Infinite limits and limits at infinity — vertical - horizontal asymptotes: Worked examples

4.1.4 · D3 · Maths › Calculus I — Limits & Derivatives › Infinite limits and limits at infinity — vertical - horizont

#	Case class	Kya cheez ise alag banati hai	Answer type	Example
A	Nonzero ÷ zero, sides disagree	left limit aur right limit ke opposite signs hain	VA, limit DNE	Ex 1
B	Nonzero ÷ zero, sides agree (squared denom)	denominator $\to 0^{+}$ dono sides se	VA, $\pm \infty$ dono sides pe	Ex 2
C	Zero ÷ zero — removable	top aur bottom mein common factor hai	hole, VA NAHI	Ex 3
D	Rational at infinity, deg top $<$ deg bottom	numerator pehle khatam ho jaata hai	HA $y = 0$	Ex 4
E	Rational at infinity, deg top $=$ deg bottom	leading terms bachte hain	HA = leaders ka ratio	Ex 4
F	Rational at infinity, deg top $>$ deg bottom	numerator jeetta hai	koi HA nahi ( $\pm \infty$ )	Ex 4
G	Root + sign care, $x \to - \infty$	$\sqrt{x^2}=	x	=-x$
H	Transcendental ( $e^{- x}$ , $arctan$ )	rational nahi, phir bhi settle karta hai	known limits se HA	Ex 6
I	Word problem (real-world settling)	asymptote = long-run value	HA with units	Ex 7
J	Exam twist — VA cancel ho jaata hai lekin doosra bachta hai	mixed hole + real asymptote	ek hole, ek VA	Ex 8
K	Slant asymptote (deg top $=$ deg bottom $+ 1$ )	numerator exactly ek degree zyada	oblique line $y = m x + b$	Ex 9

$x \to 1 lim \frac{x + 2}{x - 1}$ , aur dono one-sided limits. Forecast: $x = 1$ par Numerator $3$ hai (nonzero), denominator $\to 0$ . Toh yeh blow up hoga — lekin upar jaayega ya neeche? Parhne se pehle dono sides ka sign guess karo.

Step 1 — $x = 1$ ke paas numerator ka sign check karo. Yeh step kyun? Agar top bhi $0$ ki taraf ja raha hota toh humein $\frac{0}{0}$ hole milta (cell C) aur hum asymptote conclude NAHI kar sakte. Yahan $x + 2 \to 3 > 0$ hai, safely nonzero aur positive. Achha — yeh genuine "nonzero ÷ zero" hai (cell A/B).

Step 2 — Right side $x \to 1^{+}$ . Kyun? Answer ka sign poori tarah denominator ke sign se decide hota hai, aur woh sign left vs right mein alag hota hai. $x = 1.001$ lo: $x - 1 = 0.001 > 0$ , ek tiny positive. Toh $\frac{+ 3}{0 ^{+}} = + \infty$ .

Step 3 — Left side $x \to 1^{-}$ . $x = 0.999$ lo: $x - 1 = - 0.001 < 0$ , ek tiny negative. Toh $\frac{+ 3}{0 ^{-}} = - \infty$ .

Step 4 — Combine karo. Dono sides disagree karti hain ( $+ \infty$ vs $- \infty$ ), toh $lim_{x \to 1} \frac{x + 2}{x - 1}$ exist nahi karta. Lekin $x = 1$ phir bhi ek vertical asymptote hai (ek one-sided infinite limit kaafi hai).

Verify: $x = 1.0001$ plug karo: $\frac{3.0001}{0.0001} = 30001$ — huge positive ✓. $x = 0.9999$ plug karo: $\frac{2.9999}{- 0.0001} = - 29999$ — huge negative ✓. Signs Steps 2–3 se match karte hain.

Neeche ki figure yeh function plot karti hai: dekho cyan curve amber line $x = 1$ ke bilkul right mein upar rocket karta hai aur uske bilkul left mein neeche plunge karta hai — yeh visible disagreement exactly Steps 2–3 ka " $+ \infty$ vs $- \infty$ " hai. Dotted white line horizontal asymptote $y = 1$ hai (equal degrees, ratio $1/1$ ).

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

Ex 2 — Cell B: squared denominator, sides agree

Worked example

$x \to - 2 lim \frac{5 - x}{( x + 2 ) ^{2}}$ Forecast: Denominator ek square hai, toh woh negative nahi ja sakta. Dono sides ke baare mein yeh kya force karta hai? Parhne se pehle guess karo.

Step 1 — $x = - 2$ ke paas numerator ka sign. $5 - x \to 5 - (- 2) = 7 > 0$ , nonzero positive.

