Epsilon-delta definition of a limit — formal proofs: Visual walkthrough

4.1.9 · D2 · Maths › Calculus I — Limits & Derivatives › Epsilon-delta definition of a limit — formal proofs

Hum ek concrete limit lenge aur definition ko picture-by-picture squeeze karenge: $lim_{x \to 2} x^{2} = 4.$ Baaki sab kuch (linear case, false limits) is ek worked example ki chhaya hai.

Step 1 — Do number lines, do dots

KYUN. Limit ek baar mein do jagah ka promise hai: "agar input yahan ke paas hai, toh output wahan ke paas hoga." Aap "near" ke baare mein tab tak soch nahi sakte jab tak aap do alag rulers par do neighbourhoods nahi dekh lete. Ek ruler kaafi nahi — poora drama unke link ke baare mein hai.

PICTURE. Figure dekho. Neeche ki violet line input hai; $x = 2$ par dot woh jagah hai jahan hum aim karte hain. Upar ki magenta line output hai; $y = 4$ par dot woh jagah hai jahan hum land karna chahte hain. Orange curve $y = x^{2}$ rule $f$ hai jo ek point ko neeche ki line se upar ki line tak le jaata hai.

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

$x$ — ek chosen input, ek point jo neeche ke ruler par freely slide kar sakta hai.
$a = 2$ — woh input jiske paas hum ja rahe hain (ek dot, koi value nahi jo hum plug in karein).
$L = 4$ — woh output jiske paas hum jaana chahte hain.
Orange arrow rule $f (x) = x^{2}$ dikhata hai: yeh $x = 2$ ko $y = 4$ tak le jaata hai.

Step 2 — Adversary output band banata hai ( $ε$ )

YEH SYMBOL KYUN, AUR PEHLE KYUN? "L ke paas" ka koi matlab nahi jab tak koi kitna paas nahi bataata. Number $ε$ wahi "kitna paas" hai — yeh ek measured tolerance hai, koi mood nahi. Aur yeh pehle aata hai kyunki poori baat yeh hai ki aap tolerance choose nahi kar sakte — challenger karta hai, aur chahein toh use jitna chhota karna ho, kar sakta hai.

PICTURE. $y = 4$ ke aas-paas magenta band ki half-height $ε$ hai. Shaded strip ke andar land karne wali koi bhi cheez output side par jeet hai.

Step 3 — Band ko curve ke through wapas kheencho

KYUN. Definition mein curve par ghoomne ke baare mein kuch nahi hai — yeh sirf inputs aur outputs jaanta hai. Toh hum poochhte hain: kaunse inputs output ko band ke andar produce karte hain? Geometrically yeh bilkul waisa hi hai jaise "band ko sideways curve par shine karo, phir input ruler par shadow padho." Woh shadow unse guaranteed jeetne wale inputs ka set hai.

PICTURE. Do magenta edges orange curve par do points par milti hain; dashed navy lines input ruler par drop hoti hain aur ek chhota interval kaatti hain. Dhyan do ki yeh $x = 2$ ke baare mein symmetric nahi hai — left gap aur right gap alag hain. Yeh asymmetry Step 6 mein aane wale "min" shabd ka beej hai.

Shadow ka left edge: $x^{2} = 4 - ε$ solve karta hai, yaani $x = 4 - ε$ .
Shadow ka right edge: $x^{2} = 4 + ε$ solve karta hai, yaani $x = 4 + ε$ .
Kaunsa root? Equation $x^{2} = 4 \pm ε$ ke do solutions hain, $\pm 4 \pm ε$ . Hum sirf positive root rakhte hain kyunki hum $x = 2 > 0$ ke paas kaam kar rahe hain; negative root $x = - 2$ ke paas hai, hamare target se door, isliye yeh yahan irrelevant hai.
Kyunki curve right side par zyada steep hai, right gap thodi chhoti hai.

Step 4 — Sabse bada symmetric band fit karo jo fit ho ( $δ$ )

SYMMETRIC $δ$ -BAND KYUN, JABKI SHADOW LOPSIDED HAI? Definition $0 < ∣ x - a ∣ < δ$ ke roop mein likhi hai — har direction mein ek single distance $δ$ . Yeh ek deliberate simplification hai: ise kehna aur reuse karna aasaan hai. Price yeh hai ki hum narrower side tak shrink karein taaki symmetric band kabhi bahar na nikaale. Woh "narrower side tak shrink karna" bilkul wahi $min$ hai jo aap Step 6 mein milenge.

