Worked examples — Epsilon-delta definition of a limit — formal proofs

4.1.9 · D3 · Maths › Calculus I — Limits & Derivatives › Epsilon-delta definition of a limit — formal proofs

Agar neeche koi symbol unfamiliar lage, pehle Epsilon-delta definition of a limit — formal proofs revisit karo — lekin main har tool ko wahan explain karta hun jab woh appear hota hai.

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Scenario matrix

Ek $\epsilon$–$\delta$ problem poori tarah describe hoti hai is baat se ki kaunsa factor appear hota hai jab tum $|f(x)-L|$ ko $(\text{factor})\cdot |x-a|$ ki form mein likhte ho, aur saath mein kya weirdness $a$ ke paas hai. Yahan har cell hai:

#	Cell (kya cheez isko alag banati hai)	Core difficulty	Worked in
A	Constant function	Koi $	x-a
B	Linear, factor ek constant hai	Clean $\delta=\varepsilon/	M
C	Quadratic/polynomial, factor vary karta hai	Pehle factor ko bound karna padega	Ex 3
D	Quotient — denominator $0$ tak shrink ho sakta hai	Denominator ko $0$ se door bound karo	Ex 4
E	Square root — factor root ke andar rehta hai	$	x-a
F	Limit FALSE hai (oscillation)	Definition ko negate karo	Ex 6
G	One-sided limit / $a$ par value irrelevant	Sirf $a$ ka ek hi side count hota hai	Ex 7
H	Word problem (real tolerance)	"Within" ko $\epsilon ,\delta$ mein translate karo	Ex 8
I	Exam twist: $\delta$ depend karta hai $\epsilon$ aur point dono par	Uniform vs pointwise	Ex 9

Notation reminder (har symbol yahan earn kiya gaya hai taaki neeche kuch bhi surprise na ho):

Koi bhi algebra se pehle neeche wali figure dekho. Figure s01 woh mental image hai jo is page ke har example ke peeche hai: horizontal green strip $a$ ke aas-paas ka $\delta$-band hai; horizontal-you-look-up-into blue strip claimed limit $L$ ke aas-paas ka $\epsilon$-band hai. Limit tab exist karti hai jab — chahe adversary kitni bhi patli blue band paint kare — tum green band itna shrink kar sako ki poora curve, jahan bhi green strip cross kare, blue strip ke andar band raha ho. Ye image dhyaan mein rakho: neeche ka har proof bas compute kar raha hai ki green strip kitni narrow honi chahiye.

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Example 1 — Constant function (Cell A)

Forecast: Padhne se pehle guess karo — $\delta$ kya hoga? (Ek trick: kya $\delta$ ko $\epsilon$ par depend bhi karna zaroori hai?)

1. Output distance compute karo. $|f(x)-L|=|7-7|=0.$ Ye step kyun? Har proof vertical gap likhne se shuru hoti hai. Yahan $f$ kabhi move nahi karti, isliye gap har $x$ ke liye $0$ hai.
2. $\epsilon$ se compare karo. Kyunki $0<\epsilon$ hamesha, $|f(x)-L|=0<\epsilon$ automatically. Ye step kyun? Hum chahte hain $|f(x)-L|<\epsilon$. Ye chahe kuch bhi ho — input distance kabhi aaya hi nahi.
3. $\delta$ choose karo. Koi bhi positive number kaam karega. Lo $\delta =1$. Ye step kyun? Game mein koi na koi $\delta >0$ dena zaroori hai; value irrelevant hai kyunki step 2 ne $\delta$ use hi nahi kiya.

Proof. Let $\epsilon >0$. Choose $\delta =1$. If $0<|x-5|<1$, then $|7-7|=0<\epsilon$. $■$

Figure s01 dekho: orange line ki jagah ek flat line imagine karo. Ek flat line ka output distance har jagah $0$ hota hai, toh woh kisi bhi blue band ke andar fit ho jaati hai chahe kitni bhi patli ho — isliye $\delta$ free hai. Ye poore matrix ka degenerate corner hai.

Verify: $|7-7|=0$, aur $0<\epsilon$ har allowed $\epsilon >0$ ke liye. ✓

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Example 2 — Linear, constant factor (Cell B)

Forecast: Slope $4$ hai. Padhne se pehle $\delta$ ki form guess karo.