Step 2 — Denominator ka behaviour. Square kyun matter karta hai: $(x + 2)^{2} \geq 0$ hamesha, aur yeh $0$ sirf $x = - 2$ par hit karta hai. $- 2$ ke paas yeh ek tiny positive hai — dono sides se, kyunki squaring sign ko khatam kar deti hai. Toh $(x + 2)^{2} \to 0^{+}$ left aur right dono taraf.

Step 3 — Combine karo. $\frac{+ 7}{0 ^{+}} = + \infty$ dono sides se. Kyunki sides agree karti hain, two-sided limit $+ \infty$ ke roop mein exist karta hai: $lim_{x \to - 2} \frac{5 - x}{( x + 2 ) ^{2}} = + \infty.$ Vertical asymptote $x = - 2$ .

Verify: $x = - 1.99$ : $(0.01)^{2} = 0.0001$ , $\frac{6.99}{0.0001} = 69900$ ✓. $x = - 2.01$ : $(- 0.01)^{2} = 0.0001$ , $\frac{7.01}{0.0001} = 70100$ ✓. Dono large positive — agreement confirm hua.

Ex 3 — Cell C: $\frac{0}{0}$ removable hole

Worked example

$x \to 4 lim \frac{x ^{2} - 16}{x - 4}$ — asymptote hai ya hole? Forecast: Denominator $\to 0$ . Kya yeh explode karta hai? Guess karne se pehle pehle numerator check karo.

Step 1 — $x = 4$ par numerator test karo. $4^{2} - 16 = 0$ . Yeh kyun sab kuch badal deta hai: top aur bottom dono $\to 0$ , toh yeh indeterminate form $\frac{0}{0}$ hai — automatically asymptote NAHI. Vanishing factor cancel ho sakta hai.

Step 2 — Factor aur cancel karo. Factor kyun? Shared $(x - 4)$ ko expose aur remove karne ke liye. $\frac{x ^{2} - 16}{x - 4} = \frac{( x - 4 ) ( x + 4 )}{x - 4} = x + 4 (x \neq = 4) .$ Cancellation legal hai kyunki limit mein $x \neq = 4$ , toh hum kabhi actual zero se divide nahi karte.

Step 3 — Simplified form ka limit lo. $lim_{x \to 4} (x + 4) = 8$ . Finite! Toh koi vertical asymptote nahi — sirf point $(4, 8)$ par ek removable hole hai. (Dekho Continuity and removable discontinuities.)

Verify: $x = 4.001 \Rightarrow x + 4 = 8.001 \approx 8$ ✓. Original at $x = 3.999$ : yahan $3.99 9^{2} = 15.992001$ , toh $\frac{15.992001 - 16}{- 0.001} = \frac{- 0.007999}{- 0.001} = 7.999$ ✓ — koi blow-up nahi, confirm karta hai ki wall nahi hole hai.

Ex 4 — Cells D, E, F: rational functions at infinity (teeno degree cases)

Worked example Teeno ko

$x \to \infty$ par evaluate karo: $(D) \frac{2 x + 1}{x ^{2} + 3}, (E) \frac{4 x ^{2} - x}{2 x ^{2} + 9}, (F) \frac{x ^{3} + 1}{x ^{2}} .$ Forecast: Ek $0$ mar jaata hai, ek nonzero number par land karta hai, ek $\infty$ ki taraf bhaag jaata hai. Parhne se pehle har ek ko uski fate se match karo.

Master move — sabse badi denominator power se divide karo. Kyun: yeh har term ko $\frac{1}{x ^{k}}$ ke constant times mein badal deta hai, aur hum jaante hain $lim_{x \to \infty} \frac{1}{x ^{k}} = 0$ jab $k > 0$ ho. Woh akela fact saara kaam karta hai.

(D) deg top $1 <$ deg bottom $2$ . $x^{2}$ se divide karo: $\frac{2 x + 1}{x ^{2} + 3} = \frac{\frac{2}{x} + \frac{1}{x ^{2}}}{1 + \frac{3}{x ^{2}}} \to \frac{0 + 0}{1 + 0} = 0.$ HA $y = 0$ . Kyun: numerator mein sirf negative powers of $x$ hain — woh mar jaata hai.

(E) deg top $=$ deg bottom $= 2$ . $x^{2}$ se divide karo: $\frac{4 x ^{2} - x}{2 x ^{2} + 9} = \frac{4 - \frac{1}{x}}{2 + \frac{9}{x ^{2}}} \to \frac{4 - 0}{2 + 0} = 2.$ HA $y = 2$ — leading coefficients ka ratio $4/2$ .

