Applications — approximation, evaluating limits: Worked examples

4.3.19 · D3 · Maths › Calculus III — Sequences & Series › Applications — approximation, evaluating limits

Scenario matrix

Is topic ka har problem exactly nau boxes mein se kisi ek mein aata hai — Figure s02 inhe ek map ki tarah dikhata hai, taaki tum dekh sako ki tumhara problem kis cell mein baitha hai, bjaaye ek bhaari table padhne ke.

Figure — Applications — approximation, evaluating limits

Figure s02 — scenario map. Top row (blue): teen approximation cells A/B/C, left-to-right arrange kiye gaye hain is hisaab se ki expansion center se input kitna door hai (small $+$ , small $-$ , edge of convergence). Bottom rows (green): chhe limit cells D–I, plain single-function $\frac{0}{0}$ se lekar composed series tak. Har box apne example number ke saath labelled hai; wohi label neeche worked example ka heading bhi hai.

A — Approx, small $x > 0$ : value ko $P_{N}$ mein plug karo, error bound karo → Ex 1 ( $cos 0.3$ )
B — Approx, small $x < 0$ : negative arg, signs alternate karte hain → Ex 2 ( $e^{- 0.5}$ )
C — Approx, edge of convergence: $ln (1 + x)$ near $∣ x ∣ = 1$ , slow convergence → Ex 3 ( $ln 1.5$ )
D — Limit $\frac{0}{0}$ , ek function: lowest power cancel karo → Ex 4
E — Limit $\frac{0}{0}$ , do functions divided: leaders ka ratio → Ex 5
F — Limit $\infty - \infty$ ko $\frac{0}{0}$ mein badlo: pehle combine karo → Ex 6
G — Limit jisme gehre term ki zaroorat hai: leading terms cancel ho jaate hain → Ex 7
H — Word problem (real units): small-angle physics → Ex 8
I — Exam twist: ek series ko doosri series ke andar substitute karo → Ex 9

Yahan $P_{N}$ wohi partial sum hai jo abhi define ki; "value ko $P_{N}$ mein plug karo" ka matlab hai us polynomial ko apne number par evaluate karo.

A · Ek small positive input approximate karo

Figure s03 — Ex 1. Red curve hai $cos x$ ; yellow dashed curve hai $P_{4} (x) = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24}$ . Blue marker $x = 0.3$ par dono indistinguishable hain — vertical gap (error) lagbhag $1 0^{- 6}$ hai, hamari $1 0^{- 4}$ target se kaafi neeche.

Worked example

Ex 1 (Cell A). $cos (0.3)$ ko error $< 1 0^{- 4}$ ke saath approximate karo. Forecast: andaza lagao — kya do terms kaafi honge, ya teen? Padhne se pehle apna guess likho.

Step 1. $cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots$ use karo. Yeh step kyun? $cos$ chaar must-own Maclaurin series mein se ek hai; $0$ ke paas iske terms fast shrink hote hain, isliye kuch terms kaafi honge.

Step 2. Yeh series alternate karti hai (signs $+, -, +, -$ ) aur terms decrease karte hain, isliye Alternating Series Test ke hisaab se error $\leq$ pehla omitted term hai. $x^{2}$ tak rakhne ki koshish karo: pehla omitted term $\frac{x ^{4}}{24} = \frac{0. 3 ^{4}}{24} = \frac{0.0081}{24} = 3.4 \times 1 0^{- 4}$ hai. Yeh bahut bada hai. Yeh step kyun? Hum check karte hain ki kya do terms $1 0^{- 4}$ ki bar paar karte hain. Nahi karte — omitted term hamare target se bada hai.

Step 3. $x^{4}$ tak rakho. Pehla omitted term ab $\frac{x ^{6}}{720} = \frac{0. 3 ^{6}}{720} = \frac{0.000729}{720} = 1.0 \times 1 0^{- 6} < 1 0^{- 4}$ hai. ✓ Yeh step kyun? Ek aur term error ko target se kaafi neeche crush kar deta hai.

Step 4. Compute karo: $1 - \frac{0.09}{2} + \frac{0.0081}{24} = 1 - 0.045 + 0.0003375 = 0.9553375.$ Yeh step kyun? Polynomial $P_{4} (0.3)$ par pure arithmetic.

Verify: true $cos 0.3 = 0.95533649 \dots$ ; hamara answer $0.9553375$ lagbhag $1.0 \times 1 0^{- 6}$ se differ karta hai, jo predicted bound se match karta hai. ✓

B · Ek small negative input approximate karo

Figure s04 — Ex 2. Bars har successive term $\frac{∣ x ∣ ^{n}}{n !}$ ki size dikhate hain $x = - 0.5$ ke liye; colours alternate karte hain ( $+$ blue, $-$ red) kyunki negative number ki odd powers sign flip kar deti hain. Bars fast shrink hote hain — $x^{4}$ ke baad pehla grey (omitted) bar error bound hai.

