Limit laws — sum, product, quotient, constant multiple

WHAT are the laws?

The two atoms everything is built from: $\lim_{x\to a} c = c \qquad\text{and}\qquad \lim_{x\to a} x = a$

Figure — Limit laws — sum, product, quotient, constant multiple

WHY are they true? (Derivation from scratch)

A limit $\lim_{x\to a}f(x)=L$ means: we can force $f(x)$ as close to $L$ as we like by keeping $x$ near $a$ . Formally, for every $\varepsilon>0$ there is $\delta>0$ so that $0<|x-a|<\delta \;\Rightarrow\; |f(x)-L|<\varepsilon.$

Sum law (full proof)

We want $|(f+g)-(L+M)|$ small. Regroup and use the triangle inequality $|p+q|\le|p|+|q|$ : $|(f(x)+g(x))-(L+M)| = |(f(x)-L)+(g(x)-M)| \le |f(x)-L| + |g(x)-M|.$ Why this step? We split the total error into " $f$ 's error" + " $g$ 's error", each of which we can control.

Given $\varepsilon>0$ , choose $\delta_1$ making $|f-L|<\tfrac{\varepsilon}{2}$ and $\delta_2$ making $|g-M|<\tfrac{\varepsilon}{2}$ . Take $\delta=\min(\delta_1,\delta_2)$ . Then both hold at once: $|(f+g)-(L+M)| < \tfrac{\varepsilon}{2}+\tfrac{\varepsilon}{2}=\varepsilon.\qquad\blacksquare$ Why split $\varepsilon$ as $\varepsilon/2 + \varepsilon/2$ ? So the two halves add back exactly to $\varepsilon$ .

Constant multiple (proof)

$|cf(x)-cL| = |c|\,|f(x)-L|.$ If $c\neq0$ , make $|f-L|<\tfrac{\varepsilon}{|c|}$ , so the product is $<\varepsilon$ . (If $c=0$ both sides are $0$ .) Why divide by $|c|$ ? So that multiplying back by $|c|$ lands on $\varepsilon$ .

Product (sketch of the key trick)

The clever move is add and subtract $L\,g(x)$ : $fg - LM = f g - Lg + Lg - LM = g(f-L) + L(g-M).$ Why this step? It separates " $f$ wobbling" from " $g$ wobbling". Near $a$ , $g$ is bounded (close to $M$ ), so $g(f-L)\to 0$ , and $L(g-M)\to0$ . Both go to $0$ , so the product limit is $LM$ .

Quotient

First prove $\lim \tfrac{1}{g}=\tfrac1M$ (needs $M\neq0$ so $g$ stays away from $0$ nearby), then apply the product law to $f\cdot\tfrac1g$ . Why need $M\neq0$ ? Dividing by something heading to $0$ blows up — the result need not exist.

HOW to use them — worked examples

Common mistake Steel-manned mistakes

Mistake A: " $\lim\frac{f}{g}=\frac{L}{M}$ always." Why it feels right: the law looks unconditional. The fix: it has a fine-print clause $M\neq0$ . If $M=0$ , the law gives nothing — you must simplify, rationalize, or check one-sided/infinite behaviour.

Mistake B: " $\lim(f+g)$ exists, so $\lim f$ and $\lim g$ must each exist." Why it feels right: feels like you can run the sum law backwards. The fix: FALSE. The laws assume the parts' limits exist first. E.g. $f=\lfloor x\rfloor,\ g=-\lfloor x\rfloor$ at an integer: each jumps, but $f+g\equiv0$ . You may only combine known existing limits.

Mistake C: Cancelling $\frac{x-3}{x-3}=1$ "at $x=3$ ." Why it feels right: algebra usually lets you cancel. The fix: at exactly $x=3$ it's $\tfrac00$ ; but the limit only cares about $x$ near $3$ , where $x\neq3$ , so cancelling is valid for the limit.

