Squeeze theorem (sandwich theorem): Worked examples

4.1.5 · D3 · Maths › Calculus I — Limits & Derivatives › Squeeze theorem (sandwich theorem)

Shuru karne se pehle, ek reminder simple shabdon mein. Symbol $lim_{x \to a} f (x) = L$ ka matlab hai "jaise $x$ number $a$ ke kareebi se kareeb jaata hai (lekin kabhi equal nahi hota), value $f (x)$ $L$ ke kareebi se kareeb jaati hai." Arrow $\to$ ka matlab hai "approaches." Hum $a$ par kabhi pahunchte nahi; hum usse sneak up karte hain. Neeche sab kuch isi ek idea par based hai, aur andar se Limit definition (epsilon-delta) se power milti hai.

Scenario matrix

Har squeeze problem in cells mein se kisi ek mein aati hai. Har row difficulty ki ek class hai; last column us worked example ka naam deta hai jo use cover karta hai.

Cell	Kya cheez tricky banati hai	Covered by
A. Positive multiplier	Bounded wiggle $\times$ ek aisa factor jo hamesha $\geq 0$ hai	Example 1
B. Sign-changing multiplier	Multiplier $a$ ke paas negative ho sakta hai — inequality flip ho sakti hai	Example 2
C. Both one-sided limits	$x \to a^{+}$ aur $x \to a^{-}$ alag behave karte hain; dono sides check karni padti hain	Example 3
D. Sequence version	$x \to a$ ki jagah $n \to \infty$	Example 4
E. Degenerate / walls-don't-meet	Bounds alag values par tend karte hain → theorem kuch nahi deta	Example 5
F. Limit at infinity (function)	$x \to \infty$ , bounds $0$ par shrink hote hain	Example 6
G. Real-world word problem	Ek physical damped oscillation	Example 7
H. Exam twist	Bound obvious nahi hai; use construct karna padta hai	Example 8

Ab hum har cell ko hit karte hain.

Example 1 — Cell A: positive multiplier

Worked example

$x \to 0 lim x^{4} cos \frac{1}{x ^{2}}$ evaluate karo. Forecast: $cos (1/ x^{2})$ $0$ ke paas insanely fast oscillate karta hai. Padhne se pehle limit guess karo — kya wild wiggle survive karega, ya $x^{4}$ use crush kar dega?

Step 1. Wiggle ko bound karo. Har real input $t$ ke liye, $- 1 \leq cos t \leq 1$ . Yahan $t = 1/ x^{2}$ hai, to $- 1 \leq cos \frac{1}{x ^{2}} \leq 1.$ Ye step kyun? Hum $cos (1/ x^{2})$ ko $x = 0$ par evaluate nahi kar sakte (uska input $1/ x^{2}$ blow up ho jaata hai), lekin cosine $- 1$ aur $1$ ke beech caged rehta hai chahe input kitna bhi crazy ho. Woh cage hamaari starting wall hai.

Step 2. $x^{4}$ se multiply karo. Kyunki $x^{4} \geq 0$ har real $x$ ke liye, inequality direction preserve rehti hai: $- x^{4} \leq x^{4} cos \frac{1}{x ^{2}} \leq x^{4} .$ Ye step kyun? Ye exactly cell A hai: multiplier $x^{4}$ ek fourth power hai, to kabhi negative nahi ho sakta. Ek inequality ko non-negative number se multiply karne par $\leq$ $\leq$ hi rehta hai.

Step 3. Dono walls collapse ho jaati hain: $lim_{x \to 0} (- x^{4}) = 0$ aur $lim_{x \to 0} x^{4} = 0$ . Ye step kyun? Walls common limit $0$ share karti hain — woh ek condition jo parent note tumhe kabhi skip na karne ki warning deta hai.

Conclusion. Squeeze se, $x \to 0 lim x^{4} cos \frac{1}{x ^{2}} === 0 == .$

Verify: $x = 0.1$ lo: $x^{4} = 1 0^{- 4}$ , aur $cos (1/0.01) = cos (100) \approx 0.862$ , deta hai $\approx 8.6 \times 1 0^{- 5}$ — already tiny aur $[- 1 0^{- 4}, 1 0^{- 4}]$ ke andar pinned. Envelope wiggle ko crush kar deta hai. ✓

Ye Bounded times vanishing pattern hai: bounded $\times$ (kuch jo $\to 0$ hai) $= 0$ .

