequence ki tail ki parwah karta hai, isliye shuruaat ke finite messy terms se koi fark nahi padta. Poore $ε$ – $N$ proof mein (dekho Limit of a sequence (epsilon-N definition)) $N = max (N_{0}, N_{1}, N_{2})$ liya jaata hai taaki teeno facts ek saath hold karein.

Level 1 — Recognition

Problem 1.1

$b_{n} = \frac{cos n}{n}$ ke liye, walls ka kaun sa pair $(a_{n}, c_{n})$ squeeze karne deta hai, aur limit kya hai?

Recall Solution 1.1

Messy part ko kya bound karta hai. Kisi bhi real number ka cosine $[- 1, 1]$ mein rehta hai: yaani $- 1 \leq cos n \leq 1$ . $cos n$ ke baare mein hum bas yahi jaante hain — yeh kabhi settle nahi hota, lekin $[- 1, 1]$ se bahar bhi nahi jaata.

$n$ se divide kyun karein. Hume $\frac{c o s n}{n}$ chahiye, isliye teeno parts ko $n$ se divide karo. Kyunki $n$ ek positive integer hai ( $n \geq 1$ ), isse divide karne par $\leq$ kabhi flip nahi hota: $- \frac{1}{n} \leq \frac{c o s n}{n} \leq \frac{1}{n} .$ Walls: $a_{n} = - \frac{1}{n}$ , $c_{n} = + \frac{1}{n}$ . Dono $\to 0$ .

Conclusion: dono walls par same limit $\Rightarrow n \to \infty lim \frac{cos n}{n} = 0.$

Problem 1.2

Sahi ya galat: agar $0 \leq b_{n} \leq \frac{1}{n}$ sabhi $n \geq 5$ ke liye (lekin yeh bound $n = 1, 2, 3, 4$ ke liye fail hoti hai), tab bhi hum $b_{n} \to 0$ conclude kar sakte hain.

Recall Solution 1.2

SAHI. Limit ek sequence ki tail ke baare mein ek statement hai. Shuruaat ke finite terms ko badalna ya ignore karna yeh nahi badal sakta ki sequence kahan ja rahi hai. Yahan threshold index $N_{0} = 5$ hai: squeeze "eventually" hold karta hai, jo exactly wahi hai jo theorem maangta hai. Parent ki remark dekho: bounds sirf eventually hold karne chahiye.

Problem 1.3

Inme se kaun sa limit ke liye squeeze ho sakta hai as stated, aur kaun sa nahi? (a) $- \frac{1}{n} \leq b_{n} \leq \frac{1}{n}$ (b) $\frac{1}{n} \leq b_{n} \leq 5$ (c) $2 - \frac{1}{n} \leq b_{n} \leq 2 + \frac{1}{n}$ .

Recall Solution 1.3

(a) Dono walls $\to 0$ . Squeeze kaam karta hai, $b_{n} \to 0$ .
(b) Lower wall $\frac{1}{n} \to 0$ , upper wall $\to 5$ . Alag limits — squeeze kuch nahi kehta. $b_{n}$ kahin bhi $(0, 5]$ mein hover kar sakta hai.
(c) Dono walls $\to 2$ . Squeeze kaam karta hai, $b_{n} \to 2$ .

Level 2 — Application

Problem 2.1

$n \to \infty lim \frac{( - 1 ) ^{n}}{n ^{2} + 1}$ nikalo.

Recall Solution 2.1

Kya wiggle karta hai. $(- 1)^{n}$ hamesha $- 1$ aur $+ 1$ ke beech flip karta rehta hai, isliye $- 1 \leq (- 1)^{n} \leq 1$ .

$n^{2} + 1$ se divide kyun karein. Kyunki $n$ ek positive integer hai, $n^{2} + 1$ har $n$ ke liye positive hai, isliye divide karne par directions theek rehte hain: $- \frac{1}{n ^{2} + 1} \leq \frac{( - 1 ) ^{n}}{n ^{2} + 1} \leq \frac{1}{n ^{2} + 1} .$ Dono walls $\to 0$ (denominator bahut bada ho jaata hai). Answer: $0$ .

Problem 2.2

$n \to \infty lim \frac{3 + 2 sin ( n ^{3} )}{n}$ nikalo.