1. Output distance. $|(4x+3)-(-1)|=|4x+4|=4|x+1|.$ Ye step kyun? Hum gap ko $(\text{factor})\cdot |x-a|$ ki form mein likhte hain jahan $a=-1$, toh $|x-a|=|x+1|$. Ye pivot move hai (parent note): ek output bound ko input bound mein convert karo.
2. Factor identify karo. Factor hai constant $4$ — ye $x$ par depend nahi karta. Ye step kyun? Constant factor matlab koi pre-restriction ki zarurat nahi; $\delta$ ek clean fraction of $\epsilon$ hoga.
3. $\delta$ solve karo. Hum chahte hain $4|x+1|<\epsilon$, yaani $|x+1|<\epsilon /4$. Toh lo $\delta =\epsilon /4$. Ye step kyun? $\delta =\epsilon /4$ choose karne se last inequality automatic ho jaati hai.

Proof. Let $\epsilon >0$. Choose $\delta =\epsilon /4$. If $0<|x+1|<\delta$, then $|(4x+3)-(-1)|=4|x+1|<4\delta =4\cdot \frac{\epsilon }{4}=\epsilon .■$

Figure s02 dekho: orange line steep hai (slope $4$). Dekho ki $0.4$ height ka ek blue $\epsilon$-band sirf $0.1$ width ka green $\delta$-band allow karta hai — line jitni steep hogi, input window utni hi narrow hogi. Woh ratio $\frac{\text{width}}{\text{height}}=\frac{1}{4}$ exactly $\delta /\epsilon$ hai; slope amplifier hai.

Verify: Lo $\epsilon =0.1\Rightarrow \delta =0.025$. Lo $x=-1+0.02$ ($\delta$ ke andar). Tab $4x+3=4(-0.98)+3=-0.92$, aur $|-0.92-(-1)|=0.08<0.1$. ✓

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Example 3 — Quadratic, variable factor (Cell C)

Forecast: Factor constant nahi hoga. Guess karo kaunsa extra step aayega.

1. Difference factor karo. $|x^2-9|=|x-3||x+3|.$ Ye step kyun? $a=3$ hai, toh hum $|x-3|$ front mein chahte hain; bacha hua factor $|x+3|$ hai.
2. Danger notice karo. $|x+3|$ $x$ ke saath grow karta hai — constant nahi. Hum $\delta =\epsilon /|x+3|$ set nahi kar sakte (parent note ka steel-man mistake). Ye step kyun? $\delta$ ek fixed number hona chahiye jo $x$ se pehle choose kiya gaya ho; usmein $x$ nahi aa sakta.
3. Pre-restrict karo. Impose karo $|x-3|<1$. Tab $2<x<4$, toh $5<x+3<7$, jisse $|x+3|<7$ milta hai. Ye step kyun? $x$ ko $3$ ke paas trap karna variable factor ko constant $7$ se cap karta hai.
4. Dono conditions combine karo. $|x^2-9|<7|x-3|$ se hum chahte hain $7|x-3|<\epsilon$, yaani $|x-3|<\epsilon /7$. Hum chahte hain $|x-3|<1$ aur $|x-3|<\epsilon /7$, toh $\delta =\min \left(1,\frac{\epsilon }{7}\right).$ Ye step kyun? $\min$ guarantee karta hai ki cap (step 3) aur $\epsilon$-condition dono ek saath hold karein.

Proof. Let $\epsilon >0$. Choose $\delta =\min (1,\epsilon /7)$. If $0<|x-3|<\delta$: kyunki $\delta \le 1$, $|x+3|<7$; kyunki $\delta \le \epsilon /7$, $|x^2-9|=|x-3||x+3|<7|x-3|<7\cdot \frac{\epsilon }{7}=\epsilon .■$

Verify: $\epsilon =0.7\Rightarrow \delta =\min (1,0.1)=0.1$. Lo $x=3.1$: $x^2=9.61$, gap $=0.61<0.7$. ✓ Aur boundary $x=3.1$ par, $|x+3|=6.1<7$, cap confirm.

