Limit laws — sum, product, quotient, constant multiple: Worked examples

4.1.2 · D3 · Maths › Calculus I — Limits & Derivatives › Limit laws — sum, product, quotient, constant multiple

Scenario matrix

Yeh har case class hai jisme ek limit-law problem ho sakti hai. Har row genuinely alag tarah ka trap hai. Neeche ke worked examples us cell se tagged hain jo wo cover karte hain.

#	Case class	Kya cheez ise alag banati hai	Example
C1	Polynomial — plug in karo	Bottom ki koi chinta nahi; sum + const-mult + power	Ex 1
C2	Quotient, $M \neq = 0$	Denominator ek nonzero number ki taraf jaata hai — law cleanly fire karta hai	Ex 2
C3	Quotient $\frac{0}{0}$ , factorable	Law fail; common factor cancel karo	Ex 3
C4	Quotient $\frac{0}{0}$ , rationalising chahiye	Square roots common factor chupaate hain	Ex 4
C5	$\frac{( nonzero )}{0}$ — one-sided split	Law fail, lekin bottom ka sign decide karta hai $\pm \infty$	Ex 5
C6	Product jahan ek piece ka koi limit nahi	Sum/product law forbidden; Squeeze theorem chahiye	Ex 6
C7	Sum jahan har piece diverge kare lekin sum theek ho	Split nahi kar sakte — Mistake B live	Ex 7
C8	Real-world word problem	Units translate karo, phir law apply karo	Ex 8
C9	Exam twist — unknown constant ke liye solve karo	Reverse-engineer karo taaki law fire kar sake	Ex 9

Chalo har cell walk karte hain.

C1 — Polynomial: bas plug in karo

Worked example Example 1 (cell C1)

$x \to 2 lim (3 x^{2} - 5 x + 4)$ find karo.

Forecast: aage padhne se pehle ek number guess karo. Ek polynomial smooth hoti hai bina kisi gap ke — tumhara kya expect hai?

Sum aur constant-multiple laws se split karo: $lim_{x \to 2} (3 x^{2} - 5 x + 4) = 3 lim_{x \to 2} x^{2} - 5 lim_{x \to 2} x + lim_{x \to 2} 4.$ Yeh step kyun? Har piece ab do atoms mein se ek hai ( $lim x = a$ , $lim c = c$ ) ya unka power — jo cheezein hum directly jaante hain.
$lim_{x \to 2} x = 2$ , power law $lim x^{2} = 2^{2}$ , aur $lim 4 = 4$ use karo: $= 3 (4) - 5 (2) + 4.$ Yeh step kyun? Hum bas har atom ko uski known value se cash in kar rahe hain.
Arithmetic: $12 - 10 + 4 = 6$ .

Verify: kyunki yeh ek polynomial $p (x)$ hai, answer zaroori hai $p (2)$ ke barabar. Direct substitution: $3 \cdot 4 - 10 + 4 = 6$ . ✓ Match karta hai. Laws aur plug-in agree karte hain, jaise polynomials ke liye hamesha karte hain.

C2 — Quotient with safe (nonzero) denominator

C3 — Quotient jo $\frac{0}{0}$ deta hai lekin factor hota hai

Worked example Example 3 (cell C3)

$x \to 3 lim \frac{x ^{2} - 9}{x - 3}$ find karo.

Forecast: pehle $x = 3$ plug in karne ki koshish karo. Kya toot ta hai?

Bottom test karo: $lim_{x \to 3} (x - 3) = 0$ , aur top bhi $\to 9 - 9 = 0$ . Toh yeh indeterminate form $\tfrac00$ hai. Yeh step kyun? Quotient law yahan forbidden hai ( $M = 0$ ). $\frac{0}{0}$ ka matlab "koi limit nahi" nahi hai — iska matlab hai "law answer nahi dekh sakta; gehri khudai karo."
Top ko factor karo: $x^{2} - 9 = (x - 3) (x + 3)$ . Yeh step kyun? $\frac{0}{0}$ mein almost hamesha ek shared factor chupta hai. Ise expose karne se hum us cheez ko cancel kar sakte hain jo zero ho rahi thi.
Cancel karo — legitimately: $\frac{( x - 3 ) ( x + 3 )}{x - 3} = x + 3, for x \neq = 3.$ Yeh step kyun? Ek limit sirf $x$ ke $3$ ke paas hone ki parwah karta hai, kabhi $x = 3$ khud ki nahi, toh $x - 3 \neq = 0$ hai aur ise divide out karna honest hai.
Ab sum law: $lim_{x \to 3} (x + 3) = 6$ .

