4.1.4 · D2 · HinglishCalculus I — Limits & Derivatives

Visual walkthroughInfinite limits and limits at infinity — vertical - horizontal asymptotes

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4.1.4 · D2 · Maths › Calculus I — Limits & Derivatives › Infinite limits and limits at infinity — vertical - horizont


Step 1 — " ko infinity bhejna" actually kaisa dikhta hai

KYA. Kisi bhi algebra se pehle, function ko dekhte hain. ko baar baar bada karo: aur height plot karo.

KYUN. Infinity par limit ek trend ke baare mein sawaal hai, kisi ek value ke baare mein nahi. Hum infinity tak kabhi pahunchte nahi; hum poochte hain "height kidhar ja rahi hai?" Aur woh trend apni aankhon se dekhe bina jawab nahi de sakte — isliye hum sample lete hain.

PICTURE. Figure mein pale-yellow dots daayein taraf march kar rahe hain. Unki heights pehle bichri hui hain () lekin phir ek flat blue line se chipak jaati hain. Woh blue line hi woh horizontal asymptote hai jiske peeche hum hain.

Figure — Infinite limits and limits at infinity — vertical - horizontal asymptotes

Step 2 — Raw numbers kaam kyun nahi aate

KYA. Numerator aur denominator ko alag alag dekho jab barta hai.

KYUN. Naively tum plug karte: top , bottom . Yeh deta hai — ek symbol jiska matlab hai "abhi pata nahi", ek indeterminate form. Dono parts explode karte hain, aur do runaway cheezoon ke beech ek race ka koi obvious winner nahi hota. Hume unki speeds compare karne ka zyada sahi tarika chahiye.

PICTURE. Do chalk curves dono board ke top se upar rocket kar rahi hain — pink (top) aur blue (bottom). Do aisi cheezoon se ratio nahi padh sakte jo ceiling ki taraf bhaag rahi hoon. Yahi puri problem hai, draw karke.

Figure — Infinite limits and limits at infinity — vertical - horizontal asymptotes

Step 3 — Key move: top aur bottom ko sabse bade denominator power se divide karo

KYA. Sabse bada power downstairs hai. Har term ko, top aur bottom mein, se divide karo:

