Epsilon-delta definition of a limit — formal proofs

4.1.9 · Maths › Calculus I — Limits & Derivatives

WHY — yeh define karne ki zaroorat hi kya hai?

Yeh casual phrase ki " $f (x)$ , $L$ ke paas aata hai jab $x$ , $a$ ke paas aata hai" prove karne ke liye bahut vague hai:

"Paas" — kitna paas?
Kya $f$ ko $L$ tak pahunchna zaroori hai? (Nahi — limits point $a$ ko khud ignore karti hain.)
Kya $f (x)$ , $a$ ke paas hamesha hilta-duldta reh sakta hai? (Tab koi limit exist nahi karti, lekin yeh vague phrase isko pakad nahi sakta.)

Epsilon–delta definition fuzzy words ko ek measurable tolerances wale game se replace kar deta hai, taaki limit kuch aisa ban jaye jiske baare mein tum actually argument jeet sako.

WHAT — definition kya hai?

Har symbol ka matlab:

$∣ f (x) - L ∣ < ε$ → output $L$ ke aas-paas $ε$ half-width ki ek band ke andar hai.
$0 < ∣ x - a ∣ < δ$ → input $a$ ke aas-paas $δ$ half-width ki ek band ke andar hai. Strict $0 <$ $x = a$ ko exclude karta hai (limit ko parwah nahi ki $a$ par kya hota hai).
Quantifier ka order matter karta hai: $ε$ pehle tumhe diya jaata hai (challenge), phir tum $δ$ dhundte ho (tumhara jawab). Tumhara $δ$ , $ε$ par depend kar sakta hai.

Figure — Epsilon-delta definition of a limit — formal proofs

HOW — proof likhne ka tarika (universal recipe)

Har $ε$ – $δ$ proof ke do halve hote hain:

Scratchwork ( $δ$ dhundo): Goal $∣ f (x) - L ∣ < ε$ se shuru karo aur ulta kaam karo, $∣ x - a ∣$ ko factor out karte hue, jab tak pata na chal jaye ki $δ$ kya hona chahiye.
Clean proof (seedha): "Let $ε > 0$ . Choose $δ = \dots$ . Phir assume karo $0 < ∣ x - a ∣ < δ$ aur derive karo $∣ f (x) - L ∣ < ε$ ."

Example 1 — ek linear function (sabse saaf case)

Claim: $x \to 3 lim (2 x + 1) = 7.$

Scratchwork. Yahan $f (x) - L = (2 x + 1) - 7 = 2 x - 6 = 2 (x - 3)$ . To $∣ f (x) - L ∣ = 2∣ x - 3∣.$ Yeh step kyun? Humne $∣ x - 3∣ = ∣ x - a ∣$ , controllable input distance, ko factor out kiya. $2∣ x - 3∣ < ε$ banane ke liye humein $∣ x - 3∣ < ε /2$ chahiye. To $δ = ε /2$ lo.

Proof. Let $ε > 0$ . Choose $δ = \frac{ε}{2}$ . Suppose $0 < ∣ x - 3∣ < δ$ . Tab $∣ (2 x + 1) - 7∣ = 2∣ x - 3∣ < 2 δ = 2 \cdot \frac{ε}{2} = ε . ■$ Yeh kyun kaam karta hai: factor constant $2$ tha, isliye $δ$ , $ε$ ka ek clean fraction hai.

Example 2 — ek quadratic (factor ko pehle bound karna padega)

Claim: $x \to 2 lim x^{2} = 4.$

Scratchwork. $∣ x^{2} - 4∣ = ∣ x - 2∣ ∣ x + 2∣$ . Factor $∣ x + 2∣$ constant nahi hai — yeh $x$ par depend karta hai. Yeh problem kyun hai? Hum $δ = ε /∣ x + 2∣$ nahi pick kar sakte kyunki $δ$ ek fixed number hona chahiye, $x$ par depend nahi kar sakta.

Fix — $x$ ko pehle restrict karo. Pehle ek preliminary restriction $∣ x - 2∣ < 1$ lagao. Tab $1 < x < 3$ , to $∣ x + 2∣ = x + 2 < 5.$ Yeh step kyun? Jab $x$ , $2$ ke paas ho, to factor $∣ x + 2∣$ kabhi $5$ se zyada nahi ho sakta — humne mushkil factor ko ek constant se bound kar diya. Ab $∣ x^{2} - 4∣ = ∣ x - 2∣ ∣ x + 2∣ < 5 ∣ x - 2∣.$ $5∣ x - 2∣ < ε$ ke liye humein $∣ x - 2∣ < ε /5$ chahiye. Humein dono $∣ x - 2∣ < 1$ aur $∣ x - 2∣ < ε /5$ chahiye, to chota lo: $δ = min (1, \frac{ε}{5}) .$

Proof. Let $ε > 0$ . Choose $δ = min (1, ε /5)$ . Suppose $0 < ∣ x - 2∣ < δ$ . Kyunki $δ \leq 1$ , isliye $∣ x + 2∣ < 5$ hai. Kyunki $δ \leq ε /5$ , $∣ x^{2} - 4∣ = ∣ x - 2∣ ∣ x + 2∣ < 5∣ x - 2∣ < 5 \cdot \frac{ε}{5} = ε . ■$

Example 3 — prove karna ki limit FALSE hai (negation)

Claim: $x \to 0 lim \frac{∣ x ∣}{x}$ exist nahi karti.