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Example 4 — Quotient, denominator near zero (Cell D)

Forecast: Naya danger division hai. Kya cheez $\left|\frac{1}{x}-\frac{1}{2}\right|$ ko blow up kar sakti hai? Guess karo kahan $x$ ko forbid karna padega.

1. Output distance. $\left|\frac{1}{x}-\frac{1}{2}\right|=\left|\frac{2-x}{2x}\right|=\frac{|x-2|}{2|x|}.$ Ye step kyun? Common denominator par combine karo taaki $|x-2|=|x-a|$ upar aaye — controllable distance.
2. Danger spot karo. Factor hai $\frac{1}{2|x|}$. Agar $x$ $0$ ki taraf drift kare, ye explode ho jaayega. Hum $x$ ko $0$ se door rakhna chahte hain. Ye step kyun? Shrinking denominator factor ko unbounded bana deta hai — Cell C se bhi bura, jahan factor sirf grow karta tha.
3. Pre-restrict karo. Impose karo $|x-2|<1$, toh $1<x<3$, hence $|x|>1$, jisse $\frac{1}{2|x|}<\frac{1}{2}.$ Ye step kyun? $x$ ko $(1,3)$ mein trap karna ise $0$ se safe distance par rakhta hai, factor ko $\frac{1}{2}$ se cap karta hai.
4. Combine karo. Ab $\left|\frac{1}{x}-\frac{1}{2}\right|<\frac{1}{2}|x-2|$. Chahiye $\frac{1}{2}|x-2|<\epsilon$, yaani $|x-2|<2\epsilon$. Toh $\delta =\min (1,2\epsilon ).$ Ye step kyun? $\min$ dono conditions enforce karta hai — "$0$ se door raho" aur $\epsilon$-condition.

Proof. Let $\epsilon >0$. Choose $\delta =\min (1,2\epsilon )$. If $0<|x-2|<\delta$: kyunki $\delta \le 1$, $|x|>1$ toh $\frac{1}{2|x|}<\frac{1}{2}$; kyunki $\delta \le 2\epsilon$, $\left|\frac{1}{x}-\frac{1}{2}\right|=\frac{|x-2|}{2|x|}<\frac{|x-2|}{2}<\frac{2\epsilon }{2}=\epsilon .■$

Figure s03 dekho: orange $\frac{1}{x}$ curve $x\to 0$ ke saath infinity ki taraf plunge karti hai. Left mein red-shaded strip forbidden zone hai — agar green $\delta$-band kabhi usme pahunch jaata, toh output $\epsilon$ chahe kitna bhi chhota ho blue band se bahar nikal jaata. Note karo ki curve $x=2$ ke paas gentle hai (small slope $\approx -\frac{1}{4}$), isliye wahan modest $\delta$ kaafi hai.

Verify: $\epsilon =0.05\Rightarrow \delta =\min (1,0.1)=0.1$. Lo $x=2.1$: $\frac{1}{x}=0.47619$, gap $=|0.47619-0.5|=0.02381<0.05$. ✓

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Example 5 — Square root, rationalise (Cell E)

Forecast: Square root ke andar se $|x-4|$ kaise nikaloge? (Hint: $a-b$ times $a+b$ kya "undo" karta hai?)

