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

• $\underset{x\to 5}{\lim }7===7==$. Yeh constant atom hai ${\lim }_{x\to a}c=c$: ek function jo hamesha $7$ hai, woh $7$ par hi jaata hai chahe $x$ kahin bhi jaye.
• $\underset{x\to 5}{\lim }x===5==$. Yeh identity atom hai ${\lim }_{x\to a}x=a$: woh function jo apna input hi return karta hai, $a=5$ par jaata hai. Yahi do ek maatra facts hain jinhe hum assume kar sakte hain — baaki har limit inhi se build hoti hai.

Maano $L=4$ aur $M=-2$.

• (a) Sum law: $L+M=4+(-2)===2==$.
• (b) Constant-multiple law: $3L=3\cdot 4===12==$.
• (c) Product law: $LM=4\cdot (-2)===-8==$.
• (d) Quotient law — pehle bottom check karo: $M=-2\ne 0$, toh legal hai. $\frac{L}{M}=\frac{4}{-2}===-2==$. (d) ka denominator check kyun karein? Quotient law mein ek fine-print clause hai $M\ne 0$. Yahan $M=-2$ hai, toh hum bouncer se pass ho jaate hain.

Sum aur constant-multiple laws se split karo: $=4{\lim }_{x\to 2}{x}^3-{\lim }_{x\to 2}x+{\lim }_{x\to 2}5.$ Power law ($\lim x^n=a^n$, jo product law ko $n$ baar apply karne se milta hai) deta hai $\lim x^3=2^3=8$. $=4(8)-2+5=32-2+5===35==.$ Kyunki yeh ek polynomial hai, yeh poora detour $p(2)$ ke barabar hai — direct substitution kaam karta hai.

Pehle bottom check karo: ${\lim }_{x\to 1}(x+4)=1+4=5\ne 0$. Legal hai.

• Top: ${\lim }_{x\to 1}(x^2+3x)=1+3=4$.
• Bottom: $5$. Quotient law: $\frac{4}{5}===0.8==$.

Andar: ${\lim }_{x\to 3}(x^2+7)=9+7=16\ge 0$, toh root law apply hoti hai. $=\sqrt{16}===4==.$ Root law kyun hold karti hai? $\sqrt{\cdot }$ continuous hai, isliye limits iske through pass ho jaate hain — dekho Continuity.

Bottom check karo: ${\lim }_{x\to 4}(x-4)=0$. Quotient law forbidden hai — yeh ek $0/0$ form hai (top bhi $\to 0$ hai). Top ko difference of squares ki tarah factor karo: $\frac{x^2-16}{x-4}=\frac{(x-4)(x+4)}{x-4}=x+4(x\ne 4).$ Cancel karna legal kyun hai? Limit sirf $4$ ke paas $x$ ko dekhti hai, kabhi $x=4$ ko nahi, toh wahan $x-4\ne 0$ hai — figure dekho: graph line $y=x+4$ hai jisme $x=4$ par ek single hole hai.

Ab sum law apply karo: ${\lim }_{x\to 4}(x+4)=4+4===8==$.

Bottom $\to 0$, top $\to \sqrt{9}-3=0$ — ek aur $0/0$. Hum nicely factor nahi kar sakte, toh conjugate $\sqrt{x+9}+3$ se rationalize karte hain — ise khud se divide karte hain (yeh $1$ se multiply karna hai, toh kuch change nahi hota): $\frac{\sqrt{x+9}-3}{x}\cdot \frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}=\frac{(x+9)-9}{x(\sqrt{x+9}+3)}=\frac{x}{x(\sqrt{x+9}+3)}.$ Rationalize kyun karein? Yeh stubborn $\sqrt{}-3$ ko ek simple $x$ mein badal deta hai jo neeche ke $x$ ko cancel kar deta hai. $x\ne 0$ ke liye $x$ cancel karo: $=\frac{1}{\sqrt{x+9}+3}.$ Ab bottom ka limit $\sqrt{9}+3=6\ne 0$ hai, toh quotient law finally legal hai: $=\frac{1}{6}\approx ==0.16667==.$

