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

4.1.6 · Maths › Calculus I — Limits & Derivatives

Part 1 — $lim_{x \to 0} \frac{sin x}{x} = 1$

HOW derive karein — squeeze (sandwich) argument

Ek unit circle lo, jisme centre par ek chhota angle $x > 0$ (radians) ho.

$0 < x < \frac{π}{2}$ ke liye teen areas compare karo:

Andar wala Triangle (vertices: centre, $(1, 0)$ , circle par point): area $= \frac{1}{2} \cdot 1 \cdot sin x = \frac{s i n x}{2}$ .
Ye step kyun? Triangle area $= \frac{1}{2} \times base \times height$ ; base $= 1$ (radius), height $= sin x$ .
Circular sector: area $= \frac{1}{2} r^{2} x = \frac{x}{2}$ .
Ye step kyun? Sector area $= \frac{1}{2} r^{2} θ$ jisme $r = 1$ , $θ = x$ .
Bahar wala Triangle (tangent use karke): area $= \frac{1}{2} \cdot 1 \cdot tan x = \frac{t a n x}{2}$ .
Ye step kyun? Tangent line ek bada right triangle deta hai jiska height $tan x$ hai.

Geometrically: inner triangle $\subset$ sector $\subset$ outer triangle, isliye $\frac{s i n x}{2} \leq \frac{x}{2} \leq \frac{t a n x}{2} .$

$2$ se multiply karo aur $sin x > 0$ se divide karo: $1 \leq \frac{x}{s i n x} \leq \frac{1}{c o s x} .$ Ye step kyun? Positive quantity $sin x$ se divide karne par inequalities ki direction same rehti hai; $tan x / sin x = 1/ cos x$ .

Reciprocals lo (inequalities flip ho jaati hain): $cos x \leq \frac{s i n x}{x} \leq 1.$

Ab $x \to 0$ jaane do: $cos x \to 1$ aur right side $1$ hai. Squeeze Theorem se beech wala trap ho jaata hai: $x \to 0 lim \frac{sin x}{x} = 1.$

Part 2 — $lim_{n \to \infty} (1 + \frac{1}{n})^{n} = e$

HOW derive karein — exponent ko log mein badlo

Clean tarika: natural log lo aur Taylor/known limit use karo. $ln (1 + \frac{1}{n})^{n} = n ln (1 + \frac{1}{n}) .$ $h = \frac{1}{n} \to 0$ rakho. Tab $n ln (1 + h) = \frac{l n ( 1 + h )}{h} .$ Ye step kyun? $n = 1/ h$ , toh $n$ se multiply karna $h$ se divide karna hai — ab ye ek $0/0$ limit hai jo hum evaluate kar sakte hain.

Standard limit $lim_{h \to 0} \frac{l n ( 1 + h )}{h} = 1$ use karo (ye $ln$ ka derivative hai $1$ par, kyunki $\frac{l n ( 1 + h ) - l n 1}{h} \to \frac{d}{d x} ln x_{x = 1} = 1$ ).

Toh $ln (expression) \to 1$ , isliye expression $\to e^{1} = e$ . $■$

Common mistakes (steel-manned)

Recall

Recall Active recall — cover karo aur jawab do

Konsa theorem $\frac{s i n x}{x} \to 1$ prove karta hai, aur kaunse 3 areas compare kiye jaate hain?
$x$ radians mein kyun hona chahiye?
$\frac{1 - c o s x}{x ^{2}} \to \frac{1}{2}$ derive karo.
$lim (1 + a / x)^{x}$ kya hai aur use kaise nikaalte hain?
$1^{\infty}$ sirf $1$ kyun nahi hai?

Recall Feynman — 12 saal ke bachche ko explain karo

Sin wala part: Pizza ka ek slice lo jisme curvy crust ho. Agar slice bahut patla ho, toh seedha edge aur curvy crust basically same length ke hain — toh "slice ki height ÷ angle" exactly 1 par settle hoti hai. e wala part: Socho ek bank jo tumhe 100% interest deta hai lekin chhote-chhote chunks mein zyada zyada baar deta hai. Chahe kitni baar bhi chop karo, tumhara paisa kabhi explode nahi karta — ye lagbhag 2.718 times par ruk jaata hai. Woh magic stopping number $e$ hai.

Connections

Squeeze (Sandwich) Theorem — $\frac{s i n x}{x}$ ke peeche ka engine.
Radian Measure — kyun "1" ke liye radians chahiye.
Derivative of sin and cos — directly is limit se banta hai.
The Number e and ln — doosri limit se define hota hai.
Indeterminate Forms — $0/0$ aur $1^{\infty}$ .
L'Hôpital's Rule — ek alternative (inke liye circular hai, isliye humne geometry/log use kiya).
Taylor Series — $sin x = x - \frac{x ^{3}}{6} + \dots$ , $e^{x} = \sum x^{k} / k!$ .

Flashcards

Value of

lim_{x \to 0} \frac{s i n x}{x}

(radians)

1

Theorem used to prove

sin x / x \to 1

Squeeze (Sandwich) Theorem

Teen areas jo proof mein compare kiye jaate hain

inner triangle

\frac{s i n x}{2}

, sector

\frac{x}{2}

, outer triangle

\frac{t a n x}{2}

Squeezing ke baad inequality

cos x \leq \frac{s i n x}{x} \leq 1

Radians kyun zaroori hain

sector area

= \frac{1}{2} r^{2} θ

sirf radians mein hold karta hai; degrees mein limit

π /180

hoti hai

lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}

\frac{1}{2}

lim_{x \to 0} \frac{t a n x}{x}

1

lim_{x \to 0} \frac{s i n k x}{x}

k

e

ki definition ek limit ke roop mein

lim_{n \to \infty} (1 + 1/ n)^{n}

(1 + 1/ n)^{n}

ka indeterminate form

1^{\infty}

1^{\infty}

limits evaluate karne ki trick

natural log lo,

0/0

form milega

lim_{x \to \infty} (1 + a / x)^{x}

e^{a}

lim_{x \to 0} (1 + x)^{1/ x}

e

lim_{n \to \infty} (1 + 1/ n^{2})^{n}

1

(log

= 1/ n \to 0

)

e

ke liye use hone wala key log limit

lim_{h \to 0} \frac{l n ( 1 + h )}{h} = 1

lim_{n \to \infty} (\frac{n + 2}{n - 1})^{n}

e^{3}