1. Output distance. $|\sqrt{x}-2|.$ Variable root ke andar stuck hai, toh $|x-4|$ abhi visible nahi. Ye step kyun? Pivot use karne ke liye hum $|x-a|$ ko expose karna chahte hain.
2. Rationalise karo. Conjugate $\sqrt{x}+2$ se upar aur neeche multiply karo: $|\sqrt{x}-2|=\frac{|\sqrt{x}-2|(\sqrt{x}+2)}{\sqrt{x}+2}=\frac{|x-4|}{\sqrt{x}+2}.$ Ye tool kyun aur koi nahi? Identity $(\sqrt{x}-2)(\sqrt{x}+2)=x-4$ woh ek hi move hai jo root difference ko plain difference $x-4$ mein convert karta hai — exactly woh distance jo hum control kar sakte hain.
3. Root defined rakhne ke liye pre-restrict karo. Impose karo $|x-4|<4$, jisse $0<x<8$ milta hai, toh $x>0$ aur $\sqrt{x}$ real number hai. Tab $\sqrt{x}\ge 0$, toh $\sqrt{x}+2\ge 2$, jisse factor $\frac{1}{\sqrt{x}+2}\le \frac{1}{2}$ milta hai. Ye step kyun? Cells C aur D ki tarah growing/exploding factor nahi hai — danger ye hai ki $\sqrt{x}$ $x<0$ ke liye exist hi karna band kar deta hai. Restriction $|x-4|<4$ $x$ ko us region mein fence karti hai jahan square root defined hai, aur bonus mein already factor ko constant $\frac{1}{2}$ se cap kar deta hai.
4. $\delta$ choose karo. Tab $|\sqrt{x}-2|\le \frac{1}{2}|x-4|$. Chahiye $\frac{1}{2}|x-4|<\epsilon$, yaani $|x-4|<2\epsilon$. Step 3 ki requirement $|x-4|<4$ ke saath combine karke: $\delta =\min (4,2\epsilon ).$ $\min$ kyun? Ye dono enforce karta hai — "$\sqrt{x}$ exist karta hai" ($|x-4|<4$ se) aur $\epsilon$-condition ek saath.

Proof. Let $\epsilon >0$. Choose $\delta =\min (4,2\epsilon )$. If $0<|x-4|<\delta$: kyunki $\delta \le 4$, $x>0$ toh $\sqrt{x}$ exist karta hai aur $\sqrt{x}+2\ge 2$; hence $|\sqrt{x}-2|=\frac{|x-4|}{\sqrt{x}+2}\le \frac{|x-4|}{2}<\frac{2\epsilon }{2}=\epsilon .■$

Verify: $\epsilon =0.1\Rightarrow \delta =\min (4,0.2)=0.2$. Lo $x=4.2$: $\sqrt{4.2}=2.04939$, gap $=0.04939<0.1$. ✓

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Example 6 — Limit FALSE prove karna (Cell F)

Forecast: Disprove karne ke liye, hum definition ko negate karte hain. Pehle kaunsa quantifier flip hota hai?

Pehle forecast answer karo — quantifiers order swap karte hain. Original definition padhti hai "$\forall \epsilon \exists \delta \dots$": har challenge $\epsilon$ ke liye ek reply $\delta$ exist karta hai jo jeette hai. Ye kehne ke liye ki limit $L$ nahi hai, hum poori sentence ko negate karte hain. Negation har $\forall$ ko $\exists$ mein aur har $\exists$ ko $\forall$ mein badal deta hai, same order mein: leading $\forall \epsilon$ ban jaata hai $\exists \epsilon$ (hum ab ek clever $\epsilon$ choose kar sakte hain), aur baad wala $\exists \delta$ ban jaata hai $\forall \delta$ (woh har $\delta$ ke liye fail hona chahiye jo claimant offer kare). Inner implication $P\Rightarrow Q$ negate hoti hai $P\text{ and not }Q$, jisse milta hai $\exists x$ jo satisfy kare $0<|x-a|<\delta$ aur $|f(x)-L|\ge \epsilon$. Toh flip ye hai: outer $\forall$ pehle $\exists$ banta hai, phir inner $\exists$ $\forall$ banta hai.

1. Suppose karo limit $L$ exist karta hai. Hum har $L$ ke liye contradiction derive karte hain. Ye step kyun? Negation ko sab candidate values defeat karna chahiye, sirf ek nahi.
2. $\epsilon =1$ pick karo. $1$ kyun? Do output values $+1$ aur $-1$ aapas mein $2$ door hain. Half-width $1$ ka band kisi bhi single $L$ ke around total width $2$ ka hoga aur dono $+1$ aur $-1$ strictly andar contain nahi kar sakta.
3. Koi bhi $\delta >0$ lo. Interval $(-\delta ,\delta )$ (minus $0$) mein hamesha ek point $x_{+}=\delta /2>0$ aur $x_{-}=-\delta /2<0$ hoga. Ye step kyun? Hum har $\delta$ ke liye ek bad $x$ produce karna chahte hain jo claimant offer kare — exactly flipped "$\forall \delta$" hai ye.
4. Contradiction. $f(x_{+})=1$ aur $f(x_{-})=-1$. $|1-L|$ aur $|-1-L|$ mein se kam se kam ek $\ge 1$ hai, kyunki agar dono $<1$ hote toh $|1-(-1)|\le |1-L|+|-1-L|<2$ — lekin $|1-(-1)|=2$, impossible. Ye step kyun? Triangle inequality force karti hai ki do gaps mein se ek $\epsilon =1$ hit kare.