(F) deg top $3 >$ deg bottom $2$ . $x^{2}$ se divide karo: $\frac{x ^{3} + 1}{x ^{2}} = x + \frac{1}{x ^{2}} \to + \infty.$ Koi horizontal asymptote nahi. (Kyunki deg top, deg bottom se exactly $1$ zyada hai, yeh slant case bhi hai — Ex 9 mein poori tarah worked; general theory Slant (oblique) asymptotes via polynomial division mein.)

Verify ( $x = 1000$ plug karo): (D) $\frac{2001}{1000003} \approx 0.0020 \to 0$ ✓. (E) $\frac{3999000}{2000009} \approx 1.9995 \to 2$ ✓. (F) $\frac{1 0 ^{9} + 1}{1 0 ^{6}} = 1000000.000001 \to \infty$ ✓.

Ex 5 — Cell G: root with sign care, do alag asymptotes

Worked example

$f (x) = \frac{9 x ^{2} + 4}{2 x - 1}$ ke dono horizontal asymptotes dhundo. Forecast: $x^{2}$ ka square root ek $∣ x ∣$ chupaata hai. Kya tum dono ends par same limit expect karte ho, ya do alag? Guess karo.

Step 1 — Root ke andar se $x^{2}$ baahar nikalo. Kyun: ek $\frac{1}{x ^{2}}$ term isolate karne ke liye jise hum $0$ par bhej sakein. $9 x^{2} + 4 = x^{2} 9 + \frac{4}{x ^{2}} = ∣ x ∣ 9 + \frac{4}{x ^{2}} .$ Key fact: $x^{2} = ∣ x ∣$ , na ki $x$ .

Step 2 — $x \to + \infty$ . Yahan $x > 0$ , toh $∣ x ∣ = x$ . Top aur bottom ko $x$ se divide karo: $\frac{x 9 + 4/ x ^{2}}{2 x - 1} = \frac{9 + 4/ x ^{2}}{2 - 1/ x} \to \frac{9}{2} = \frac{3}{2} .$ HA $y = \frac{3}{2}$ right side par.

Step 3 — $x \to - \infty$ . Sign kyun flip hota hai: ab $x < 0$ , toh $∣ x ∣ = - x$ . Woh minus sign poora exam trap hai. $\frac{- x 9 + 4/ x ^{2}}{2 x - 1} = \frac{- 9 + 4/ x ^{2}}{2 - 1/ x} \to \frac{- 3}{2} = - \frac{3}{2} .$ HA $y = - \frac{3}{2}$ left side par.

Step 4 — Conclusion. Do alag horizontal asymptotes: $y = \frac{3}{2}$ (right) aur $y = - \frac{3}{2}$ (left).

Verify: $x = 1000$ : $\frac{9 , 000 , 004}{1999} \approx \frac{3000.0007}{1999} \approx 1.5008 \to 1.5$ ✓. $x = - 1000$ : $\frac{3000.0007}{- 2001} \approx - 1.4993 \to - 1.5$ ✓.

Figure sign flip ko visible banati hai: cyan curve right jaate waqt amber line $y = \frac{3}{2}$ ki taraf level off hota hai, lekin left jaate waqt white line $y = - \frac{3}{2}$ ki taraf. Yeh mirror-image split poori tarah $∣ x ∣ = - x$ ki wajah se hai jo left side par appear hota hai.

Ex 6 — Cell H: transcendental functions

Worked example

$g (x) = \frac{e ^{x} - 1}{e ^{x} + 1}$ ke horizontal asymptotes, dono ends par. Forecast: Jaise $x$ badhta hai, $e^{x}$ constants ko dwarf karta hai; jaise $x \to - \infty$ , $e^{x} \to 0$ . Do alag ceilings? Guess karo.

Step 1 — $x \to + \infty$ . $e^{x}$ se divide kyun karo: yeh sabse fast-growing term hai, "highest power" ka analogue. $\frac{e ^{x} - 1}{e ^{x} + 1} = \frac{1 - e ^{- x}}{1 + e ^{- x}} .$ Kyunki $e^{- x} = 1/ e^{x} \to 0$ : $\to \frac{1 - 0}{1 + 0} = 1.$ HA $y = 1$ right par.

Step 2 — $x \to - \infty$ . Yahan $e^{x}$ se divide kyun NAHI karo: jaise $x \to - \infty$ , $e^{x} \to 0$ , toh seedha plug in karo. $\frac{e ^{x} - 1}{e ^{x} + 1} \to \frac{0 - 1}{0 + 1} = - 1.$ HA $y = - 1$ left par.