PICTURE. Half-width $δ$ ka violet band shadow ke andar baitha hai. $x = 2$ par chhota hollow circle excluded point hai — hum kabhi target par actually khade nahi hote.

Step 5 — Kyun " $∣ x - a ∣$ factor out karna" honest bookkeeping hai

YEH KYUN AUR BRUTE SQUARE-ROOTS KYUN NAHI? Hume ek aisa rule chahiye jo "input $δ$ ke andar hai" ko har $x$ ke liye, cleanly, "output $ε$ ke andar hai" mein convert kare. Factoring machine expose karta hai: output distance input distance hai jo stretch factor $∣ x + 2∣$ se multiply hoti hai. Input distance control karo aur stretch control karo, aur output distance automatically maanta hai.

PICTURE. Figure $x = 2$ ke paas curve aur local "stretch" dikhata hai: input ko tiny step $∣ x - 2∣$ hilane par output lagbhag $∣ x + 2∣$ times itna bada step move hota hai. Curve jitna steep, stretch utna bada — yahi $∣ x + 2∣$ measure karta hai.

$∣ x - 2∣$ — aapka input target se kitna door gaya; woh cheez jo aap $δ$ ke saath directly control karte hain.
$∣ x + 2∣$ — amplification: $x = 2$ ke paas yeh lagbhag $4$ hai, isliye $x$ mein size $d$ ka wobble $y$ mein lagbhag $4 d$ ka wobble ban jaata hai.

Step 6 — Stretch ko tame karo, phir $δ = min$ choose karo

EK NAZAR MEIN $min$ KYUN. Do constraints, do candidate widths. $δ$ ko dono satisfy karna hai, isliye woh either se bada nahi ho sakta — yahi "minimum" ka matlab hai. Yeh Steps 3–4 ka symmetric-band-inside-the-lopsided-shadow idea hai, ab algebra mein.

PICTURE. Figure do candidate half-widths ( $1$ aur $ε /5$ ) ko do bars ke roop mein stack karta hai; chhoti bar jeetti hai aur $δ$ ban jaati hai. Chhote $ε$ ke liye $ε /5$ bar chhoti hai ( $ε$ -condition karti hai); bade $ε$ ke liye constant $1$ bar jeetti hai (cage karti hai).

Step 7 — Degenerate & edge cases (kuch nahi chhoota)

Case A — bada $ε$ (loose demand). Agar $ε = 100$ , toh $ε /5 = 20$ , aur $min (1, 20) = 1$ . Cage jeetta hai; $δ = 1$ . Kyun theek hai: loose demand chhota $δ$ force nahi kar sakta — aap apna comfortable margin rakhte hain.

Case B — chhota $ε$ (brutal demand). Agar $ε = 0.05$ , toh $ε /5 = 0.01$ , aur $min (1, 0.01) = 0.01$ . $ε$ -rule jeetta hai; $δ = 0.01$ . Kyun theek hai: output band jaise-jaise shrink hota hai, aapka input band saath shrink hota hai — aur yeh hamesha kar sakta hai, kyunki $ε /5 > 0$ har $ε > 0$ ke liye.

Case C — linear function, jahan stretch constant hai. $lim_{x \to 3} (2 x + 1) = 7$ ke liye hume $∣ f (x) - L ∣ = 2∣ x - 3∣$ milta hai: stretch constant $2$ hai, koi cage zaroori nahi, aur $δ = ε /2$ exactly. Yahi wajah hai ki straight lines easy case hain — stretch mein koi $x$ chhupa nahi hai, isliye koi $min$ nahi.

PICTURE. Left panel: parabola ka shadow lopsided hai (stretch vary karta hai) → $min$ zaroori. Right panel: line ka shadow perfectly symmetric hai (stretch constant) → clean $δ = ε /2$ .