Worked example

Ex 2 (Cell B). $e^{- 0.5}$ ko error $< 1 0^{- 3}$ ke saath approximate karo. Forecast: negative argument ke saath $e^{x}$ ke terms sign flip karte hain. Guess karo kitne terms chahiye.

Step 1. $e^{x} = 1 + x + \frac{x ^{2}}{2} + \frac{x ^{3}}{6} + \frac{x ^{4}}{24} + \dots$ , $x = - 0.5$ rakho: $e^{- 0.5} = 1 - 0.5 + \frac{0.25}{2} - \frac{0.125}{6} + \frac{0.0625}{24} - \dots$ Yeh step kyun? Negative number ki odd powers negative hoti hain, isliye series alternating ban jaati hai — yeh humein free mein easy alternating-error rule de deta hai.

Step 2. Kyunki ab yeh shrinking terms ke saath alternate karti hai, error $\leq$ pehla omitted term. $x^{3}$ tak rakho; pehla omitted term $\frac{x ^{4}}{24} = \frac{0.0625}{24} = 2.6 \times 1 0^{- 3}$ hai. Ab bhi bahut bada. Yeh step kyun? Wohi alternating-series logic jaise Ex 1, bound check karne ke liye apply kiya.

Step 3. $x^{4}$ tak rakho; pehla omitted term $\frac{∣ x ∣ ^{5}}{120} = \frac{0.03125}{120} = 2.6 \times 1 0^{- 4} < 1 0^{- 3}$ hai. ✓ Yeh step kyun? Hum ek aur term jodte hain taaki first-omitted term required $1 0^{- 3}$ se neeche aa jaaye; Alternating Series Test ke hisaab se woh omitted term hi error ceiling hai, isliye jab woh target se neeche aa jaaye toh hum guaranteed accurate hain aur ruk sakte hain.

Step 4. $x^{4}$ tak sum karo: $1 - 0.5 + 0.125 - 0.0208333 + 0.0026042 = 0.6067708.$ Yeh step kyun? Polynomial $P_{4} (- 0.5)$ par arithmetic.

Verify: true $e^{- 0.5} = 0.60653066 \dots$ ; hamara $0.6067708$ ka fark $2.4 \times 1 0^{- 4} < 1 0^{- 3}$ hai. ✓ (Sanity check: $e^{- 0.5} < 1$ kyunki exponent negative hai — sahi hai.)

C · Convergence interval ka edge

Figure s05 — Ex 3. $ln (1.5)$ ke running partial sums (blue dots) true value (yellow dashed line) ki taraf bahut dheere crawl karte hain, uske upar aur neeche ping-pong karte hue kyunki series alternate karti hai. Figure s03 ki fast $cos$ convergence se contrast karo — yeh hai "edge ke paas" ka cost.

Worked example

Ex 3 (Cell C). $ln (1.5)$ estimate karo aur explain karo yeh hard case kyun hai. Forecast: $ln (1 + x)$ mein $x = 0.5$ $∣ x ∣ < 1$ ke andar hai, lekin series slowly converge karti hai. Guess karo: 4 terms, ya 8?

Step 1. $ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \frac{x ^{4}}{4} + \dots$ , valid hai $- 1 < x \leq 1$ ke liye (Radius & Interval of Convergence). Yahan $x = 0.5$ legal hai. Yeh step kyun? Pehle check karna zaroori hai ki $x$ interval mein hai — is series ko $x = 2$ par use karna classic blunder hoga jo parent note mein bataya gaya tha.

Step 2. Yeh alternate karti hai, isliye error $\leq$ pehla omitted term $\frac{0. 5 ^{n + 1}}{n + 1}$ hai. 6 decimals ke liye bahut saare terms chahiye kyunki $0. 5^{n + 1}$ har baar sirf half shrink hota hai. $x^{4}$ tak rakho: pehla omitted term $\frac{0. 5 ^{5}}{5} = \frac{0.03125}{5} = 6.25 \times 1 0^{- 3}$ hai. Yeh step kyun? Yeh dikhata hai kyun yeh hard case hai — edge ke paas terms dheere decay karte hain, isliye bahut saare chahiye.