Toh $\epsilon =1$ ke liye, har $\delta$ fail karta hai. Limit exist nahi karta. $■$

Figure s04 dekho: orange graph $x=0$ par $-1$ se $+1$ par jump karta hai. Half-width $1$ ka blue band most generous choice $L=0$ ke around draw kiya gaya hai; phir bhi do open circles ($0$ ke bilkul left aur right ki values) band se bahar jhankti hain. Band ki koi bhi vertical slide dono ko swallow nahi kar sakti — woh visual impossibility hi proof hai.

Verify: Arbitrary $L$ ke saath, $\max (|1-L|,|-1-L|)\ge 1$ sab real $L$ ke liye. Numeric spot-check "best" $L=0$ par: $|1-0|=1\ge 1$. ✓

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Example 7 — One-sided limit (Cell G)

Forecast: $+$ matlab "$x$ sirf right se approach karta hai." Input condition kaise badlegi?

1. Input band rewrite karo. $0<|x-0|<\delta$ ki jagah, right-hand limit use karta hai $0<x<\delta$ (sirf positive $x$, dekho One-sided limits). Ye step kyun? $\sqrt{x}$ $x<0$ ke liye real nahi, toh sirf right side meaningful hai.
2. Output distance. $|\sqrt{x}-0|=\sqrt{x}.$ Ye step kyun? $L=0$ se gap simply $\sqrt{x}$ ban jaati hai.
3. Solve karo. Chahiye $\sqrt{x}<\epsilon$. Kyunki $x>0$, dono sides square karo: $x<{\epsilon }^2$. Toh lo $\delta ={\epsilon }^2$. Squaring legal kyun hai: dono sides non-negative hain, toh squaring inequality preserve karti hai — ye woh tool hai jo root ko doosri direction mein "undo" karta hai.

Proof. Let $\epsilon >0$. Choose $\delta ={\epsilon }^2$. If $0<x<\delta ={\epsilon }^2$, then $\sqrt{x}<\sqrt{{\epsilon }^2}=\epsilon$. $■$

Verify: $\epsilon =0.3\Rightarrow \delta =0.09$. Lo $x=0.08$: $\sqrt{0.08}=0.28284<0.3$. ✓ Note karo $\delta ={\epsilon }^2$ $\epsilon$ se faster shrink karta hai — curve $0$ ke paas bahut steep hai.

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Example 8 — Word problem, real tolerance (Cell H)

Forecast: $\epsilon$ kaunsi quantity hai? $\delta$ kaunsi hai? Compute karne se pehle window guess karo.

1. Translate karo. "$8.5$ ke $0.02$ L andar" matlab $|V(t)-8.5|<\epsilon$ jahan $\epsilon =0.02$. Timing window woh $\delta$ hai jo hum find karna chahte hain. Ye step kyun? Output tolerance demand hai; input (time) tolerance humara control knob hai.
2. Output distance. $|V(t)-8.5|=|2t+0.5-8.5|=|2t-8|=2|t-4|.$ Ye step kyun? Pivot: $|t-4|=|t-a|$ factor out karo jahan $a=4$.
3. Factor identify karo aur solve karo. Factor hai constant $2$ (litres per second). Chahiye $2|t-4|<0.02$, yaani $|t-4|<0.01$. Toh $\delta =0.01$ s. Ye step kyun? Constant factor matlab $\delta =\epsilon /2=0.02/2=0.01$ — same clean division as Cell B.
4. Answer ek window ke roop mein state karo. Volume spec mein rehti hai jab bhi timer $4$ s ke $\pm 0.01$ s ke andar ho — total window of $0.02$ s. Ye step kyun? $|t-4|<0.01$ exactly interval $(3.99,4.01)$ hai; uski full width $2\delta =0.02$ s hai.

