Visual walkthrough — Limit laws — sum, product, quotient, constant multiple

4.1.2 · D2 · Maths › Calculus I — Limits & Derivatives › Limit laws — sum, product, quotient, constant multiple

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Step 0 — Ek word jo pehle define karna zaroori hai: "limit"

Kisi bhi law se pehle, humein agree karna hoga ki ${\lim }_{x\to a}f(x)=L$ mein arrow ka actually matlab kya hai. Koi bhi symbol apni jagah tab tak nahi banata jab tak hum usse point na kar sakein.

Term by term, neeche di gayi picture mein:

• $a$ — woh input jiske taraf hum creep kar rahe hain (horizontal axis pe).
• $L$ — woh height jiske taraf outputs ja rahe hain (vertical axis pe).
• $\epsilon$ (epsilon) — $L$ ke aas-paas ek horizontal band ki half-height. Yeh aapka challenge hai: "is band ke andar aao."
• $\delta$ (delta) — $a$ ke aas-paas ek vertical strip ki half-width. Yeh mera jawab hai: "is strip mein raho aur tum apni band ke andar rahoge."

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Step 1 — Do functions set up karo jo hum add kar rahe hain

KYA. Hum assume karte hain ki do limits pehle se exist karti hain: ${\lim }_{x\to a}f(x)=L{\lim }_{x\to a}g(x)=M.$

• $f$ height $L$ ki taraf jaata hai; $g$ height $M$ ki taraf jaata hai.
• Hum unke sum $f+g$ ki destination chahte hain, jis ki height har $x$ pe $f(x)+g(x)$ hai.

KYUN. Poora trick yeh hai ki hum $f$ aur $g$ ko already tame assume kar sakte hain. (Agar nahi hote — One-sided limits aur parent mein Mistake B dekho — toh law kuch nahi kehta.)

PICTURE. Do curves, har ek apni band mein slide karti hai jaise $x\to a$. Red curve sum $f+g$ hai; dekho uski height kahan jaati hai.

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Step 2 — Error ko name karo, aur use do mein split karo

KYA. Woh quantity jo hume shrink karni hai woh hai sum ka error — $f(x)+g(x)$ target $L+M$ se kitna door hai: $\underset{\text{sum ka total error}}{\underbrace{|(f(x)+g(x))-(L+M)|}}.$ Yahan $|\cdot |$ ka matlab distance hai (hamesha $\ge 0$; sign bhool jaata hai). Andar ko regroup karo: $(f(x)+g(x))-(L+M)=\underset{f\text{ ka error}}{\underbrace{(f(x)-L)}}+\underset{g\text{ ka error}}{\underbrace{(g(x)-M)}}.$

KYUN. Hum sum ke error ko directly control nahi kar sakte. Lekin hum $f$ ka error aur $g$ ka error alag alag control kar sakte hain, kyunki un dono ki existing limits hain. Toh hum bade unknown ko do jaane-paehchaane ki combination ke roop mein rewrite karte hain.

PICTURE. Total error ek vertical gap hai; uske neeche wahi gap do stacked pieces ke roop mein draw ki gayi hai — $f$-gap aur $g$-gap.

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Step 3 — Woh ek hi inequality jo chahiye: triangle inequality

KYA. Kisi bhi do numbers $p$ aur $q$ ke liye: $|p+q|\le |p|+|q|.$

• $|p+q|$ — $0$ se woh distance jahan aap dono steps saath lene ke baad pahuunchte ho.
• $|p|+|q|$ — total distance agar aap har step alag lete aur unhe cancel nahi hone dete.

Isse $p=f(x)-L$ aur $q=g(x)-M$ ke saath apply karo: $|(f-L)+(g-M)|\le \underset{f\text{ ka error}}{\underbrace{|f-L|}}+\underset{g\text{ ka error}}{\underbrace{|g-M|}}.$

KYUN YEH TOOL, koi aur nahi? Hume combined error pe ek upper bound chahiye. Khatre ki baat yeh hai ki cancellation galat ho jaaye — do errors add up ho jaayein. Triangle inequality exactly yeh statement hai ki "combined distance kabhi bhi alag alag distances ke sum se zyada nahi hoti": yeh hume worst case free mein deti hai, toh agar worst case chhota hai, hum safe hain.