Step 3. $x^{4}$ tak sum karo: $0.5 - \frac{0.25}{2} + \frac{0.125}{3} - \frac{0.0625}{4} = 0.5 - 0.125 + 0.0416667 - 0.015625 = 0.4010417.$ Yeh step kyun? Arithmetic; answer sirf $\sim 6 \times 1 0^{- 3}$ tak accurate hai — deliberately mediocre taaki baat prove ho sake.

Verify: true $ln 1.5 = 0.4054651 \dots$ ; hamara $0.4010417$ ka fark $4.4 \times 1 0^{- 3}$ hai, safely $6.25 \times 1 0^{- 3}$ bound ke andar. ✓ (Lesson: center se door, accuracy bahut saare terms se kharido — ya recenter karo.)

D · Ek $\frac{0}{0}$ limit, single function

Figure s01 — Ex 4. Red curve hai $1 - cos x$ ; dashed yellow curve hai uska leading parabola $\frac{x ^{2}}{2}$ . $x = 0$ ke paas dono almost perfectly overlap karte hain, isliye unka ratio $\frac{1}{2}$ ki taraf approach karta hai. Axes: horizontal $x$ , vertical value; blue dot origin mark karta hai jahan dono $0$ hain.

Worked example

Ex 4 (Cell D). $x \to 0 lim \frac{1 - cos x}{x ^{2}}$ . Forecast: raw $x = 0$ plug karo aur milta hai $\frac{1 - 1}{0} = \frac{0}{0}$ . Finite answer guess karo.

Step 1. $cos x = 1 - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \dots$ , toh $1 - cos x = \frac{x ^{2}}{2} - \frac{x ^{4}}{24} + \dots$ Yeh step kyun? $cos$ ko uski series se replace karne par sabse chhota surviving power expose hota hai — $\frac{0}{0}$ ka source yehi hai.

Step 2. $x^{2}$ se divide karo: $\frac{1}{2} - \frac{x ^{2}}{24} + \dots$ Yeh step kyun? Numerator aur denominator mein $x^{2}$ same size ke hain — woh exactly cancel ho jaate hain, $\frac{0}{0}$ theek ho jaata hai.

Step 3. $x \to 0$ jaane do: bacha hua har term mein $x$ hai aur woh mar jaata hai. Limit $= \frac{1}{2}$ . Yeh step kyun? $x \to 0$ jaane dena leftover $x$ -terms ko khatam karta hai, sirf constant bachta hai.

Figure dekho: red curve $1 - cos x$ aur yellow parabola $\frac{x ^{2}}{2}$ $0$ ke paas ek doosre se chipke hain — unka ratio exactly wohi hai jo $\frac{1}{2}$ ki taraf tend karta hai.

Verify: L'Hôpital's Rule do baar: $\frac{s i n x}{2 x} \to \frac{c o s x}{2} = \frac{1}{2}$ . Wohi answer. ✓

E · Ek $\frac{0}{0}$ limit, do functions divided

Figure s06 — Ex 5. Red curve hai $tan x - x$ ; yellow dashed curve hai uska leading cubic $\frac{x ^{3}}{3}$ . Dono $0$ ke paas flat-then-rising hain (ek cubic, parabola nahi) — $x^{3}$ se divide karne par constant $\frac{1}{3}$ bachta hai, blue horizontal line jis taraf ratio approach karta hai.

F · Ek $\infty - \infty$ limit ko $\frac{0}{0}$ mein recast karo

Figure s07 — Ex 6. Left: $\frac{1}{x}$ (blue) aur $\frac{1}{s i n x}$ (red) dono $x \to 0$ par $\infty$ ki taraf rocket karte hain — do infinities subtract karna meaningless hai. Right: unka difference (green) ek tame curve hai jo smoothly $0$ se guzarti hai — yahi woh value hai jo limit find karti hai.

Worked example

Ex 6 (Cell F). $x \to 0 lim (\frac{1}{x} - \frac{1}{sin x})$ . Forecast: dono pieces blow up karte hain ( $\infty - \infty$ ). Finite answer guess karo.

Step 1. Ek denominator par combine karo: $\frac{sin x - x}{x sin x}$ . Yeh step kyun? $\infty - \infty$ directly meaningless hai; ek single fraction mein merge karne par yeh ek $\frac{0}{0}$ ban jaata hai jis par hum series se attack kar sakte hain.

Step 2. Numerator: $sin x - x = - \frac{x ^{3}}{6} + \dots$ . Denominator: $x sin x = x (x - \frac{x ^{3}}{6} + \dots) = x^{2} - \frac{x ^{4}}{6} + \dots$ Yeh step kyun? Dono ko unke leading powers tak expand karo taaki hum sizes compare kar sakein.