Bottom: $\lim (x^2-4)=4-4=0$ — quotient law forbidden. Top check karo: $\lim (x^2+5x+6)=4-10+6=0$. Toh yeh $0/0$ hai; dono ko factor karo. $\frac{x^2+5x+6}{x^2-4}=\frac{(x+2)(x+3)}{(x+2)(x-2)}=\frac{x+3}{x-2}(x\ne -2).$ Dono mein $(x+2)$ kyun tha? Kyunki $x=-2$ dono ka root hai — yeh shared factor hi $0/0$ ka cause hai. Ab naye bottom ka limit $-2-2=-4\ne 0$ hai, toh quotient law apply hoti hai: $=\frac{-2+3}{-2-2}=\frac{1}{-4}===-0.25==.$

One-sided limits mein split karo:

• $x>0$ ke liye, $|x|=x$, toh $f(x)=\frac{x}{x}=1$. Isliye right limit hai ${\lim }_{x\to 0^{+}}f===1==$.
• $x<0$ ke liye, $|x|=-x$, toh $f(x)=\frac{-x}{x}=-1$. Isliye left limit hai ${\lim }_{x\to 0^{-}}f===-1==$.

Dono sides agree nahi karte ($1\ne -1$), toh two-sided limit exist nahi karta. Yeh kis law ke baare mein warn karta hai? Quotient law ko pehle dono — $\lim$(top) aur $\lim$(bottom) — exist karne chahiye. Yahan ${\lim }_{x\to 0}x=0$ bottom ko $0$ par le jaata hai, aur assembled function ka koi single limit nahi hai — ek live reminder ki laws sirf unhi limits ko combine karti hain jo already exist karti hain.

Product law sirf forward chalti hai: pehle dono parts ke limits exist karne chahiye. Ise backward chalana galat hai. Counterexample at $a=0$: maano $f(x)=\{\begin{array}{ll}1& x\ge 0\\ -1& x<0\end{array},g(x)=f(x).$ Na ${\lim }_{x\to 0}f$ exist karta hai na ${\lim }_{x\to 0}g$ (left value $-1$, right value $+1$ — wahi jump jaisa Exercise 4.2 mein). Lekin unka product hai $f(x)\cdot g(x)=f(x{)}^2=(\pm 1{)}^2=1\text{ for all }x,$ toh $\underset{x\to 0}{\lim }(fg)===1==$ exist karta hai. Product ka limit exist karna parts ke baare mein kuch nahi bataata. $■$

Yeh Derivative as a limit hai — aur yeh disguise mein ek $0/0$ form hai, toh quotient law pehle blocked hai. $\frac{h(x)-h(3)}{x-3}=\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3(x\ne 3),$ difference of squares use karke factor kiya aur phir cancel kiya (legal kyunki limit point ke paas $x\ne 3$ hota hai). Ab sum law finish karta hai: $h^{\prime }(3)={\lim }_{x\to 3}(x+3)=3+3===6==.$ Pattern notice karo: har derivative-from-definition ek $0/0$ quotient hai jise limit laws $(x-a)$ cancel karke rescue karte hain. Isliye differentiation ke sum/product/quotient rules inhi chaar limit laws se neeche aate hain.

Kis law mein side-condition hai, aur woh kya hai? ::: Quotient law; $M\ne 0$ chahiye. Kisi bhi quotient limit par pehla reflex kya hai? ::: Kuch bhi karne se pehle denominator ka limit evaluate karo. $0/0$ ka matlab actually kya hai? ::: Indeterminate — plain law chup hai; factor/cancel/rationalize karo.  ​-type $0/0$ ke liye rescue tool? ::: Conjugate se multiply karo (rationalize karo). Kya laws backward chalte hain? ::: Nahi — unhe pehle pieces ke limits exist karne chahiye. Har derivative ek limit-law problem kyun hai? ::: Yeh ek $0/0$ quotient hai jo $(x-a)$ cancel karke rescue hota hai.