Recall Solution 2.2

Kya bound karna hai. $sin (n^{3})$ unpredictable hai lekin $- 1 \leq sin (n^{3}) \leq 1$ , isliye $2 sin (n^{3}) \in [- 2, 2]$ , aur isliye numerator $3 + 2 sin (n^{3}) \in [1, 5]$ .

Kyun yeh constant walls kaam aati hain. $n > 0$ se divide karo: $\frac{1}{n} \leq \frac{3 + 2 s i n ( n ^{3} )}{n} \leq \frac{5}{n} .$ Lower wall $\frac{1}{n} \to 0$ , upper wall $\frac{5}{n} \to 0$ . Kyun $\frac{5}{n} \to 0$ ? $0$ ki taraf jaane wali sequence ka ek fixed constant se product bhi $0$ ki taraf jaata hai: kisi bhi target band ke half-width $ε$ ke liye, jab ek baar $\frac{1}{n} < \frac{ε}{5}$ ho jaata hai toh $\frac{5}{n} < ε$ . (Yeh Algebra of limits (sum, product, quotient) ka product rule hai.) Answer: $0$ .

Problem 2.3

$n \to \infty lim \frac{n + cos n}{n ^{2}}$ nikalo.

Recall Solution 2.3

Numerator split karo. $\frac{n + c o s n}{n ^{2}} = \frac{n}{n ^{2}} + \frac{c o s n}{n ^{2}} = \frac{1}{n} + \frac{c o s n}{n ^{2}}$ .

Sirf wiggly piece ko bound karo. $- \frac{1}{n ^{2}} \leq \frac{c o s n}{n ^{2}} \leq \frac{1}{n ^{2}}$ , isliye $\frac{1}{n} - \frac{1}{n ^{2}} \leq \frac{n + c o s n}{n ^{2}} \leq \frac{1}{n} + \frac{1}{n ^{2}} .$ Dono walls $\to 0 + 0 = 0$ . Answer: $0$ .

Level 3 — Analysis

Problem 3.1

$n \to \infty lim k = 1 \sum n \frac{1}{n ^{2} + k}$ nikalo.

Recall Solution 3.1

Har denominator ko kyun bound karein. $n$ terms hain, index $k$ $1$ se $n$ tak jaata hai. Har aise $k$ ke liye: $n^{2} + 1 \leq n^{2} + k \leq n^{2} + n$ . Bada denominator $\Rightarrow$ chhota fraction, isliye $\frac{1}{n ^{2} + n} \leq \frac{1}{n ^{2} + k} \leq \frac{1}{n ^{2} + 1} .$ $k$ par sum karo (sab $n$ terms identically bounded hain): $\frac{n}{n ^{2} + n} \leq \sum_{k = 1}^{n} \frac{1}{n ^{2} + k} \leq \frac{n}{n ^{2} + 1} .$ Walls ki limits: $\frac{n}{n ^{2} + n} = \frac{1}{n + 1} \to 0$ , aur $\frac{n}{n ^{2} + 1} \to 0$ . Answer: $0$ .

(Parent ke Example 4 se compare karo, jisme $\frac{n}{n ^{2} + k}$ tha aur limit $1$ thi — wahan numerator mein extra factor $n$ hone se sab kuch badal jaata hai.)

Problem 3.2

$n \to \infty lim n 2^{n} + 3^{n}$ nikalo.

Recall Solution 3.2

Dominant term ko squeeze karo. Kyunki $3^{n} \leq 2^{n} + 3^{n} \leq 3^{n} + 3^{n} = 2 \cdot 3^{n}$ , $n$ th roots lo. Kyun le sakte hain? $n$ th-root map $t \mapsto n t$ , $t \geq 0$ par increasing hai: agar $t_{1} \leq t_{2}$ toh $n t_{1} \leq n t_{2}$ (chhote number ko same positive power par uthane par woh chhota hi rehta hai). Isliye inequality directions survive karti hain: $n 3^{n} \leq n 2^{n} + 3^{n} \leq n 2 \cdot 3^{n} .$ Yaani $3 \leq n 2^{n} + 3^{n} \leq 3 \cdot 2^{1/ n}$ .