=\frac{\dfrac{3x^2}{x^2}-\dfrac{x}{x^2}}{\dfrac{5x^2}{x^2}+\dfrac{7}{x^2}} =\frac{\;3-\dfrac{1}{x}\;}{\;5+\dfrac{7}{x^2}\;}.$$ Term by term: - $\dfrac{3x^2}{x^2}=3$ — top ka leading term ek plain constant ban jaata hai. - $\dfrac{x}{x^2}=\dfrac1x$ — ek shrinking piece. - $\dfrac{5x^2}{x^2}=5$ — bottom ka leading term ek plain constant ban jaata hai. - $\dfrac{7}{x^2}$ — ek aur shrinking piece. **KYUN.** Top aur bottom ko *ek hi cheez* $x^2$ se divide karna fraction ki value nahi badalta (yeh $\tfrac{x^2}{x^2}=1$ se multiply karne jaisa hai). Lekin yeh **har growing term ko ya toh ek constant mein ya $\tfrac{1}{x^p}$ form ke fraction mein convert kar deta hai** — aur un chote fractions ko hum pehle se khatam karna jaante hain. Hum ek unreadable race ko aise cheezoon ke tidy sum mein badal rahe hain jo ya toh *khadi rehti hain* ya *zero ho jaati hain*. **PICTURE.** Figure mein chaar terms alag alag chalk bars ke roop mein hain. Do constant bars ($3$ aur $5$) tall rehte hain; do "$1/x^p$" bars clearly shrink karte hain jab $x$ barta hai — neeche arrow "$\to 0$" ke saath annotated. ![[deepdives/dd-maths-4.1.04-d2-s03.png]] --- ## Step 4 — Shrinking terms ko master fact se khatam karo **KYA.** $x\to\infty$ bhejo. Har $\tfrac{1}{x^p}$ jahan $p>0$ hai woh mar jaata hai: $$\frac1x\to0,\qquad \frac{7}{x^2}\to0.$$ **KYUN.** Yahi woh engine hai jo parent note ne $\varepsilon$–$N$ definition se build kiya tha: kisi bhi tininess $\varepsilon$ ke liye jo tum maango, jab $x>\varepsilon^{-1/p}$ ho toh term $\varepsilon$ se chhoti ho jaati hai. Yeh **zaroor** zero tak pahuchega. Hum ise isliye use karte hain kyunki yahi *ek maatra tool* hai jo "$x$ bahut bada ho jaata hai" ko ek exact number mein convert karta hai jise hum likh sakein. **PICTURE.** Step 3 ke do shrinking chalk bars kuch nahi rah jaate (board par dotted stubs), jabki constants $3$ aur $5$ solid rehte hain. ![[deepdives/dd-maths-4.1.04-d2-s04.png]] > [!recall]- Ek breath mein $1/x^p\to 0$ kyun > $\tfrac{1}{x^p}<\varepsilon$ chahiye? ::: $x$ ke liye solve karo: $x>\varepsilon^{-1/p}$ chahiye; us point ke baad term har us $\varepsilon$ se chhoti rehti hai jo tum name karo, aur yahi exactly "$\to 0$" ka matlab hai. --- ## Step 5 — Limit padho aur ceiling draw karo **KYA.** Shrinking terms gone hone ke baad: $$\lim_{x\to\infty}\frac{3-\frac1x}{5+\frac{7}{x^2}}=\frac{3-0}{5+0}=\frac{3}{5}.$$ Har bacha hua symbol: $3$ = numerator ka leading coefficient, $5$ = denominator ka leading coefficient. Limit unka **ratio** hai. **KYUN.** Sirf highest-power terms mein $x^2$ se divide hone ke baad survive karne ki taqat thi; baki sab outrun ho gaye aur gayab ho gaye. Toh *sirf leading coefficients* hi destination dictate karte hain — yahi degree-equal rule hai ("$m=n\Rightarrow y=a_m/b_n$") *derive* hoke, yaad kiya nahi. **PICTURE.** Blue horizontal line $y=\tfrac35$ draw ki gayi hai, aur curve dikhaya gaya hai ki woh usse bina touch kiye hug karta hua aa raha hai — ek ceiling. ![[deepdives/dd-maths-4.1.04-d2-s05.png]] > [!formula] Degree-equal shortcut, ab kamaya hua > Agar $\deg(\text{top})=\deg(\text{bottom})$, toh $\displaystyle\lim_{x\to\pm\infty}\frac{a_mx^m+\cdots}{b_nx^n+\cdots}=\frac{a_m}{b_n}$. > Yahan $\dfrac35=0.6$ — Step 1 ke dots se check karo, jo bilkul $0.6$ ke aaspaas flat hue the. ✓ --- ## Step 6 — Teeno degree cases, side by side **KYA.** Wahi divide-by-$x^n$ trick, lekin badlo ki kaun sa side jeette hai. - **Top chhota** ($m<n$), jaise $\dfrac{x}{5x^2+7}=\dfrac{1/x}{5+7/x^2}\to\dfrac{0}{5}=0$: ceiling $y=0$ par. - **Top bada** ($m>n$), jaise $\dfrac{2x^3+1}{x^2}=2x+\dfrac1{x^2}\to+\infty$: **koi** ceiling nahi; yeh ud jaata hai (shayad ek [[Slant (oblique) asymptotes via polynomial division|slant asymptote]]). **KYUN.** Division hamesha winner ko constant chhod deta hai aur loser ko $1/x^p$ bana deta hai. Agar *top* loser hai toh woh $0$ mar jaata hai; agar top winner hai toh $x$ ka ek honest positive power survive karta hai aur hamesha ke liye barta hai. Teenon cases cover karne se reader koi bhi rational function classify kar sakta hai. **PICTURE.** Teen mini-panels: top-heavy curve upar blast kar rahi hai (koi ceiling nahi), bottom-heavy curve $y=0$ par flat ho rahi hai, aur equal-degree curve $y=\tfrac35$ par flat ho rahi hai. ![[deepdives/dd-maths-4.1.04-d2-s06.png]] > [!mnemonic] BOTU (parent se) > **B**ottom bada → $0$. **O**dds even, yaani equal degrees → leading coefficients ka ratio. **T**op bada → b**U**st, koi horizontal asymptote nahi. --- ## Step 7 — $x\to-\infty$ mein sign trap (roots ke saath) **KYA.