Step 3. Leaders ka ratio: $\frac{- x ^{3} /6}{x ^{2}} = - \frac{x}{6} \to 0$ . Yeh step kyun? Numerator ( $x^{3}$ ) denominator ( $x^{2}$ ) se higher power hai, isliye poori cheez $0$ ki taraf shrink hoti hai.

Verify: $\frac{s i n x - x}{x s i n x}$ par L'Hôpital eventually $0$ deta hai; numerically $x = 0.1$ par: $\frac{1}{0.1} - \frac{1}{s i n 0.1} = 10 - 10.01668 = - 0.01668 \approx - x /6 = - 0.01667$ . ✓

G · Jab leading terms cancel ho jaayein — gehraai se dhundho

Figure s08 — Ex 7. Numerator $e^{x^{2}} - 1 - x^{2}$ ka term-by-term ledger: constant term aur $x^{2}$ term (red, struck through) $- 1 - x^{2}$ ke against cancel ho jaate hain; pehla survivor $\frac{x ^{4}}{2}$ (green) hai. $x^{4}$ se pehle ruko toh tum galti se " $0$ " dekhoge — isliye gehraai mein dhundho.

Worked example

Ex 7 (Cell G). $x \to 0 lim \frac{e ^{x^{2}} - 1 - x ^{2}}{x ^{4} cos x}$ . Forecast: trap — agar bahut jaldi ruke toh galti se phir $\frac{0}{0}$ milega. Guess karo.

Step 1. $e^{u} = 1 + u + \frac{u ^{2}}{2} + \dots$ mein $u = x^{2}$ rakho: $e^{x^{2}} = 1 + x^{2} + \frac{x ^{4}}{2} + \frac{x ^{6}}{6} + \dots$ Yeh step kyun? Series mein series substitute karna legal aur instant hai — koi chain-rule jhanjhat nahi.

Step 2. $e^{x^{2}} - 1 - x^{2} = \frac{x ^{4}}{2} + \frac{x ^{6}}{6} + \dots$ — $1$ aur $x^{2}$ cancel ho jaate hain, $x^{4}$ leader ban ke bacha. Yeh step kyun? Yahi mistake-#1 warning concrete roop mein hai: terms $x^{4}$ tak rakhni hongi warna numerator "vanish" ho jaata hai.

Step 3. Denominator: $x^{4} cos x = x^{4} (1 - \frac{x ^{2}}{2} + \dots) = x^{4} - \frac{x ^{6}}{2} + \dots$ , leader $x^{4}$ hai. Yeh step kyun? Denominator ki leading degree ko numerator ki degree se match karo.

Step 4. Leaders ka ratio: $\frac{x ^{4} /2}{x ^{4}} = \frac{1}{2} \to \frac{1}{2}$ . Yeh step kyun? Numerator aur denominator ab same leading power $x^{4}$ share karte hain, isliye woh $x^{4}$ exactly cancel ho jaata hai aur surviving quotient do leading coefficients ka ratio hai, $\frac{1/2}{1} = \frac{1}{2}$ ; baaki terms sab $x$ ke extra factors carry karte hain aur $x \to 0$ par vanish ho jaate hain.

Verify: $x = 0.05$ par: numerator $= e^{0.0025} - 1 - 0.0025 = 3.1276 \times 1 0^{- 6}$ , denom $= 0.0 5^{4} cos 0.05 = 6.2492 \times 1 0^{- 6}$ , ratio $= 0.5003 \approx \frac{1}{2}$ . ✓

H · Real-world word problem (units rakho)

Figure s09 — Ex 8. Ek pendulum bob angle $θ_{0}$ par swing kiya gaya. Bob apne lowest point se jitna upar uthta hai woh $1 - cos θ_{0}$ ke proportional hai; yellow bracket us chhoti height ko mark karta hai, jise series $\frac{θ _{0}^{2}}{2}$ se approximate karti hai.

Worked example

Ex 8 (Cell H). Ek pendulum ke liye, bob apne lowest point se jitna upar uthta hai woh $1 - cos θ_{0}$ ke proportional hai, jahan $θ_{0}$ radians mein swing angle hai. $θ_{0} = 0.2 rad$ swing ke liye $1 - cos θ_{0}$ evaluate karo. (Units: $θ_{0}$ radians mein, $1 - cos θ_{0}$ dimensionless hai.) Forecast: small angle — order of magnitude guess karo.

Step 1. $1 - cos θ_{0} \approx \frac{θ _{0}^{2}}{2}$ ( $cos θ = 1 - \frac{θ ^{2}}{2} + \dots$ se). Yeh step kyun? Radians dimensionless hote hain, isliye $θ_{0}^{2}$ ek pure number hai — small-angle series physics approximations ke liye exactly sahi tool hai.

