Visual walkthrough — Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e

4.1.6 · D2 · Maths › Calculus I — Limits & Derivatives › Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e

Hum ek line prove kar rahe hain: ${\lim }_{x\to 0}\frac{\sin x}{x}=1,x\text{ measured in }==\text{radians}==.$

Neeche diya gaya har cheez us line ka har piece earn karta hai.

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Step 1 — Stage draw karo: ek unit circle aur ek angle

KYA. Ek circle banao jiska radius $1$ ho aur jo origin $O$ par centred ho. Woh point $B=(1,0)$ mark karo jahan circle positive $x$-axis ko cross karti hai. Angle $x$ counter-clockwise sweep karo aur circle par point $P$ par land karo.

KYUN. Exactly $1$ radius wala circle isliye choose kiya jata hai taaki lengths aur angles ek hi number ban jaayein — yahi poora trick hai, aur Step 2 ise explain karta hai. Jo bhi hum compare karenge woh sab ek hi picture ke andar hoga, isliye koi bhi symbol bina kisi jagah ko point kiye appear nahi karega.

PICTURE. Blue circle dekho. $O$ centre hai, $B$ axis par red dot hai, $P$ rim par orange dot hai. $B$ se $P$ tak ki choti arc (green) woh hai jo angle "cut out" karta hai.

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Step 2 — "Radian" ka matlab kya hai, aur $\text{arc}=x$ kyun hota hai

KYA. $x$ ki measure ko green arc $BP$ ki length ke roop mein define karo. Yahi definition hai jo "radians" ka matlab hai.

KYUN. Hume $x$ (ek angle, ek rotation) aur ek length (kuch aisa jo hum $\sin x$ se compare kar sakein) ko same quantity banana hai. Radian measure yeh literally true banata hai: radius $r$ wale circle par, arc length $=r\theta$ hoti hai, aur $r=1$ ke saath hume arc $=x$ milta hai. Agar hum degrees use karte, toh arc $=\frac{\pi }{180}x$ hota — ek alag number — aur clean "$=1$" break ho jaata. Yahi akela reason hai ki poora result radians ki zaroorat kyun hai.

PICTURE. Green arc apni khud ki length ke saath labelled hai: woh hai $x$. Same curve, do readings — "angle" aur "length."

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Step 3 — $\sin x$ aur $\tan x$ ko heights ke roop mein padhna

KYA. $P$ se seedha $x$-axis tak ek vertical line drop karo; uski height $\sin x$ hai. Ab doosri taraf: circle par $B$ pe vertical tangent line banao, aur ray $OP$ ko extend karo jab tak woh us tangent line ko point $T$ par hit na kare. $T$ ki axis se upar ki height $\tan x$ hai.

KYUN. Yeh do heights hi woh do "competitors" hain jinke beech hum $x$ ko squeeze karenge. Hume geometric objects chahiye, formulas nahi, toh dono ko directly picture par define karte hain:

$\sin x=\underset{\text{vertical drop from the rim}}{\underbrace{\text{height of }P}},\tan x=\underset{\text{ray extended to the tangent}}{\underbrace{\text{height of }T}}.$

PICTURE. Orange segment $=\sin x$ (chota, circle par khatam hota hai). Red segment $=\tan x$ (zyada lamba, bahar wali tangent line par khatam hota hai). Aankhon se notice karo: orange height arc se choti hai, aur arc red height se choti hai. Yahi ordering poora proof hai — Steps 4–6 ise exact banate hain.

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Step 4 — Teen nested regions (sandwich ki bread)

KYA. Teen regions dekho jo ek doosre ke andar baithi hain:

1. inner triangle $OBP$ (vertices $O$, $B$, $P$),
2. circular sector $OBP$ (pie slice jo do radii aur arc se bounded hai),
3. outer triangle $OBT$ (vertices $O$, $B$, $T$).

Kyunki $P$ circle par hai aur $T$ uske bahar hai, region 1 $\subset$ region 2 $\subset$ region 3.