Step 3 — $arctan$ par sanity check (parent ka forecast drill). Comparison ke liye: $lim_{x \to \infty} arctan x = \frac{π}{2}$ aur $lim_{x \to - \infty} arctan x = - \frac{π}{2}$ — same two-ceiling pattern.

Verify: $g (5) = \frac{e ^{5} - 1}{e ^{5} + 1} = \frac{147.41}{149.41} \approx 0.9866 \to 1$ ✓. $g (- 5) = \frac{e ^{- 5} - 1}{e ^{- 5} + 1} = \frac{- 0.9933}{1.0067} \approx - 0.9866 \to - 1$ ✓.

Ex 7 — Cell I: word problem (long-run value)

Worked example

$t$ ghante baad blood mein ek drug ki concentration (mg/L) $C (t) = \frac{40 t}{t + 8}$ hai jab $t \geq 0$ . Patient bahut lambe time baad kis concentration ki taraf approach karta hai, aur woh number kya mean karta hai? Forecast: Jaise ghante barhte hain, kya $C$ hamesha badhta rehta hai ya level off ho jaata hai? Ceiling guess karo.

Step 1 — Case pehchano. $C$ ek rational function hai, deg top $=$ deg bottom $= 1$ → yeh cell E hai. Ek finite horizontal asymptote expect karo = leading coefficients ka ratio.

Step 2 — Infinity par limit lo. $t$ se divide karo: $C (t) = \frac{40 t}{t + 8} = \frac{40}{1 + \frac{8}{t}} \to \frac{40}{1 + 0} = 40.$

Step 3 — Units ke saath interpret karo. Horizontal asymptote $C = 40$ mg/L hai. Physically: concentration ek plateau of 40 mg/L ki taraf badhti hai aur kabhi isse exceed nahi karti — long-run steady level. Yeh kabhi bilkul $40$ nahi pahunchti (asymptotes approach kiye jaate hain, chhuye nahi jaate), yeh match karta hai ki graph ceiling se kaise chipakta hai.

Verify: $C (100) = \frac{4000}{108} \approx 37.04$ ; $C (10000) = \frac{400000}{10008} \approx 39.97 \to 40$ ✓. $40$ ki taraf badh raha hai, usse neeche reh raha hai ✓. Units: $\frac{mg/L \cdot h}{h} =$ mg/L ✓.

Figure mein $C (t)$ pehle steeply climb karta hai, phir amber ceiling $C = 40$ ke bilkul neeche flat run karne ke liye bend karta hai — curve usse approach karta hai lekin amber arrow mark karta hai ki woh kabhi nahi pahunchta.

Ex 8 — Cell J: exam twist (ek hole AUR ek real asymptote)

Worked example

$h (x) = \frac{x ^{2} - 2 x - 3}{x ^{2} - 9}$ ke liye har vertical asymptote aur har hole dhundo. Forecast: Denominator $x^{2} - 9 = (x - 3) (x + 3)$ $x = 3$ aur $x = - 3$ par vanish karta hai. Kya dono vertical asymptotes hain? Dhyan se — har ek par numerator check karo.

Step 1 — Dono ko factor karo. Kyun: yeh dekhne ke liye ki bottom ke kaunse zeros top ke saath shared hain (holes) aur kaunse nahi (true asymptotes). $h (x) = \frac{( x - 3 ) ( x + 1 )}{( x - 3 ) ( x + 3 )} .$

Step 2 — $x = 3$ par: shared factor $(x - 3)$ cancel ho jaata hai. Yeh hole kyun mean karta hai: top aur bottom dono wahan $0$ hit karte hain — ek $\frac{0}{0}$ form jo cancel hoti hai, toh koi blow-up nahi. Simplified: $h (x) = \frac{x + 1}{x + 3}$ jab $x \neq = 3$ . Hole ki height: simplified form mein $x = 3$ plug karo, $\frac{3 + 1}{3 + 3} = \frac{4}{6} = \frac{2}{3}$ . Hole at $(3, \frac{2}{3})$ .

Step 3 — $x = - 3$ par: factor cancel NAHI hota. $\frac{x + 1}{x + 3}$ mein denominator $x = - 3$ par phir bhi vanish karta hai jabki numerator $- 3 + 1 = - 2 \neq = 0$ deta hai. Nonzero ÷ zero → vertical asymptote $x = - 3$ . (Odd power → sides disagree karengi: left $\frac{- 2}{0 ^{-}} = + \infty$ hai, right $\frac{- 2}{0 ^{+}} = - \infty$ hai.)