Proof. Let $\epsilon =0.02$. Choose $\delta =0.01$. If $0<|t-4|<0.01$, then $|V(t)-8.5|=2|t-4|<2\cdot 0.01=0.02=\epsilon .■$

Verify: $t=4.005$ par (window ke andar): $V=2(4.005)+0.5=8.51$, gap $=|8.51-8.5|=0.01<0.02$. ✓ Boundary $t=4.01$ par: $V=8.52$, gap $=0.02$ — exactly $\epsilon$, toh strictly andar rehne ke liye $|t-4|<0.01$ chahiye, strict window confirm. Units: rate $2$ L/s $\times$ $0.01$ s $=0.02$ L — dimensionally consistent.

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Example 9 — Exam twist: pointwise vs uniform (Cell I)

Forecast: Ex 3 mein humein mila tha $\delta =\min (1,\epsilon /7)$ $a=3$ par. Guess karo general $a$ ke liye $7$ ki jagah kya aayega.

1. General factor. $|x^2-a^2|=|x-a||x+a|.$ Pre-restriction $|x-a|<1$ se $|x+a|<|2a|+1$ milta hai. Ye step kyun? Same Cell-C machinery, lekin $a$ symbolic rakhte hain: $|x-a|<1$ se $|x|<|a|+1$ milta hai, toh $|x+a|\le |x|+|a|<|2a|+1$.
2. General $\delta$. $\delta =\min \left(1,\frac{\epsilon }{|2a|+1}\right).$ Ye step kyun? Constant $7$ (jo $2\cdot 3+1$ tha) ki jagah point-dependent cap $|2a|+1$ rakh do. Notice karo ki $\delta$ ab dono $\epsilon$ aur point $a$ par depend karta hai.
3. Trend padho. Jaise $a\to \infty$, denominator $|2a|+1\to \infty$, toh $\delta \to 0$. Ye kyun matter karta hai: wohi $\epsilon$ ko bahut door ever-smaller $\delta$ chahiye. Koi single positive $\delta$ sab $a$ ke liye simultaneously kaam nahi karta — exactly isliye $x^2$ har jagah continuous hai lekin $\mathbb{R}$ par uniformly continuous nahi hai.

Figure s05 dekho: blue curve plot karta hai largest usable $\delta$ versus point $a$ (fixed $\epsilon$ ke liye). $a=3$ par (red dot) woh $0.1$ par hai; $a=10$ par (orange dot) $0.033$ tak gir gaya hai. Curve hamesha $0$ ki taraf slide karta rehta hai — koi horizontal floor exist nahi karta, ye hai "koi single $\delta$ sab $a$ ke liye kaam nahi karta" ka visual meaning.

Verify (Ex 3 match karta hai): $a=3$ par, $|2a|+1=7$ — $\delta =\min (1,\epsilon /7)$ recover hota hai. ✓ $a=10$ par, denominator $=21$, toh $\epsilon =0.7$ ke liye, $\delta =\min (1,0.0333)=0.0333$ — $a=3$ ke mukable $3\times$ se zyada chhota (jahan $0.1$ tha).

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Coverage check

Recall Kya humne har matrix cell cover kiya?

A::: Ex 1 constant. B::: Ex 2 linear. C::: Ex 3 quadratic. D::: Ex 4 quotient. E::: Ex 5 root. F::: Ex 6 non-existence. G::: Ex 7 one-sided. H::: Ex 8 word problem. I::: Ex 9 uniform-vs-pointwise. Har cell covered.

Active recall

Connections

• Epsilon-delta definition of a limit — formal proofs — parent; ye page uska worked-example tournament hai.
• Limit Laws (sum, product, quotient) — Ex 4 ka quotient handling quotient law ka seed hai.
• Continuity — yahan har example $f$ ko $a$ par continuous bhi prove karta hai (kyunki $L=f(a)$ har baar jahan defined hai).
• One-sided limits — Ex 7 one-sided input band use karta hai.
• Uniform continuity — Ex 9 gateway hai.
• Definition of the derivative — same $\epsilon$–$\delta$ machinery difference quotients par.
• Sequences and their limits — $\delta$ ki jagah threshold index $N$ lo.