Exercises — Squeeze theorem (sandwich theorem)

Rule (from Squeeze theorem (sandwich theorem)): dono walls ko same value $L$ ki taraf tend karna chahiye.

• (a) ${\lim }_{x\to 0}(-x^2)=0$ aur ${\lim }_{x\to 0}{x}^2=0$. Same $\Rightarrow$ YES, $f\to 0$.
• (b) Walls constants $2$ aur $4$ hain; woh kabhi nahi milti. NO conclusion — $f$ $2$ aur $4$ ke beech kahin bhi wander kar sakti hai.
• (c) ${\lim }_{x\to 0}(-|x|)=0$ aur ${\lim }_{x\to 0}|x|=0$. Same $\Rightarrow$ YES, $f\to 0$.

"Bounded" = aise numbers $m,M$ hain jahan $m\le f(x)\le M$ har jagah ho.

• (a) $\sin$ of kuch bhi $[-1,1]$ mein rehta hai. Bounded.
• (b) $\cos$ of kuch bhi $[-1,1]$ mein rehta hai. Bounded.
• (c) $x$ bina limit ke grow karta hai, isliye $x\sin x$ bounded nahi hai (lo $x=\frac{\pi }{2}+2\pi n$, value $=x\to \infty$).
• (d) Kyunki $\sin x\in [-1,1]$ hai, denominator $2+\sin x\in [1,3]$ hai, isliye $\frac{1}{2+\sin x}\in [\frac{1}{3},1]$. Bounded.

Step 1 — wild part ko bound karo. $\cos$ of anything $[-1,1]$ mein rehta hai: $-1\le \cos \frac{1}{x^3}\le 1.$ Kyun: cosine kabhi $[-1,1]$ se bahar nahi ja sakta, chahe $1/x^3$ kitni bhi tez spin kare.

Step 2 — tame part se multiply karo. Dono taraf $x^4$ se multiply karo. Kyunki $x^4\ge 0$ hamesha rehta hai, inequality directions preserved rehti hain (yeh Bounded times vanishing pattern hai): $-x^4\le x^4\cos \frac{1}{x^3}\le x^4.$

Step 3 — squeeze. ${\lim }_{x\to 0}(-x^4)=0$ aur ${\lim }_{x\to 0}{x}^4=0$. Same wall-limit $\Rightarrow$ ${\lim }_{x\to 0}{x}^4\cos \frac{1}{x^3}=\boxed{0}.$

Step 1 — numerator ko bound karo. Har real $x$ ke liye, $-2\le \sin x+\cos x\le 2.$ Kyun: $|\sin x|\le 1$ aur $|\cos x|\le 1$, isliye unka sum size mein zyada se zyada $2$ hoga.

Step 2 — $x$ se divide karo. $x>0$ ke liye (hum $+\infty$ ki taraf ja rahe hain, isliye $x>0$ safely hai), positive $x$ se divide karne par directions same rehti hain: $-\frac{2}{x}\le \frac{\sin x+\cos x}{x}\le \frac{2}{x}.$

Step 3 — squeeze. ${\lim }_{x\to \infty }\frac{2}{x}=0$ aur ${\lim }_{x\to \infty }(-\frac{2}{x})=0$. Isliye ${\lim }_{x\to \infty }\frac{\sin x+\cos x}{x}=\boxed{0}.$

Yeh Limits of sequences wala flavour hai: $x\to a$ ban jaata hai $n\to \infty$.

Step 1. $(-1{)}^n$ ya toh $+1$ hai ya $-1$, isliye $-1\le (-1{)}^n\le 1$.

Step 2 — positive $n^2+1$ se divide karo: $-\frac{1}{n^2+1}\le \frac{(-1{)}^n}{n^2+1}\le \frac{1}{n^2+1}.$

Step 3 — squeeze. Dono walls $\pm \frac{1}{n^2+1}\to 0$. Toh limit $\boxed{0}$ hai.

Squeeze karne ka mann karta hai — lekin pehle tame factor ki limit check karo. Jab $x\to 0$, $\sqrt{x^2+4}\to \sqrt{4}=2$, 0 nahi. Tame part vanish nahi karta, isliye Bounded times vanishing apply nahi hoti.

$0$ ke paas product behave karta hai $2\sin (1/x)$ ki tarah, jo hamesha ke liye roughly $-2$ aur $2$ ke beech oscillate karta rehta hai — kabhi settle nahi hota. Koi bhi candidate walls roughly $-2\le f\le 2$ hongi: woh nahi milti, isliye squeeze fire nahi kar sakta.

Conclusion: limit exist nahi karti. (Dekho Oscillating functions.) Lesson yeh hai: squeeze tab hi kaam karta hai jab walls close in hoti hain; yahan woh $4$ apart rehti hain.