Recall Arithmetic khud check karo

$ε = 0.05$ aur $δ = 0.01$ ke saath, sabse bura input $x = 2.01$ hai. Tab $x^{2} = 4.0401$ , isliye $∣ x^{2} - 4∣ = 0.0401 < 0.05$ . ✓ $ε = 100$ , $δ = 1$ , sabse bura input $x = 3$ : $x^{2} = 9$ , $∣9 - 4∣ = 5 < 100$ . ✓

Formal statement, ab poori tarah earned

Ek-picture summary

Recall Feynman retelling — ise ek story ki tarah bolo

Aap aur ek dost $y = x^{2}$ ki tarah shaped ek magic funnel ke through catch khelate hain. Aapka dost "4" marked spot ke aas-paas wall par ek chhota hoop draw karta hai — us hoop ki tightness $ε$ hai. Aapka kaam: "2" ke paas aim karte hue apna throwing hand kitna steady rakhna hai, yeh dhundho, taaki ball hamesha hoop se guzre. Aap notice karte hain ki funnel ek side par zyada bend karta hai, isliye safe hand-wobble left aur right alag hai — isliye aap dono taraf ki chhoti wobble ka promise karte hain; wahi aapka $δ$ hai, aur "chhota lo" ka matlab shabd min hai. Kyunki funnel target ke paas throws ko sirf lagbhag char-guna stretch karta hai, hoop ko aadha karne se aapko sirf apni wobble aadhi karni padti hai — kabhi zero nahi. Kyunki aap koi bhi hoop jo aapka dost draw kare use meet kar sakte hain, ball ki sacchi destination "4" hai. Yahi exactly $lim_{x \to 2} x^{2} = 4$ hai.

Active recall

Do rulers kaunse hain par limit rehti hai?

input line (

a

ke aas-paas) aur output line (

L

ke aas-paas); rule/curve unhe link karta hai.

ε

kya measure karta hai, aur kaun use pick karta hai?

output band ki half-height; adversary use pehle pick karta hai.

y = x^{2}

ke liye

ε

-band ka pulled-back shadow asymmetric kyun hai?

curve right par zyada steep hai, isliye same output tolerance wahan chhota input gap allow karta hai.

x^{2} = 4 \pm ε

ka sirf positive root kyun rakhte hain?

hum

x = 2 > 0

ke paas hain; negative root

x = - 2

ke paas hai, target se door, isliye yeh irrelevant hai.

δ

ko shadow ki narrower side tak kyun shrink karna chahiye?

0 < ∣ x - a ∣ < δ

ek symmetric width use karta hai, isliye woh tighter side se zyada nahi ho sakta — isliye

min

∣ x^{2} - 4∣ = ∣ x - 2∣ ∣ x + 2∣

mein

∣ x + 2∣

visually kya represent karta hai?

local stretch factor: curve kitna ek input wobble ko output wobble mein amplify karta hai (

x = 2

ke paas lagbhag

4

δ

choose karne se pehle

x

∣ x - 2∣ < 1

se cage kyun karte hain?

moving stretch

∣ x + 2∣

ko constant bound

5

se replace karne ke liye, taaki

δ

ek fixed number ho sake.

ε = 0.05

ke liye

δ

kya hai aur kya yeh kaam karta hai?

δ = min (1, 0.01) = 0.01

; worst

x = 2.01

deta hai

∣ x^{2} - 4∣ = 0.0401 < 0.05

. ✓

Linear case mein

min

kyun zaroori nahi?

stretch constant

2

hai,

x

se independent, isliye

δ = ε /2

directly kaam karta hai.

Connections

Epsilon-delta definition of a limit — formal proofs — parent: definition, recipe, aur in pictures ke peeche ki algebra.
Definition of the derivative — derivative wahi $∣ x + 2∣$ -style slopes ke limit hai jo stretch factor hai.
Continuity — same two-band picture $L = f (a)$ ke saath: shadow ab target point include karta hai.
One-sided limits — shadow ka sirf left ya right half use karo.
Limit Laws (sum, product, quotient) — har law Step 5 ka "factor and bound" move reuse karta hai.
Uniform continuity — ek $δ$ -band demand karo jo shadow ko har $a$ par ek saath fit kare.

Visual walkthrough — Epsilon-delta definition of a limit — formal proofs