KYUN. Areas compute karna easy hai aur woh containment respect karte hain: agar ek shape doosre ke andar baithi hai, toh uska area bada nahi hoga. Toh containment hume ek inequality muft mein deta hai, bina kisi trig identity ke. Woh inequality $x$ ko trap kar legi.

PICTURE. Same figure par teen shaded layers: filled orange (inner triangle), uske upar green pie slice, peeche pale red outer triangle. Nesting visible hai — har bada region pichle ko contain karta hai.

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Step 5 — Teen regions ko teen areas mein badalna

KYA. Har ek area compute karo.

$$\underset{\text{inner }△}{\underbrace{\frac{1}{2}\sin x}}\le \underset{\text{sector}}{\underbrace{\frac{1}{2}x}}\le \underset{\text{outer }△}{\underbrace{\frac{1}{2}\tan x}}.$$

Term by term:

KYUN. Hum pictures (nested shapes) ko numbers (areas) se swap kar rahe hain, Step 4 ki same $\le$ ordering rakhte hue. Ab hum algebra kar sakte hain.

PICTURE. Same teen regions, har ek apne area ke saath labelled. Right side par bars teen magnitudes ko ek chote angle par compare karte hain taaki tum dekh sako $\frac{\sin x}{2}<\frac{x}{2}<\frac{\tan x}{2}$.

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Step 6 — Algebra: sandwich ko $\frac{\sin x}{x}$ ke around fold karo

KYA. Step 5 se shuru karo aur clean up karo.

Teeno parts ko $2$ se multiply karo: $\sin x\le x\le \tan x.$

$\sin x$ se divide karo. $0<x<\frac{\pi }{2}$ ke liye $\sin x>0$ hai, toh positive number se divide karne par har $\le$ same direction mein rehta hai: $1\le \frac{x}{\sin x}\le \frac{\tan x}{\sin x}=\frac{1}{\cos x}.$ (Yahan $\frac{\tan x}{\sin x}=\frac{\sin x/\cos x}{\sin x}=\frac{1}{\cos x}$ — $\sin x$ cancel ho jaata hai.)

Reciprocals lo. Positive numbers ke $a\le b\le c$ ko $\frac{1}{a}\ge \frac{1}{b}\ge \frac{1}{c}$ mein flip karna har $\le$ ko reverse karta hai, toh: $\cos x\le \frac{\sin x}{x}\le 1.$

KYUN. Hum chahte the ki $\frac{\sin x}{x}$ middle mein akela trapped ho, un do cheezón ke beech jinhe hum samajhte hain. Har move (multiply, positive $\sin x$ se divide, reciprocate) ek legal inequality step hai, aur humne clearly bataya kyun direction same raha ya flip hua.

PICTURE. Ek number line: value $\frac{\sin x}{x}$ (orange tick) floor $\cos x$ (blue) aur ceiling $1$ (gray) ke beech caught hai. Jaise $x$ chhota hota hai floor $1$ ki taraf utha aur gap slam shut ho jaata hai.

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Step 7 — $x\to 0$ jaane do: Squeeze Theorem trap band karta hai

KYA. $x\to 0$ bhejo. Ceiling constant $1$ hai. Floor $\cos 0=1$ hai, toh $\cos x\to 1$ hota hai. Dono walls $1$ par converge karti hain: $\underset{\to 1}{\underbrace{\cos x}}\le \frac{\sin x}{x}\le \underset{=1}{\underbrace{1}}.$

Squeeze (Sandwich) Theorem ke by, koi bhi cheez jo do ऐसी cheezón ke beech trapped ho jo dono $1$ par jaati hain, khud bhi $1$ par jaani chahiye: $\boxed{\underset{x\to 0^{+}}{\lim }\frac{\sin x}{x}=1.}$

KYUN. Do known limits ke beech sandwiched limit unka common value inherit karti hai — yahi exactly woh kaam hai jiske liye Squeeze Theorem hai, aur precisely woh tool hai jo yeh shape hume deti hai.

PICTURE. Floor $\cos x$ aur ceiling $1$ curves ke roop mein plotted; orange curve $\frac{\sin x}{x}$ unke beech se thread karti hai aur teeno single point $(0,1)$ par milte hain.