Kyun $2^{1/ n} \to 1$ . Yeh ek standard limit hai: kisi bhi fixed $a > 0$ ke liye, $n a \to 1$ . Intuition: $2 = x^{n}$ hold karne ke liye jab $n$ bahut bada ho, toh $x$ must just barely $1$ se upar hona chahiye — $1$ se thoda bhi bada number, bade power par uthane par $2$ se zyada ho jaata hai. Isliye upper wall $\to 3 \cdot 1 = 3$ , lower wall constant $3$ hai. Answer: $3$ (sabse bada base jeetta hai).

Problem 3.3

Dikhao ki $n \to \infty lim \frac{⌊ n x ⌋}{n} = x$ kisi bhi fixed real $x$ ke liye, jahan $⌊ \cdot ⌋$ floor hai (sabse bada integer $\leq \cdot$ ).

Recall Solution 3.3

Floor kya guarantee karta hai. Floor ki definition se, kisi bhi real $y$ ke liye: $y - 1 < ⌊ y ⌋ \leq y$ . $y = n x$ rakkho: $n x - 1 < ⌊ n x ⌋ \leq n x .$ $n > 0$ se divide kyun karein. $n$ ek positive integer hai, isliye directions preserve hote hain: $x - \frac{1}{n} < \frac{⌊ n x ⌋}{n} \leq x .$ Walls: lower $x - \frac{1}{n} \to x$ , upper constant $x$ . Dono $\to x$ . Answer: $x$ . $■$

Level 4 — Synthesis

Problem 4.1

$n \to \infty lim n 3 n^{2} + 5$ nikalo.

Recall Solution 4.1

Polynomial ko constant ki powers ke beech trap karo. Hum ek sandwich banate hain jo explicit threshold index $N_{0} = 8$ se valid hai.

Lower side (sabhi $n \geq 1$ ke liye hold karta hai): clearly $3 n^{2} + 5 \geq 1$ .
Upper side: hum claim karte hain ki $3 n^{2} + 5 \leq n^{3}$ sabhi $n \geq 8$ ke liye. Check: $n^{3} - 3 n^{2} - 5 = n^{2} (n - 3) - 5$ ; $n \geq 8$ ke liye $n^{2} (n - 3) \geq 64 \cdot 5 = 320 > 5$ , isliye $n^{3} - 3 n^{2} - 5 > 0$ . Isliye yeh bound exactly $N_{0} = 8$ se aage hold karta hai.

Isliye sabhi $n \geq N_{0} = 8$ ke liye: $1 \leq 3 n^{2} + 5 \leq n^{3} .$ $n$ th roots lo (increasing map, directions hold): $1 \leq n 3 n^{2} + 5 \leq n n^{3} = (n n)^{3} .$

Kyun $n n \to 1$ . Standard limit (parent ke Example 3 mein logs lekar prove kiya: $\frac{l n n}{n} \to 0$ ). Isliye upper wall $\to 1^{3} = 1$ , lower wall constant $1$ . Kyunki sandwich sabhi $n \geq N_{0}$ ke liye hold karta hai, squeeze apply hota hai. Answer: $1$ .

Problem 4.2

$n \to \infty lim \frac{n !}{n ^{n}} cos (n!)$ nikalo. Figure dekho.

Recall Solution 4.2

Kya dangerous hai. $cos (n!)$ oscillate karta hai aur $\frac{n !}{n ^{n}}$ ek positive shrinking cheez hai. "Wiggle" aur "shrink" ka product.

Wiggle ko bound karo, shrink par sawaar ho. Kyunki $- 1 \leq cos (n!) \leq 1$ aur $\frac{n !}{n ^{n}} > 0$ (positive hai!), multiply karne par directions preserve hote hain: $- \frac{n !}{n ^{n}} \leq \frac{n !}{n ^{n}} cos (n!) \leq \frac{n !}{n ^{n}} .$ Kyun dono walls mar jaati hain. Parent se (Example 2), $\frac{n !}{n ^{n}} \leq \frac{1}{n} \to 0$ , isliye $- \frac{n !}{n ^{n}} \to 0$ aur $\frac{n !}{n ^{n}} \to 0$ . Figure mein yahi yellow band ka blue line par collapse karna hai; pink sequence, andar trapped, bhi $0$ par force hoti hai. Answer: $0$ .

Problem 4.3

$n \to \infty lim \frac{1}{n} k = 1 \sum n sin (\frac{k}{n ^{2}})$ nikalo.