Step 2. $\frac{0. 2 ^{2}}{2} = \frac{0.04}{2} = 0.02$ . Yeh step kyun? Leading term par arithmetic.

Step 3. Hum agla term $\frac{θ _{0}^{4}}{24} = \frac{0.0016}{24} = 6.7 \times 1 0^{- 5}$ drop karte hain kyunki yeh $0.02$ se hazaar guna chhota hai — do significant figures par invisible hai. Toh $1 - cos (0.2) \approx 0.02$ , yaani bob ka rise lagbhag $2%$ effect hai. Yeh step kyun? Sirf leading term rakhna tab justified hota hai jab pehla dropped term iske against negligible ho; yahan $6.7 \times 1 0^{- 5} ≪ 0.02$ , isliye ek term honest hai, aur $0.02$ fraction ke roop mein $2%$ hai.

Verify: true $1 - cos 0.2 = 0.0199667$ ; hamara $0.02$ ka fark $3.3 \times 1 0^{- 5}$ hai, omitted-term size se match karta hai. ✓ (Dimensionless in, dimensionless out — units consistent hain.)

I · Exam twist — ek series ke andar ek series

Figure s10 — Ex 9. Red curve hai $ln (1 + sin x) - x$ ; yellow dashed curve hai uska leading term $- \frac{x ^{2}}{2}$ . Woh $0$ ke paas coincide karte hain (dono neeche khulete hain), isliye $x^{2}$ se divide karne par constant $- \frac{1}{2}$ milta hai, blue horizontal line.

Worked example

Ex 9 (Cell I). $x \to 0 lim \frac{ln ( 1 + sin x ) - x}{x ^{2}}$ . Forecast: do series compose karni padengi. Guess karo.

Step 1. Inner: $sin x = x - \frac{x ^{3}}{6} + \dots$ . $ln (1 + u) = u - \frac{u ^{2}}{2} + \dots$ mein $u = sin x$ ke saath feed karo. Yeh step kyun? Series ka composition $0$ ke paas allowed hai kyunki $sin x \to 0$ , jo hamein $ln (1 + u)$ ke convergence region ke andar rakhta hai.

Step 2. $x^{2}$ tak rakho: $u = x - \frac{x ^{3}}{6} + \dots, u^{2} = x^{2} + \dots$ $ln (1 + sin x) = (x - \frac{x ^{3}}{6}) - \frac{1}{2} (x^{2}) + \dots = x - \frac{x ^{2}}{2} + \dots$ Yeh step kyun? Hume sirf $x^{2}$ tak chahiye kyunki denominator $x^{2}$ hai; higher powers crawl mein survive nahi karenge.

Step 3. Numerator: $ln (1 + sin x) - x = - \frac{x ^{2}}{2} + \dots$ . $x^{2}$ se divide karo: $- \frac{1}{2} + \dots \to - \frac{1}{2}$ . Yeh step kyun? $ln (1 + sin x)$ ka $x$ akele $- x$ ke saath cancel ho jaata hai, $- \frac{x ^{2}}{2}$ leader ban ke bachta hai; $x^{2}$ se divide karne par degrees match hote hain isliye $x^{2}$ cancel ho jaata hai, aur $x \to 0$ jaane dene par bacha hua har $x$ -term khatam ho jaata hai, sirf constant $- \frac{1}{2}$ bachta hai.

Verify: $x = 0.01$ par: $ln (1 + sin 0.01) - 0.01 = - 5.00 \times 1 0^{- 5}$ , $x^{2} = 1 0^{- 4}$ se divide karne par $- 0.500$ milta hai. ✓

Recall Quick self-test (answers reveal karo)

$lim_{x \to 0} \frac{1 - c o s x}{x ^{2}}$ ::: $\frac{1}{2}$ $lim_{x \to 0} \frac{t a n x - x}{x ^{3}}$ ::: $\frac{1}{3}$ $lim_{x \to 0} (\frac{1}{x} - \frac{1}{s i n x})$ ::: $0$ $lim_{x \to 0} \frac{l n ( 1 + s i n x ) - x}{x ^{2}}$ ::: $- \frac{1}{2}$ Kis series ka interval $- 1 < x \leq 1$ hai? ::: $ln (1 + x)$

Connections

Parent: Applications — approximation & limits
Taylor & Maclaurin Series
Lagrange Remainder & Error Estimation
Alternating Series Test
Radius & Interval of Convergence
L'Hôpital's Rule
Power Series — Operations (add, multiply, differentiate)