Example 2 — Cell B: multiplier sign change karta hai

Worked example

$x \to 0 lim x sin \frac{1}{x}$ evaluate karo. Forecast: almost identical Example 1 se, lekin multiplier $x^{4}$ nahi balki $x$ hai. $0$ ke paas, $x$ left par negative, right par positive hota hai. Kya isse squeeze toot jaata hai?

Step 1. Pehle ki tarah, $- 1 \leq sin \frac{1}{x} \leq 1$ . Ye step kyun? Same cage — sine $[- 1, 1]$ mein rehta hai chahe uska input kuch bhi kare.

Step 2. Yahan trap hai. Agar main naively $x$ se multiply karun, to left side par ( $x < 0$ ) inequality flip ho jaati hai (ek inequality ko negative number se multiply karne par $\leq$ $\geq$ ban jaata hai), to main safely $- x \leq x sin (1/ x) \leq x$ nahi likh sakta. Clean fix: $∣ x ∣$ se multiply karo, jo hamesha $\geq 0$ hota hai. Pehle note karo $sin \frac{1}{x} \leq 1$ , to $x sin \frac{1}{x} = ∣ x ∣ sin \frac{1}{x} \leq ∣ x ∣.$ Ye step kyun? Absolute values ke saath kaam karna sign question ko completely sidestep kar deta hai — $∣ x ∣ \geq 0$ kabhi kuch flip nahi karta. Ye exactly woh sign-flip trap hai jiske baare mein parent note warn karta hai ( $x$ se, jo negative ho sakta hai, $0$ ke paas inequality multiply karna); $∣ x ∣$ use karne se ye dodge ho jaata hai.

Step 3. Bound ko baahri absolute value ke bina rewrite karo: $- ∣ x ∣ \leq x sin \frac{1}{x} \leq ∣ x ∣.$ Ye step kyun? $∣ A ∣ \leq B$ same statement hai jaise $- B \leq A \leq B$ . Ab hamare paas honest walls hain jo $0$ ke dono sides par valid hain.

Step 4. $lim_{x \to 0} (- ∣ x ∣) = 0$ aur $lim_{x \to 0} ∣ x ∣ = 0$ .

Conclusion. $x \to 0 lim x sin \frac{1}{x} === 0 == .$

Verify: $x = - 0.05$ : $∣ x ∣ = 0.05$ , $sin (1/ (- 0.05)) = sin (- 20) \approx 0.913$ , product $\approx - 0.0456$ , $[- 0.05, 0.05]$ ke andar. ✓ Negative $x$ se koi trouble nahi hua kyunki $∣ x ∣$ ne sign absorb kar liya.

Example 3 — Cell C: one-sided limits, alag alag check kiye

Figure — Squeeze theorem (sandwich theorem)

Step 1 (right side, $x \to 0^{+}$ ). Figure dekho: unit circle, ek angle $x$ radians mein measure kiya gaya (radius-1 circle par arc length). Teen regions nest karte hain: $area of inner triangle \leq area of sector \leq area of outer triangle .$ Unke areas hain $\frac{1}{2} sin x$ , $\frac{1}{2} x$ , $\frac{1}{2} tan x$ respectively. To $sin x \leq x \leq tan x$ . Ye step kyun? Hume $sin x$ ko $x$ se compare karna hai, aur geometry hume woh comparison free mein deti hai — sector do triangles ke beech squeezed hai jo picture mein dikh rahe hain.

Step 2. Chain $sin x \leq x \leq tan x$ ko $sin x$ se divide karo ( $0 < x < \frac{π}{2}$ ke liye positive), phir reciprocals lo (jo $\leq$ ko $\geq$ mein flip karta hai): $cos x \leq \frac{s i n x}{x} \leq 1.$ Ye step kyun? Hum rearrange karte hain jab tak middle exactly woh quantity $\frac{s i n x}{x}$ na ho jo hum chahte hain. Positives ki chain ko reciprocate karne se woh reverse ho jaati hai, isliye flip.

Step 3. Jaise $x \to 0^{+}$ : $cos x \to 1$ (walls meet). Squeeze $\Rightarrow lim_{x \to 0^{+}} \frac{s i n x}{x} = 1$ .