Step 1 — floor ko bound karo. Floor ki definition se, kisi bhi real $t$ ke liye, $t-1<\lfloor t\rfloor \le t.$ Kyun: greatest integer $\le t$, $t$ se ek unit neeche ke andar rehta hai. $t=1/x^2$ rakho: $\frac{1}{x^2}-1<\lfloor \frac{1}{x^2}\rfloor \le \frac{1}{x^2}.$

Step 2 — $x^2\ge 0$ se multiply karo (direction preserved): $x^2\left(\frac{1}{x^2}-1\right)<x^2\lfloor \frac{1}{x^2}\rfloor \le x^2\cdot \frac{1}{x^2},$ jo simplify hoke ban jaata hai $1-x^2<x^2\lfloor \frac{1}{x^2}\rfloor \le 1.$

Step 3 — squeeze. Lower wall $1-x^2\to 1$, upper wall $1\to 1$. Same limit $\Rightarrow$ ${\lim }_{x\to 0}{x}^2\lfloor \frac{1}{x^2}\rfloor =\boxed{1}.$ Notice karo: walls $1$ par mili, $0$ par nahi — squeeze target necessarily zero nahi hona chahiye.

Step 1 — ek friendly window tak restrict karo. $0<x<1$ lo. Yahan $\ln x<0$ hai, isliye $x\ln x<0$, matlab $x^x=e^{x\ln x}<e^0=1$. Yeh hamari upper wall hai: $x^x\le 1$.

Step 2 — lower wall. $0<x<1$ ke liye $x\ln x\ge$ (kuch cheez $\to 0$) hai. Concretely, kyunki $x\ln x\to 0$, $0^{+}$ ke paas $x$ ke liye exponent ek shrinking interval $[-\epsilon ,0]$ mein rehta hai, isliye $e^{-\epsilon }\le x^x\le 1.$ Jaise $x\to 0^{+}$ hota hai hum $\epsilon \to 0$ le sakte hain, jisse lower wall ki limit $e^0=1$ milti hai.

Step 3 — squeeze. Dono walls $\to 1$, isliye ${\lim }_{x\to 0^{+}}{x}^x=\boxed{1}$. (Cleaner route: $x^x=e^{x\ln x}$ aur Continuity of $e^t$ se, $\lim =e^{\lim (x\ln x)}=e^0=1$. Squeeze view dikhata hai kyun $e^{x\ln x}$ ke paas $1$ ke alawa kahin jaane ki jagah nahi hai.)

Idea: expression ko (ek aisa piece jo ki known limit rakhta hai) $\times$ (ek bounded piece) mein peel karo. $\frac{\sin (x^2)\cos (1/x)}{x}=\underset{\to 1}{\underbrace{\frac{\sin (x^2)}{x^2}}}\cdot \underset{\to 0}{\underbrace{x}}\cdot \underset{\text{bounded}}{\underbrace{\cos \frac{1}{x}}}.$ Kyun aise split karein: $\frac{\sin t}{t}\to 1$ standard trig limit hai (Limits of trigonometric functions) jahan $t=x^2\to 0$; bacha hua $x$ vanish ho jaata hai aur $\cos (1/x)$ $[-1,1]$ ke andar rehta hai.

Step 1 — (vanishing)$\times$(bounded) ke roop mein regroup karo. $B(x)=\frac{\sin (x^2)}{x^2}\cos \frac{1}{x}$ lo. Kyunki $\frac{\sin (x^2)}{x^2}\to 1$ hai toh yeh $1$ ke paas rehta hai (maano $[0,2]$ ke andar) $0$ ke close, aur $\cos (1/x)\in [-1,1]$ hai, isliye $B(x)$ bounded hai: $|B(x)|\le 2$ near $0$.

Step 2 — poori cheez ko squeeze karo. Expression $x\cdot B(x)$ ke barabar hai, aur $-2|x|\le xB(x)\le 2|x|.$ Kyun $|x|$: $|B|\le 2$ ko $|x|\ge 0$ (safe direction) se multiply karne par product ka size bound ho jaata hai.

Step 3. Dono walls $\pm 2|x|\to 0$, isliye ${\lim }_{x\to 0}\frac{\sin (x^2)\cos (1/x)}{x}=\boxed{0}.$

Step 1 — absolute value ko do walls mein unfold karo. $|f(x)-3|\le 5|x-2{|}^2$ ka matlab hai $3-5|x-2{|}^2\le f(x)\le 3+5|x-2{|}^2.$

Step 2 — squeeze. Jaise $x\to 2$, $|x-2{|}^2\to 0$, isliye dono walls $\to 3$. Isliye ${\lim }_{x\to 2}f(x)=\boxed{3}.$

Step 3 — continuity. Kyunki ${\lim }_{x\to 2}f(x)=3=f(2)$ hai, Continuity ki definition se, $f$ $2$ par continuous hai. (Yeh bound-and-squeeze bilkul wahi hai jo differentiability-style estimates prove karne mein use hota hai.)