Definition ko negate karne ke liye, quantifiers swap karo:

Yahan $f (x) = 1$ jab $x > 0$ aur $f (x) = - 1$ jab $x < 0$ . Maan lo koi claim kare ki limit $L$ hai. $ε = 1$ lo. Kisi bhi $δ > 0$ ke liye, interval $(- δ, δ)$ mein dono ek point $x_{+} > 0$ (jahan $f = 1$ ) aur $x_{-} < 0$ (jahan $f = - 1$ ) hote hain. Yeh dono $2$ se alag hain, isliye dono ek saath $L$ ke $1$ ke andar nahi ho sakte. Isliye koi bhi $L$ kaam nahi karta. $ε = 1$ kyun? Yeh itna chota hai ki $ε$ -band ek saath $+ 1$ aur $- 1$ dono ko straddle nahi kar sakta.

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho tum ek paani ki pipe se ek target par nishana laga rahe ho. Ek dost kehta hai "mujhe paani is chote ring ke andar chahiye." Woh ring hai $ε$ . Tum jawab dete ho: "Tab mera haath sahi angle se itni choti wobble ke andar rakho" — woh wobble hai $δ$ . Pipe ki real limit tab hoti hai jab chahiye kitni bhi choti ring dost draw kare, tum hamesha ek itni choti hand-wobble dhundh sako jo paani ko andar rakhe. Agar woh itni choti ring pick kare ki tum kabhi paani andar na rakh sako — wahan koi limit nahi hai.

Active recall

lim_{x \to a} f (x) = L

ka formal matlab kya hai?

\forall ε > 0 \exists δ > 0

s.t.

0 < ∣ x - a ∣ < δ \Rightarrow ∣ f (x) - L ∣ < ε

0 < ∣ x - a ∣

(strict) kyun hona chahiye?

x = a

ko exclude karne ke liye; limit

f

ki value ko (ya non-value ko)

a

par hi ignore karti hai.

Pehle kya choose hota hai,

ε

δ

ε

(challenge) pehle;

δ

ε

par depend kar sakta hai.

∣ f (x) - L ∣

ko kis general shape mein rewrite karna chahiye?

(koi factor)

\times ∣ x - a ∣

, taaki input bound output bound ban jaye.

lim_{x \to 3} (2 x + 1) = 7

ke liye kaunsa

δ

kaam karta hai?

δ = ε /2

, kyunki

∣ (2 x + 1) - 7∣ = 2∣ x - 3∣

Quadratic ke liye

δ = min (1, ε /5)

kyun use karte hain?

min

dono conditions enforce karta hai — preliminary bound (taaki variable factor

\leq

constant rahe) aur

ε

-condition.

Limit definition ka negation kya hai?

\exists ε > 0 \forall δ > 0 \exists x : 0 < ∣ x - a ∣ < δ

and

∣ f (x) - L ∣ \geq ε

δ

x

par kyun depend nahi kar sakta?

δ

x

choose hone se pehle commit ho jaata hai;

x

-dependent

δ

valid response nahi hai.

Kaunsa

ε

prove karta hai ki

∣ x ∣/ x

ki 0 par koi limit nahi?

ε = 1

: 0 ke aas-paas kisi bhi interval mein aise points hote hain jo

+ 1

aur

- 1

dete hain, jo

2

apart hain.

Connections

Limit Laws (sum, product, quotient) — har law $ε$ – $δ$ se prove hoti hai.
Continuity — $f$ , $a$ par continuous hoti hai iska matlab hai $lim_{x \to a} f (x) = f (a)$ (same definition, $L = f (a)$ ke saath).
One-sided limits — $0 < ∣ x - a ∣ < δ$ ko $a < x < a + δ$ ya $a - δ < x < a$ se replace karo.
Definition of the derivative — derivative ek $ε$ – $δ$ limit hai difference quotients ki.
Uniform continuity — same game, lekin $δ$ ko sabhi $a$ ke liye ek saath kaam karna hoga.
Sequences and their limits — discrete analogue jisme $δ$ ki jagah $N$ hota hai.