Step 4 (left side, $x \to 0^{-}$ ). Function $\frac{s i n x}{x}$ even hai: $x$ ki jagah $- x$ rakhne par $\frac{s i n ( - x )}{- x} = \frac{- s i n x}{- x} = \frac{s i n x}{x}$ , unchanged. Ye step kyun? Evenness ka matlab hai graph vertical axis ke baare mein mirror-symmetric hai. Concretely, kisi bhi $x < 0$ ke liye $x = - u$ likho jahan $u > 0$ ; phir $\frac{s i n x}{x} = \frac{s i n u}{u}$ , aur jaise $x \to 0^{-}$ hota hai, $u \to 0^{+}$ hota hai, to left-hand values literally same numbers hain jaise right-hand wale. Isliye $lim_{x \to 0^{-}} \frac{s i n x}{x} = lim_{u \to 0^{+}} \frac{s i n u}{u} = 1$ bhi.

Step 5 (dono sides jodna). Two-sided limit tab exist karta hai aur $L$ ke equal hota hai jab exactly dono one-sided limits exist karte hain aur same $L$ ke equal hote hain — ye two-sided limit ki unpacked definition hai. Yahan dono sides ne $1$ diya, to loop close ho gaya. Ye step kyun? Steps 3 aur 4 ne sirf right aur left limits alag alag prove ki; ye step woh rule state karta hai jo humein unhe ek single two-sided statement mein combine karne deta hai.

Conclusion. Dono one-sided limits $1$ ke equal hain, to two-sided $x \to 0 lim \frac{sin x}{x} === 1 == .$ (Dekho Limits of trigonometric functions.)

Verify: $x = 0.01$ : $\frac{s i n 0.01}{0.01} \approx 0.99998$ . $x = - 0.01$ : evenness se same value $\approx 0.99998$ . Dono $1$ ke karib. ✓

Example 4 — Cell D: sequence version

Worked example Positive integers

$n$ ke liye $n \to \infty lim \frac{( - 1 ) ^{n} + sin n}{n}$ evaluate karo. Forecast: numerator unpredictably jittery hai — $(- 1)^{n}$ $\pm 1$ flip karta hai, $sin n$ wander karta hai. Kya denominator ka $n$ jeetta hai?

Step 1. Numerator ko bound karo. $- 1 \leq (- 1)^{n} \leq 1$ aur $- 1 \leq sin n \leq 1$ , to add karne par: $- 2 \leq (- 1)^{n} + sin n \leq 2.$ Ye step kyun? Har piece $[- 1, 1]$ mein caged hai; do caged cheezein ka sum $[- 2, 2]$ mein caged hoga. Ab hamare paas pure jittery top par ek constant bound hai.

Step 2. $n > 0$ se divide karo (positive, koi flip nahi): $- \frac{2}{n} \leq \frac{( - 1 ) ^{n} + s i n n}{n} \leq \frac{2}{n} .$ Ye step kyun? Ek inequality ko positive number se divide karne par direction same rehti hai. Ye "positive factor se multiply karne" ka sequence analogue hai.

Step 3. $lim_{n \to \infty} \frac{2}{n} = 0$ aur $lim_{n \to \infty} (- \frac{2}{n}) = 0$ . Ye step kyun? Dono bounding sequences vanish ho jaate hain — ye Limits of sequences version hai "walls meet" ka.

Conclusion. Squeeze $\Rightarrow n \to \infty lim \frac{( - 1 ) ^{n} + sin n}{n} === 0 == .$

Verify: $n = 100$ : numerator $= 1 + sin 100 \approx 1 - 0.506 = 0.494$ ; $100$ se divide karne par $\approx 0.0049$ , jo $[- 0.02, 0.02]$ ke andar snug hai. ✓ Sequences ke liye " $x \to a$ " simply " $n \to \infty$ " ban jaata hai.

Example 5 — Cell E: degenerate case, walls DON'T meet

Step 1. Bounds honest hain: $- 1 \leq sin \frac{1}{x} \leq 1$ sabhi $x \neq = 0$ ke liye. Ye step kyun? Same cage jaise hamesha. Inequality mein koi gadbad nahi.

Step 2. Walls ke limits check karo. $lim_{x \to 0} (- 1) = - 1$ aur $lim_{x \to 0} (1) = 1$ . Ye step kyun? Parent ka golden rule: theorem conclude karta hai sirf tab jab dono walls same $L$ par tend karein. Yahan lower wall $- 1$ par hai, upper $+ 1$ par. Ye kabhi nahi milte.