Step 1 — wiggle ko bound karo: $-1\le \sin \frac{1}{x}\le 1$.

Step 2 — $x$ se multiply karo, lekin $x$ ka sign change hota hai. Yahi toh poora point hai.

• Agar $x>0$: multiply karna direction preserve karta hai: $-x\le x\sin \frac{1}{x}\le x$.
• Agar $x<0$: ek negative se multiply karna directions flip karta hai: $-x\ge x\sin \frac{1}{x}\ge x$, matlab $x\le x\sin \frac{1}{x}\le -x$.

Dono cases mein do walls $x$ aur $-x$ hain — bas swapped roles mein. Isse uniformly likhne ka tarika (sign split se bachte hue) $|x|$ use karna hai: $-|x|\le x\sin \frac{1}{x}\le |x|,\text{true for all }x\ne 0.$ Kyun yeh legit hai: $\left|x\sin \frac{1}{x}\right|=|x|\left|\sin \frac{1}{x}\right|\le |x|\cdot 1=|x|$.

Step 3 — squeeze. $\pm |x|\to 0$, isliye ${\lim }_{x\to 0}x\sin \frac{1}{x}=\boxed{0}.$ Neeche wali figure dekho: graph puri tarah "cone" $\pm |x|$ (chalk blue) ke andar rehta hai, kisi bhi wall ke escape ko touch nahi karta.

Conclusion sahi hai, lekin squeeze ki wajah se nahi. Squeeze theorem sirf kabhi ek limit produce kar sakta hai; yeh kabhi prove nahi kar sakta ki limit fail hoti hai. Walls jo nahi milti simply koi information nahi deti — woh limit ko forbid nahi karti.

Sahi reasoning: one-sided limits directly compute karo. $x>0$ ke liye, $\frac{|x|}{x}=\frac{x}{x}=1$; $x<0$ ke liye, $\frac{|x|}{x}=\frac{-x}{x}=-1$. Left limit $-1\ne 1$ right limit, isliye ${\lim }_{x\to 0}\frac{|x|}{x}\text{does not exist.}$ Answer: conclusion sahi, reasoning galat — ek non-meeting squeeze kuch prove nahi karta.

Step 1 — lower wall. $n\ge 1$ ke liye, $n\ge 1\Rightarrow \sqrt[n]{n}\ge 1$. Toh lower wall constant $1$ hai.

Step 2 — upper wall banao. $\sqrt[n]{n}=1+t_n$ likho jahan $t_n\ge 0$ ho. $n$-th power tak raise karo: $n=(1+t_n{)}^n.$ Binomial expansion se, saare terms $\ge 0$ hain, isliye sirf $k=2$ wala term rakhte hain (for $n\ge 2$): $n=(1+t_n{)}^n\ge (\genfrac{}{}{0ex}{}{n}{2})t_n^2=\frac{n(n-1)}{2}{t}_n^2.$ Rearrange karne par: $t_n^2\le \frac{2}{n-1}$, hence $0\le t_n\le \sqrt{\frac{2}{n-1}}$. Isliye upper wall hai $1\le \sqrt[n]{n}\le 1+\sqrt{\frac{2}{n-1}}.$

Step 3 — squeeze. Jaise $n\to \infty$, $\sqrt{\frac{2}{n-1}}\to 0$, isliye upper wall $\to 1$; lower wall $1$ hai. Dono $1$ par milti hain: ${\lim }_{n\to \infty }\sqrt[n]{n}=\boxed{1}.$ Yeh mastery-level hai kyunki tumne upper wall manufacture ki ek aisi inequality se jo kisine tumhe di nahi thi.

Kya meri do walls $f$ ko $a$ ke ek poore neighborhood mein sandwich karti hain? ::: Haan — inequality $a$ ke paas hold karni chahiye, sirf $a$ par nahi. Kya dono walls same number $L$ ki taraf converge karti hain? ::: Haan — alag alag limits koi conclusion nahi deti. Jab maine multiply/divide kiya, kya factor guaranteed non-negative tha (ya maine sign se split kiya / $|x|$ use kiya)? ::: Haan — warna inequality flip ho sakti thi. Kya main kisi limit ka exist karna prove karne ki koshish kar raha hoon, na ki disprove karne ki? ::: Squeeze sirf existence prove karta hai.