Recall Feynman: explain to a 12-year-old

Imagine two friends walking toward a meeting spot. One ends up at chair $L$ , the other at chair $M$ . If you measure the distance between them, it heads to $L+M$ apart; if you stack them, $L\cdot M$ . Knowing where each friend is going tells you where any combination is going — unless you try to divide by a friend who walks all the way to zero, because then "share per person" explodes. That's the whole game: limits slide through $+,-,\times$ , and through $\div$ only if the bottom isn't heading to zero.

Flashcards

Sum law statement

\lim(f+g)=\lim f+\lim g=L+M

, when both limits exist.

Constant multiple law

\lim(c\,f)=c\lim f=cL

Product law

\lim(fg)=(\lim f)(\lim g)=LM

Quotient law + condition

\lim(f/g)=L/M

, valid only if

M\neq0

Two atomic limits everything builds on

\lim_{x\to a}c=c

and

\lim_{x\to a}x=a

Key trick in the product-law proof

Add & subtract

Lg(x)

fg-LM=g(f-L)+L(g-M)

Why split

\varepsilon

into

\varepsilon/2

in the sum proof

So the two bounded errors add back to exactly

\varepsilon

What to do when the quotient gives

0/0

Law fails; factor/cancel/rationalize, then take the limit.

Does

\lim(f+g)

existing imply each limit exists

No — counterexample

\lfloor x\rfloor + (-\lfloor x\rfloor)

Limit of a polynomial at

a

p(a)

— direct substitution (from sum + const-mult + power laws).

Connections

Epsilon-Delta definition of a limit — the foundation every law is proved from.
One-sided limits — laws apply identically to left/right limits.
Indeterminate forms 0 over 0 — what to do when the quotient law fails.
Continuity — a function is continuous at $a$ iff $\lim_{x\to a}f=f(a)$ ; limit laws ⇒ sums/products of continuous functions are continuous.
Squeeze theorem — the tool for limits the algebra laws can't reach.
Derivative as a limit — every derivative rule (sum/product/quotient rule) descends from these laws.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, limit ka matlab simple hai: jaise-jaise $x$ kisi point $a$ ke paas jaata hai, $f(x)$ kis value ki taraf bhaag raha hai. Ab limit laws kehte hain ki agar tumhe $f$ aur $g$ dono ki limit pata hai ( $L$ aur $M$ ), to inke sum, difference, product, aur quotient ki limit bhi turant nikal jaati hai — bas $L$ aur $M$ ko usi operation se jod do. Yani limit andar ghus jaati hai operations ke through. Isliye humein har baar epsilon-delta wala lamba kaam nahi karna padta; bade function ko chhote-chhote known tukdo me toR kar reassemble kar lete hain.

Sabse important baat: quotient law me ek shart hai — neeche wali limit $M$ zero nahi honi chahiye. Agar $M=0$ aa gaya, to law fail, aur tumhe factor karna padega (jaise $\frac{x^2-9}{x-3}=x+3$ wala trick). Yahan $x\to3$ ka matlab hai $x$ exactly $3$ nahi, sirf paas, isliye cancel karna bilkul legal hai.

Ek aur galti se bachna: agar $\lim(f+g)$ exist karti hai, iska matlab ye nahi ki $f$ aur $g$ ki alag-alag limit bhi exist karti hai. Laws tabhi lagao jab dono pieces ki limit pehle se exist karti ho. Mnemonic yaad rakho: "Same Operation, Same Result — par Zero se divide mat karo." Bas itna pakad lo, to polynomial aur rational functions ki saari limits seconds me ho jayengi.

Limit laws — sum, product, quotient, constant multiple

WHAT are the laws?

WHY are they true? (Derivation from scratch)

Sum law (full proof)

Constant multiple (proof)

Product (sketch of the key trick)

Quotient

HOW to use them — worked examples

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus I — Limits & Derivatives

Connections