Step 3. Kyunki walls $2$ ki poori distance par alag rehti hain, $sin \frac{1}{x}$ ke liye roam karne ka wide corridor hai. Figure dekho: jaise $x \to 0$ , graph hamesha ke liye faster aur faster $- 1$ aur $+ 1$ ke beech oscillate karta rehta hai. Ye kabhi settle nahi karta. Ye step kyun? Common $L$ nahi hone ka matlab squeeze koi conclusion nahi deta. Aur sach mein limit genuinely exist nahi karta — ye ek Oscillating functions failure hai.

Conclusion. Squeeze theorem: no conclusion (walls differ). Actual limit: $x \to 0 lim sin \frac{1}{x}$ exist nahi karta.

Verify: $x_{n} = \frac{1}{2 π n + π /2}$ sample karo to $sin \frac{1}{x _{n}} = sin (2 π n + \frac{π}{2}) = + 1$ milta hai; $x_{n} = \frac{1}{2 π n - π /2}$ sample karo to $- 1$ milta hai. Do subsequences $0$ approach karte hain lekin function ek par $+ 1$ aur doosre par $- 1$ jaata hai → koi single limit nahi. ✓ Isliye tumhe zaroori hai ki walls tight hon.

Example 6 — Cell F: function ka limit at infinity

Worked example

$x \to \infty lim \frac{cos x}{x ^{2} + 1}$ evaluate karo. Forecast: ab $x$ finite point ki taraf nahi balki $\infty$ ki taraf jaata hai. Kya squeeze idea phir bhi apply hota hai? Limit guess karo.

Step 1. $- 1 \leq cos x \leq 1$ sabhi $x$ ke liye. Ye step kyun? Cosine caged hai. Trick identical hai chahe $x \to$ finite number ho ya $x \to \infty$ — hum sirf bound aur shrink karte hain.

Step 2. $x^{2} + 1$ se divide karo, jo hamesha positive hai (ek square plus $1$ ), to koi flip nahi: $- \frac{1}{x ^{2} + 1} \leq \frac{c o s x}{x ^{2} + 1} \leq \frac{1}{x ^{2} + 1} .$ Ye step kyun? $x^{2} + 1 > 0$ har jagah hai, to ye safe, sign-free division hai — koi case split needed nahi.

Step 3. Jaise $x \to \infty$ , $x^{2} + 1 \to \infty$ , to $\frac{1}{x ^{2} + 1} \to 0$ aur $- \frac{1}{x ^{2} + 1} \to 0$ . Ye step kyun? Dono walls $0$ par squeeze ho jaati hain — "at infinity" limit ki definition same closing-in idea hai.

Conclusion. $x \to \infty lim \frac{cos x}{x ^{2} + 1} === 0 == .$

Verify: $x = 10$ : $cos 10 \approx - 0.839$ , $x^{2} + 1 = 101$ , ratio $\approx - 0.0083$ , $[- 0.0099, 0.0099]$ ke andar. ✓

Example 7 — Cell G: real-world word problem (damped vibration)

Step 1. Buzz ko cage karo: $- 1 \leq cos (50 t) \leq 1$ sabhi $t$ ke liye. Ye step kyun? String kitni bhi fast vibrate kare, cosine $[- 1, 1]$ mein rehta hai. Frequency $50$ cage ke liye irrelevant hai.

Step 2. $e^{- t}$ se multiply karo. Exponentials hamesha positive hote hain ( $e^{- t} > 0$ har $t$ ke liye), to direction preserve rehti hai: $- e^{- t} \leq e^{- t} cos (50 t) \leq e^{- t} .$ Ye step kyun? Ye disguise mein phir se cell A hai — $e^{- t}$ hamaara non-negative multiplier hai. Yahan exponential kyun hai aur, say, $1/ t$ kyun nahi? Kyunki physical damping (friction, air resistance) har unit time mein energy ka ek fixed fraction remove karta hai, aur woh function jo har step par constant fraction se shrink karta hai exactly exponential $e^{- t}$ hai.

Step 3. Jaise $t \to \infty$ , $e^{- t} \to 0$ aur $- e^{- t} \to 0$ (walls, ek envelope, collapse ho jaati hain — figure mein do dashed curves dekho jo wiggle ko hug kar rahi hain). Ye step kyun? Dono walls same value $0$ par tend karti hain.

Conclusion. $t \to \infty lim e^{- t} cos (50 t) === 0 ==$ metres — string apni equilibrium position par rest mein aa jaati hai.