Recall Solution 4.3

Sine par tame bound use karo. $x \geq 0$ ke liye: $0 \leq sin x \leq x$ (sine apne input ko $[0, \infty)$ par kabhi nahi overtake karta; aur $sin x \geq 0$ yahan isliye hai kyunki har $\frac{k}{n ^{2}} \leq \frac{n}{n ^{2}} = \frac{1}{n}$ ek chhota non-negative angle hai). Isliye har term ke liye $0 \leq sin (\frac{k}{n ^{2}}) \leq \frac{k}{n ^{2}} .$ Sum karo phir $n$ se divide karo: $0 \leq \frac{1}{n} \sum_{k = 1}^{n} sin (\frac{k}{n ^{2}}) \leq \frac{1}{n} \sum_{k = 1}^{n} \frac{k}{n ^{2}} = \frac{1}{n} \cdot \frac{1}{n ^{2}} \cdot \frac{n ( n + 1 )}{2} = \frac{n + 1}{2 n ^{2}} .$ Upper wall $\frac{n + 1}{2 n ^{2}} \to 0$ , lower wall $0$ . Answer: $0$ .

Level 5 — Mastery

Problem 5.1

Ek student likhta hai: " $0 \leq b_{n} \leq c_{n}$ aur $c_{n} \to 3$ , isliye squeeze se $b_{n} \to 3$ ." Precisely explain karo yeh galat kyun hai, aur ek explicit $b_{n}$ do jo hypotheses satisfy kare par jiska limit $3$ na ho.

Recall Solution 5.1

Kami: lower wall constant $0$ hai, jiska limit $0 \neq = 3$ hai. Squeeze maangta hai ki dono walls same limit $L$ share karein. Yahan walls aapas mein agree nahi karti ( $0$ vs $3$ ), isliye theorem chup hai.

Counterexample. Maano $c_{n} = 3$ (isliye $c_{n} \to 3$ ) aur $b_{n} = 0$ sabhi $n$ ke liye. Tab $0 \leq b_{n} \leq c_{n}$ hold karta hai, phir bhi $b_{n} \to 0 \neq = 3$ . Galat "conclusion" refute ho gayi. $■$

Problem 5.2

Prove karo: agar $∣ b_{n} - L ∣ \leq c_{n}$ sabhi $n \geq N_{0}$ ke liye aur $c_{n} \to 0$ , tab $b_{n} \to L$ . (Yeh "absolute-value squeeze" workhorse form hai.)

Recall Solution 5.2

Absolute value ko do walls mein badlo. Statement $∣ b_{n} - L ∣ \leq c_{n}$ exactly yeh kehta hai: $- c_{n} \leq b_{n} - L \leq c_{n} ⟹ L - c_{n} \leq b_{n} \leq L + c_{n} .$ Squeeze apply karo. Lower wall $L - c_{n} \to L - 0 = L$ ; upper wall $L + c_{n} \to L + 0 = L$ . Same limit $L$ , isliye $b_{n} \to L$ . $■$

(Yahi wajah hai ki " $∣ b_{n} - L ∣ \to 0$ show karo" standard route hai: yeh squeeze chhupe hue roop mein hai, jo $\varepsilon$–$N$ definition par tika hai.)

Problem 5.3

$n \to \infty lim \frac{2 ^{n}}{n !}$ squeeze use karke nikalo. (Ratio test nahi.)

Recall Solution 5.3

Product likho aur ek fixed head alag karo. $n \geq 2$ ke liye: $\frac{2 ^{n}}{n !} = \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \dots \frac{2}{n} .$ Last factor $\frac{2}{n}$ rakho aur beech ke factors ko $1$ se bound karo: $0 < \frac{2 ^{n}}{n !} = 2 \cdot (\frac{2}{3} \cdot \frac{2}{4} \dots \frac{2}{n - 1}) \cdot \frac{2}{n} \leq 2 \cdot 1 \cdot \frac{2}{n} = \frac{4}{n} .$ (Har factor $\frac{2}{3}, \dots, \frac{2}{n - 1}$ $\leq 1$ hai.) Squeeze: $0 < \frac{2 ^{n}}{n !} \leq \frac{4}{n} \to 0$ . Answer: $0$ .