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Step 8 — Doosri side aur exact centre (koi case chhuta nahi)

KYA. Humne sirf right limit ($x\to 0^{+}$) prove ki. Do baaki scenarios check karo.

Negative side, $x\to 0^{-}$. Function $\frac{\sin x}{x}$ even hai: $x$ ko $-x$ se replace karo, $\frac{\sin (-x)}{-x}=\frac{-\sin x}{-x}=\frac{\sin x}{x}.$ Toh graph vertical axis ke across mirror image hai — left limit right limit ke barabar hai. Isliye two-sided limit $1$ hai.

Exactly $x=0$. Expression hai $\frac{\sin 0}{0}=\frac{0}{0}$: undefined, graph mein ek hole. Limit $1$ hai, lekin $0$ par value exist nahi karti. Limits approach describe karte hain, landing point nahi.

KYUN. Parent claim ek two-sided limit hai; one-sided proof sirf aadha kaam hai. $x<0$ aur degenerate $x=0$ point ko cover karna hi boxed result ko airtight banata hai.

PICTURE. $\frac{\sin x}{x}$ ka full graph negative aur positive $x$ ke across: symmetric hump jo $1$ par peak karta hai, exactly $(0,1)$ par ek open circle (hole) drawn ke saath.

Recall Cases par quick check

Kya $\frac{\sin x}{x}$ $x=0$ par defined hai? ::: Nahi — yeh $0/0$ hai, ek hole. Sirf limit $1$ hai. $x\to 0^{-}$ same answer kyun deta hai? ::: Function even hai, toh uska graph $y$-axis ke baare mein mirror-symmetric hai; left aur right limits match karte hain.

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Ek picture mein summary

Sab kuch compressed: nested inner triangle $\subset$ sector $\subset$ outer triangle deta hai $\sin x\le x\le \tan x\Rightarrow \cos x\le \frac{\sin x}{x}\le 1\stackrel{x\to 0}{\to }1.$

Recall Feynman retelling — poora walkthrough simple words mein

Radius one ka ek circle banao aur ek chota angle kholo. Kyunki hum angles ko is circle par jo arc woh cut karte hain usse measure karte hain, angle number exactly arc length hai — isliye hum radians par insist karte hain. Ab teen shapes ko ek doosre ke saath khadha karo: pie slice ke andar ek patla triangle, pie slice khud, aur uske bahar ek mota triangle. Kyunki har shape agli ke andar packed hai, unke areas line up hote hain $\frac{\sin x}{2}\le \frac{x}{2}\le \frac{\tan x}{2}$ ki tarah. Algebra clean karo — do se multiply karo, positive $\sin x$ se divide karo, reciprocals flip karo — aur $\frac{\sin x}{x}$ $\cos x$ aur $1$ ke beech pin ho jaata hai. Angle chhotaa karo: floor $\cos x$ $1$ ki taraf uthta hai, ceiling pehle se $1$ hai, toh beech wali bechare cheez ke paas $1$ ke siwa kahin jaane ki jagah nahi hai. Graph left pe mirror image hai, aur zero par sirf ek chota sa hole hai — lekin woh value jis taraf jaati hai dono sides se $1$ hai. Ho gaya.

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Connections

• Squeeze (Sandwich) Theorem — Step 7 ka closing move.
• Radian Measure — woh reason ki arc $=x$ (Step 2) aur sector $=\frac{1}{2}x$ (Step 5).
• Derivative of sin and cos — yeh limit hai $\frac{d}{dx}\sin x$ at $0$.
• Indeterminate Forms — kyun $\frac{\sin 0}{0}=\frac{0}{0}$ ko ek real argument chahiye.
• L'Hôpital's Rule — ${\sin }^{\prime }$ ki zaroorat padti, jo is limit ki zaroorat rakhta hai, toh yahan circular hai.
• Taylor Series — $\sin x=x-\frac{x^3}{6}+\cdots$ se $\frac{\sin x}{x}=1-\frac{x^2}{6}+\cdots \to 1$ milta hai.