Verify: $t = 5$ par: $e^{- 5} \approx 0.00674$ , $cos (250) \approx - 0.087$ , to $y \approx - 0.00059$ m — size mein ek millimetre se bhi kam, $[- 0.00674, 0.00674]$ ke andar trapped. Units poore mein metres hain. ✓ Vibration ruka nahi hai, lekin uska size kuch nahi rah gaya.

Example 8 — Cell H: exam twist (tumhe bound construct karni padegi)

Worked example

$x \to 0 lim \frac{x ^{2}}{2 + sin \frac{1}{x}}$ evaluate karo. Forecast: wiggle ab denominator mein hai, upar multiply nahi hua. Yahan obvious "bounded $\times$ vanishing" nahi hai. Guess karo ki ye phir bhi $0$ hai ya nahi.

Step 1. Denominator control karo. Kyunki $- 1 \leq sin \frac{1}{x} \leq 1$ , $2$ add karne par $1 \leq 2 + sin \frac{1}{x} \leq 3.$ Ye step kyun? Exam trick ye notice karna hai ki denominator, wobbly hone ke baavajood, hamesha kam se kam $1$ hai (to kabhi zero nahi, kabhi negative nahi). Ye humein bachata hai: hum poore fraction ko uske denominator se bound kar sakte hain.

Step 2. Kyunki denominator $[1, 3]$ mein hai aur $x^{2} \geq 0$ , fraction satisfy karta hai $\frac{x ^{2}}{3} \leq \frac{x ^{2}}{2 + s i n \frac{1}{x}} \leq \frac{x ^{2}}{1} = x^{2} .$ Ye step kyun? Bada denominator positive fraction ko chota banata hai; chota denominator use bada banata hai. $x^{2}$ ko largest denominator ( $3$ ) se divide karne par lower wall milti hai, smallest ( $1$ ) se divide karne par upper wall milti hai. Dono walls non-negative hain kyunki $x^{2} \geq 0$ hai.

Step 3. Dono walls collapse ho jaati hain: $lim_{x \to 0} \frac{x ^{2}}{3} = 0$ aur $lim_{x \to 0} x^{2} = 0$ . Ye step kyun? Dono constructed walls common limit $0$ share karti hain — woh ek condition jo squeeze theorem demand karta hai.

Conclusion. Squeeze se, $x \to 0 lim \frac{x ^{2}}{2 + sin \frac{1}{x}} === 0 == .$

Verify: $x = 0.1$ : $sin (10) \approx - 0.544$ , denominator $\approx 1.456$ , fraction $\approx 0.01/1.456 \approx 0.00687$ , $[\frac{0.01}{3}, 0.01] = [0.00333, 0.01]$ ke andar. ✓

Recall Matrix par quick self-test

Kis cell mein koi conclusion nahi hai, aur kyun? ::: Cell E — dono walls alag values ( $- 1$ aur $+ 1$ ) par tend karti hain, to squeeze kuch nahi kehta; limit truly exist nahi karti. Example 2 mein $x$ ki jagah $∣ x ∣$ kyun use kiya? ::: $x$ $0$ ke left par negative hota hai aur inequality flip kar deta; $∣ x ∣ \geq 0$ kabhi flip nahi karta. Example 8 mein kya cheez denominator ko divide karne ke liye safe banati thi? ::: $2 + sin (1/ x) \geq 1 > 0$ hamesha, to ye kabhi zero ya negative nahi hota. Example 7 mein $e^{- t}$ (na ki $1/ t$ ) sahi damping wall kyun hai? ::: Physical damping har unit time mein energy ka fixed fraction remove karta hai — woh constant-fraction decay exactly exponential hai. Example 3 mein dono one-sided limits ka $1$ ke equal hona two-sided limit ke liye enough kyun hai? ::: Two-sided limit exist karta hai aur $L$ ke equal hota hai exactly tab jab dono one-sided limits exist karein aur same $L$ ke equal hoon.

Connections

Squeeze theorem (sandwich theorem) — parent rule jise ye examples exercise karte hain.
Bounded times vanishing — Examples 1, 2, 6, 7 sab ye pattern hain.
Limits of trigonometric functions — Example 3 $\frac{s i n x}{x} \to 1$ rebuild karta hai.
Limits of sequences — Example 4 $n \to \infty$ form hai.
Oscillating functions — Example 5 failure mode dikhata hai.
Continuity — yahan found har limit ek continuity check seed kar sakta hai.
Limit definition (epsilon-delta) — engine jo har "walls meet" step ke neeche hai.