Problem 5.4

Ek sequence $(b_{n})$ do jo bounded ho ( $- 1 \leq b_{n} \leq 1$ ) lekin jise squeeze se limit assign na ki ja sake — aur batao kyun, convergence se connect karte hue.

Recall Solution 5.4

$b_{n} = (- 1)^{n}$ lo. Yeh walls $- 1 \leq b_{n} \leq 1$ ke andar rehta hai, lekin woh walls alag-alag numbers par converge karti hain ( $- 1$ aur $+ 1$ ), isliye squeeze kuch nahi deta. Gehri wajah: $(- 1)^{n}$ genuinely diverge karta hai — iske terms hamesha $- 1$ aur $+ 1$ dono ke paas cluster karte rehte hain, isliye koi bhi squeeze ise kabhi pin nahi kar sakta (pin karne ke liye kuch hai hi nahi). Boundedness akele convergence force nahi karta; tumhe ya to matching walls chahiye (squeeze) ya monotonicity (dekho Monotone convergence theorem).

Recall the whole page

Recall Rapid review (cover the right side)

$\frac{c o s n}{n}$ ke liye walls (P1.1) ::: $- \frac{1}{n}$ aur $+ \frac{1}{n}$ , dono $\to 0$ , isliye limit $0$ . Kya $\frac{1}{n} \leq b_{n} \leq 5$ ek limit pin karta hai? (P1.3b) ::: Nahi — walls ki alag limits hain, isliye squeeze chup hai. $lim \frac{( - 1 ) ^{n}}{n ^{2} + 1}$ (P2.1) ::: $0$ , walls $\pm \frac{1}{n ^{2} + 1}$ . $lim \frac{3 + 2 s i n ( n ^{3} )}{n}$ (P2.2) ::: $0$ , walls $\frac{1}{n}$ aur $\frac{5}{n}$ . $lim \frac{n + c o s n}{n ^{2}}$ (P2.3) ::: $0$ , walls $\frac{1}{n} \pm \frac{1}{n ^{2}}$ . $lim \sum_{k = 1}^{n} \frac{1}{n ^{2} + k}$ (P3.1) ::: $0$ , walls $\frac{n}{n ^{2} + n}$ aur $\frac{n}{n ^{2} + 1}$ . $lim n 2^{n} + 3^{n}$ (P3.2) ::: $3$ (sabse bada base dominate karta hai). $lim \frac{⌊ n x ⌋}{n}$ (P3.3) ::: $x$ , via $x - \frac{1}{n} < \frac{⌊ n x ⌋}{n} \leq x$ . $lim n 3 n^{2} + 5$ (P4.1) ::: $1$ , $1$ aur $(n n)^{3}$ ke beech trapped, $n \geq N_{0} = 8$ ke liye. $lim \frac{n !}{n ^{n}} cos (n!)$ (P4.2) ::: $0$ , envelope $\pm \frac{n !}{n ^{n}}$ . $lim \frac{1}{n} \sum_{k = 1}^{n} sin (k / n^{2})$ (P4.3) ::: $0$ , upper wall $\frac{n + 1}{2 n ^{2}}$ . Absolute-value squeeze (P5.2) ::: $∣ b_{n} - L ∣ \leq c_{n}$ jab $c_{n} \to 0$ toh $b_{n} \to L$ . $lim \frac{2 ^{n}}{n !}$ (P5.3) ::: $0$ (upar $\frac{4}{n}$ se bounded). $(- 1)^{n}$ squeeze kyun resist karta hai (P5.4) ::: iski do walls $\pm 1$ ki alag limits hain; yeh diverge karta hai. Poore page ka ek-line moral ::: do walls ek SAME limit ke saath trapped sequence ko us limit par force karte hain; alag-limit walls kuch prove nahi karte.

Connections

Parent: Squeeze theorem — statement, proof, base examples.
Limit of a sequence (epsilon-N definition) — har " $\to 0$ " ke peeche ki machinery.
Bounded sequences — jahan se $\pm 1$ walls aati hain.
Standard limits ( n-th roots, n!/n^n, ln n / n ) — $n a \to 1$ , $\frac{n !}{n ^{n}} \to 0$ upar use kiye.
Monotone convergence theorem — limit ka doosra raasta jab walls match nahi karte.
Squeeze theorem for functions — continuous cousin (use kiya $0 \leq sin x \leq x$ ke liye).