Step 4 — Bonus horizontal asymptote. Equal degrees $2 = 2$ , leading coefficients $1/1$ → HA $y = 1$ .

Verify: hole height $\frac{x + 1}{x + 3}$ at $x = 3$ gives $\frac{4}{6} = 0.6667$ ✓. $x = - 3$ ke paas: $x = - 2.99 \Rightarrow \frac{- 1.99}{0.01} = - 199$ (right side, $- \infty$ ) ✓; $x = - 3.01 \Rightarrow \frac{- 2.01}{- 0.01} = 201$ (left side, $+ \infty$ ) ✓. HA: $h (1000) = \frac{997997}{999991} \approx 0.998 \to 1$ ✓.

Ex 9 — Cell K: slant (oblique) asymptote

Worked example

$f (x) = \frac{x ^{2} + 3 x + 2}{x - 1}$ ka slant asymptote dhundo. Forecast: deg top $= 2$ , deg bottom $= 1$ , toh deg top exactly deg bottom se ek zyada hai. Koi horizontal asymptote nahi hai (top bigger → bust), lekin curve door jaake ek slanted straight line se align ho jaata hai. Guess karo woh line rise karti hai ya fall.

Step 1 — Polynomial division kyun? Yeh tool kyun: jab deg top $=$ deg bottom $+ 1$ ho, polynomials ko divide karne se fraction $(line) + (small remainder)$ ke roop mein rewrite ho jaata hai. Remainder ek proper fraction hai jo infinity par $\to 0$ ho jaata hai, toh line exactly woh hai jise curve hug karta hai. Division woh tool hai jo line ko top se peel karta hai.

Step 2 — $x^{2} + 3 x + 2$ ko $x - 1$ se long division karo: $\frac{x ^{2} + 3 x + 2}{x - 1} = x + 4 + \frac{6}{x - 1} .$ (Check: $(x + 4) (x - 1) = x^{2} + 3 x - 4$ , plus remainder $6$ deta hai $x^{2} + 3 x + 2$ ✓.)

Step 3 — $x \to \pm \infty$ bhejo. Kyun: term $\frac{6}{x - 1}$ nonzero over a huge number hai → $0$ . Jo bachta hai woh line $y = x + 4$ hai. Toh slant asymptote $y = x + 4$ hai dono ends par (yeh rise karta hai, slope $+ 1$ ).

Step 4 — Vertical asymptote bhi note karo. Denominator $x - 1 = 0$ at $x = 1$ jabki numerator $1 + 3 + 2 = 6 \neq = 0$ hai: nonzero ÷ zero → VA $x = 1$ . Toh is function mein ek slant aur ek vertical asymptote hai, koi horizontal nahi.

Verify: $x = 1000$ par, $f (1000) = \frac{1003002}{999} \approx 1004.006$ , jabki $x + 4 = 1004$ — gap $\approx 0.006 \to 0$ ✓. $x = - 1000$ par, $f = \frac{997002}{- 1001} \approx - 996.006$ aur $x + 4 = - 996$ , gap $\approx - 0.006 \to 0$ ✓. Remainder identity $(x + 4) (x - 1) + 6 = x^{2} + 3 x + 2$ ✓.

Recall Matrix-cover self-test

One-sided signs jab denominator mein odd-power zero ho ::: opposite (limit DNE, phir bhi ek VA) Denominator mein even-power zero ::: dono sides same sign Top aur bottom dono $\to 0$ ::: possible removable hole — pehle factor karo deg top $<$ deg bottom at infinity ::: HA $y = 0$ deg top $=$ deg bottom ::: HA = leading coefficients ka ratio deg top $>$ deg bottom ::: koi HA nahi deg top $=$ deg bottom $+ 1$ ::: slant asymptote (polynomial division se milta hai) $x^{2}$ jaise $x \to - \infty$ ::: $- x$ , sign flip karta hai — do HAs possible

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
Curve sketching using first and second derivatives
Limits — formal epsilon-delta definition

Worked examples — Infinite limits and limits at infinity — vertical - horizontal asymptotes

Scenario matrix

Ex 1 — Cell A: sides disagree

Ex 2 — Cell B: squared denominator, sides agree

Ex 3 — Cell C: $\frac{0}{0}$ removable hole

Ex 4 — Cells D, E, F: rational functions at infinity (teeno degree cases)

Ex 5 — Cell G: root with sign care, do alag asymptotes

Ex 6 — Cell H: transcendental functions

Ex 7 — Cell I: word problem (long-run value)

Ex 8 — Cell J: exam twist (ek hole AUR ek real asymptote)

Ex 9 — Cell K: slant (oblique) asymptote

Connections

Case-class map