** Daayein ki bajay **baayein** chalo, aur ek square root add karo: $$\lim_{x\to-\infty}\frac{\sqrt{4x^2+1}}{x}.$$ Root mein se $x^2$ factor karo — lekin dhyan se: $\sqrt{x^2}=|x|$, aur $x<0$ ke liye, $|x|=-x$. $$\sqrt{4x^2+1}=|x|\sqrt{4+\tfrac1{x^2}}=-x\sqrt{4+\tfrac1{x^2}} \;\Rightarrow\;\frac{-x\sqrt{4+1/x^2}}{x}=-\sqrt{4+\tfrac1{x^2}}\to-2.$$ **KYUN.** Square root kabhi negative nahi hoti, lekin baayein taraf $x$ *negative* hota hai. Agar tumne blindly $\sqrt{x^2}=x$ likhte toh ek galat sign galat ceiling tak le jaata. Root ko $-x$ ke roop mein bahar nikalna arithmetic ko sahi rakhta hai. Isliye ek curve ke **do alag** horizontal ceilings ho sakte hain — daayein $+2$, baayein $-2$. **PICTURE.** Board dono branches dikhata hai: right branch $y=+2$ par flat ho rahi hai (blue), left branch $y=-2$ par flat ho rahi hai (pink), sign flip circle kiya hua. ![[deepdives/dd-maths-4.1.04-d2-s07.png]] > [!mistake] $\sqrt{x^2}=x$ likhna jab $x\to-\infty$ > **Kyun sahi lagta hai:** positive $x$ ke liye yeh sach hai jisse hum usually milte hain. > **Fix:** $\sqrt{x^2}=|x|$, aur $|x|=-x$ jab $x<0$ ho. Yeh bhool jaane se tumhare poore answer ka sign palat jaata hai. --- ## Step 8 — *Vertical* wall: wahi idea, denominator zero hit karta hai **KYA.** Question badlo: $x$ ko **finite** rakho lekin denominator ko $0$ push karo nonzero top ke saath, jaise $x=3$ ke paas $\dfrac{x+1}{(x-3)^2}$. **KYUN.** Infinity par limit ne pucha tha "jab $x$ bhaag jaata hai toh kya hota hai?" Vertical asymptote mirror sawaal puchta hai: "jab *height* bhaag jaati hai jabki $x$ wahi rehta hai toh kya hota hai?" $0$ tak shrink hoti kisi cheez par ek nonzero number ka size $\infty$ tak explode karta hai; **sign** top aur bottom ke dono taraf ke signs se fix hoti hai (isliye [[One-sided limits]] matter karte hain). Yahan $(x-3)^2\ge0$ hai, toh bottom dono taraf se $0^+$ hai aur top $+4$ hai: dono taraf $+\infty$. **PICTURE.** $x=3$ par ek vertical dashed wall; curve dono taraf $+\infty$ tak rocket karta hai, $1/x$ ke ek chote inset ke saath contrast mein jahan dono sides *disagree* karti hain ($+\infty$ vs $-\infty$). ![[deepdives/dd-maths-4.1.04-d2-s08.png]] > [!mistake] "Bottom $=0$ matlab hamesha ek vertical wall" > **Fix:** agar top bhi $\to0$ ho toh tumhare paas $\tfrac00$ hai — pehle factor aur cancel karo; tumhe [[Continuity and removable discontinuities|removable hole]] mil sakta hai, wall nahi. Example: $\dfrac{x^2-9}{x-3}=x+3\to6$, $(3,6)$ par ek hole. --- ## Ek picture mein summary ![[deepdives/dd-maths-4.1.04-d2-s09.png]] Final board dono kahaniyaan ek saath rakhta hai: **horizontal ceiling** $y=\tfrac35$ jise $\tfrac{3x^2-x}{5x^2+7}$ hug karta hai jab $x$ bhaag jaata hai, aur ek **vertical wall** jahan denominator zero hit karta hai — wo do tarike jinse ek graph ka skeleton fix hota hai. > [!recall]- Feynman retelling — plain words mein poora walkthrough > Hum jaanna chahte the ki curve ki height kahan jaati hai jab $x$ hamesha ke liye daayein chala jaata hai. "Infinity" plug karne se sirf $\tfrac{\infty}{\infty}$ mila — ek shrug. Toh humne fair khela: humne har piece ko, top aur bottom mein, $x$ ke sabse bade power se divide kiya jo downstairs tha. Isse har term ya toh ek *number ban gayi jo khadi rehti hai* (leading coefficients $3$ aur $5$) ya ek *fraction jo kuch nahi rah jaati* ($\tfrac1x,\tfrac{7}{x^2}$). Ek baar fading pieces mare, sirf $\tfrac{3}{5}$ bacha — ceiling. Wahi recipe har rational function classify karti hai: bottom jeete ⇒ ceiling $0$ par, top jeete ⇒ koi ceiling nahi, tie ⇒ leaders ka ratio. *Baayein* $-\infty$ jaate waqt hume yaad rakhna pada ki square root kabhi negative nahi hoti, toh wahan $\sqrt{x^2}=-x$ hai, jo ceiling ka sign flip kar sakta hai. Aur mirror-image sawaal — height explode hoti hai jabki $x$ wahi rehta hai — vertical walls deta hai, agar top bhi zero na ho (warna woh sirf ek fixable hole hai). ## Connections - [[Infinite limits and limits at infinity — vertical - horizontal asymptotes]] - [[One-sided limits]] - [[Continuity and removable discontinuities]] - [[Slant (oblique) asymptotes via polynomial division]] - [[L'Hôpital's rule for indeterminate forms]] - [[Limits — formal epsilon-delta definition]] - [[Curve sketching using first and second derivatives]]