PICTURE. Number line pe do steps $p$ phir $q$. Straight-shot distance $|p+q|$ (red) kabhi $|p|$ phir $|q|$ chalne (black) se lambi nahi hoti, aur tabhi barabar hoti hai jab dono steps ek hi direction mein hon.

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Step 4 — Apni tolerance spend karo: har function ko aadha do

KYA. Aap mujhe sum ke liye target tolerance $\epsilon$ dete ho. Main use do mein split karta hoon aur har function pe $\frac{\epsilon }{2}$ spend karta hoon.

• Kyunki $\lim f=L$ exist karta hai, main ek strip half-width ${\delta }_1$ dhundh sakta hoon jo $|f(x)-L|<\frac{\epsilon }{2}$ force karta hai.
• Kyunki $\lim g=M$ exist karta hai, main ek strip half-width ${\delta }_2$ dhundh sakta hoon jo $|g(x)-M|<\frac{\epsilon }{2}$ force karta hai.

$\frac{\epsilon }{2}+\frac{\epsilon }{2}$ mein split kyun? Kyunki jab main do guaranteed errors wapas add karta hoon, main chahta hoon ki woh total exactly $\epsilon$ banein — isse zyada nahi. Halves natural budget hain: $\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon$.

PICTURE. Sum ka band (height $\epsilon$) do half-bands mein split hota hai jinki height $\epsilon /2$ hai — ek budget $f$ ke liye, ek $g$ ke liye.

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Step 5 — Dono strips ek saath: chhota wala lo

KYA. ${\delta }_1$ $f$ ko budget mein rakhta hai; ${\delta }_2$ $g$ ko budget mein rakhta hai. Dono ko simultaneously budget mein rakhne ke liye, chhoti strip ke andar khade raho: $\delta =\min ({\delta }_1,{\delta }_2).$

• $\min$ — "dono mein se chhota." Chhoti strip ke andar rehne ka matlab hai tum automatically badi strip mein bhi ho.

KYUN minimum? Har guarantee sirf apni strip ke andar hold karti hai. Unka overlap chhoti strip hai. $\delta =\min$ choose karna hume overlap mein rakhta hai, jahan dono promises ek saath active hain.

PICTURE. Do strips widths ${\delta }_1$ aur ${\delta }_2$ ki overlaid hain; shaded overlap chosen $\delta$-strip hai.

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Step 6 — Pieces ko saath snap karo

KYA. Har $x$ ke liye jis mein $0<|x-a|<\delta$ hai, sab kuch chain karo:

$$|(f+g)-(L+M)|\underset{\text{Step 3}}{\underbrace{\le }}\underset{<\epsilon /2}{\underbrace{|f-L|}}+\underset{<\epsilon /2}{\underbrace{|g-M|}}<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .$$

Left to right padhne par: error split karo (Step 2), use do alag errors se bound karo (Step 3), jinmein se har ek $\epsilon /2$ se kam hai kyunki hum $\delta$-strip mein hain (Steps 4–5), aur halves $\epsilon$ rebuild karte hain.

KYUN yeh finish karta hai. Aapne koi bhi $\epsilon$ name kiya; maine ek $\delta$ produce kiya jo sum ko aapki band ke andar rakhta hai. Yahi precisely Step 0 ki definition hai. Toh: $\boxed{\underset{x\to a}{\lim }(f(x)+g(x))=L+M}■$

PICTURE. Poori chain ek flow ke roop mein: total error → do half-errors → har ek $\epsilon /2$ se kam → sum $\epsilon$ se kam.

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Step 7 — Edge cases (taaki koi scenario surprise na kare)

Proof tab tak complete nahi jab tak har input cover na ho.

Case A — ek limit exist nahi karti. Tab $L$ ya $M$ ek number nahi hai aur poora argument kabhi start hi nahi hota. Law chup hai, galat nahi. Classic trap: $f=\lfloor x\rfloor$, $g=-\lfloor x\rfloor$ kisi integer pe — dono jump karte hain (koi limit nahi), phir bhi $f+g\equiv 0$ ki limit $0$ hai. Sum law ko backwards nahi chala sakte. (One-sided limits dekho.)

Case B — difference $f-g$. Same proof $g$ ki jagah $-g$ ke saath: constant-multiple idea deta hai $\lim (-g)=-M$, toh $\lim (f-g)=L-M$. Koi naya kaam nahi.

Case C — kai functions add karna. Ek teesra function $h\to N$ add karo: do-function law ko $(f+g)$ aur $h$ pe apply karo. Kyunki $(f+g)\to L+M$ pehle se hai, $(f+g)+h\to (L+M)+N$. Induction isse kisi bhi finite number of pieces tak extend karta hai — yahi wajah hai ki polynomials plug-in ke samne jhuk jaate hain.

Case D — $\delta$-smallness kahan fail hoti hai? Kabhi nahi, ek finite sum ke liye. Lekin yeh functions ke infinite sum ke liye fail ho sakti hai — woh ek alag theory hai. Hamara law ek finite-sum law hai.

PICTURE. Floor-function counterexample: do functions jo opposite direction mein jump karti hain, jinका sum ek flat red line hai — proof ki "sum ka limit hai" parts ko nahi bachata.

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Ek-picture summary

Upar ki sab cheez, compressed: aap $L+M$ ke aas-paas height $\epsilon$ ki ek band name karte ho; main use do half-bands mein split karta hoon, $f$ ko ek mein aur $g$ ko doosre mein pakadta hoon unki apni limits ke guaranteed strips use karke, chhoti strip leta hoon, aur triangle inequality dono captures ko ek capture mein glue kar deti hai aapki poori band ke andar.

Recall Feynman retelling — poora walkthrough ek dost ko explain karo

Do log do kursiyon ki taraf chal rahe hain. Ek height $L$ pe baithega, doosra height $M$ pe. Main claim karta hoon ki unki combined height $L+M$ ki taraf jaati hai. Aap skeptical ho, toh aap $L+M$ ke aas-paas ek patli band draw karte ho aur mujhe challenge karte ho: "inhe total yahan land karao." Main aapki band ko do patli bands mein split karta hoon — har walker ke liye ek — aur main pehle se jaanta hoon ki har walker eventually apni patli band ke andar rehta hai (yahi "limit hai" ka matlab hai). Main tab tak wait karta hoon jab tak dono walkers settle na ho jaayein — matlab dono waits mein se lambe wale ka wait karna, yani chhoti strip. Jab dono apni half-bands mein hote hain, unka total aapki poori band mein squeeze ho jaata hai, kyunki do chhoti errors jab add hoti hain toh worst case mein bhi chhoti rehti hain (triangle inequality). Kyunki main kisi bhi band ko beat kar sakta hoon jo aap draw karo, total ki destination $L+M$ pe nail ho jaati hai. Aur fine print: yeh sirf tabhi kaam karta hai jab har walker ke paas jaane ke liye ek kursi ho — agar ek bhatakta raha, toh sab bets off hain.

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Connections

• Epsilon-Delta definition of a limit — Step 0 yahi definition hai; poora page ek $\epsilon$–$\delta$ game hai.
• One-sided limits — edge cases (floor function) yahan rehte hain.
• Continuity — continuous functions ke sums continuous hote hain, directly is proof se.
• Squeeze theorem — sibling tool jab algebra laws reach nahi kar paate.
• Derivative as a limit — derivatives ke liye sum rule yahi law hai ek costume mein.