Verify: $x = 2.999 \to 5.999$ , $x = 3.001 \to 6.001$ ke paas table. Dono $6$ par squeeze karte hain. ✓ Graph $y = x + 3$ line hai jisme $(3, 6)$ par ek single hole hai — figure dekho.

Red open circle dekho: function ki wahan koi value nahi hai, lekin limit woh height hai jis taraf line aim kar rahi hai — exactly $6$ .

C4 — $\frac{0}{0}$ jisme rationalising chahiye

Worked example Example 4 (cell C4)

$x \to 0 lim \frac{x + 4 - 2}{x}$ find karo.

Forecast: $x = 0$ plug in karo: top hai $4 - 2 = 0$ , bottom hai $0$ . Wahi $\frac{0}{0}$ trap — lekin square root ki wajah se cancel karne ke liye koi obvious factor nahi hai.

Form confirm karo: top $\to 0$ , bottom $\to 0$ , toh quotient law off-limits hai.
Top aur bottom ko conjugate $x + 4 + 2$ se multiply karo: $\frac{x + 4 - 2}{x} \cdot \frac{x + 4 + 2}{x + 4 + 2} .$ Yeh step kyun? $(u - 2) (u + 2) = u - 4$ square root ko khatam kar deta hai aur hidden common factor manufacture karta hai. Yeh "rationalising" hai — factoring ka root version.
Top ban jaata hai $(x + 4) - 4 = x$ : $\frac{x}{x ( x + 4 + 2 )} = \frac{1}{x + 4 + 2}, x \neq = 0.$ Yeh step kyun? Jo $x$ $\frac{0}{0}$ force kar raha tha woh cancel ho jaata hai, ek aisa piece chhod ke jiska bottom ab safely nonzero hai.
Quotient law apply karo (bottom $\to 4 + 2 = 4 \neq = 0$ ): $\frac{1}{0 + 4 + 2} = \frac{1}{4} .$

Verify: $x = 0.0001$ : $\frac{4.0001 - 2}{0.0001} \approx 0.24999 \dots \approx \frac{1}{4}$ . ✓

C5 — $\frac{nonzero}{0}$ : bottom ka sign rule karta hai

Worked example Example 5 (cell C5 — sab sub-cases)

$x \to 2 lim \frac{1}{x - 2}$ investigate karo.

Forecast: top $1$ ki taraf jaata hai (nonzero), bottom $0$ ki taraf jaata hai. Yeh $\frac{0}{0}$ nahi hai — yeh $\frac{1}{0}$ hai. Guess karo: kya ek single limit exist karta hai?

Bottom $\to 0$ , toh quotient law fail. Lekin top $\to 1 \neq = 0$ , toh yeh kisi acche number tak cancel nahi ho sakta — yeh blow up zaroor karega. Bacha ek hi sawal: $+ \infty$ ki taraf ya $- \infty$ ki taraf? Yeh step kyun? Jab sirf bottom zero hit karta hai, magnitude explode hoti hai; do sides sign mein disagree kar sakti hain, toh hume split karna hoga.
Right side $x \to 2^{+}$ (toh $x > 2$ ): $x - 2$ ek tiny positive number hai, jaise $0.001$ . Tab $\frac{1}{0.001} = + 1000$ . $lim_{x \to 2^{+}} \frac{1}{x - 2} = + \infty.$
Left side $x \to 2^{-}$ (toh $x < 2$ ): $x - 2$ ek tiny negative number hai, jaise $- 0.001$ . Tab $\frac{1}{- 0.001} = - 1000$ . $lim_{x \to 2^{-}} \frac{1}{x - 2} = - \infty.$
Do one-sided limits agree nahi karte, toh two-sided limit exist nahi karta.

Verify: $x = 2.001 \to + 1000$ , $x = 1.999 \to - 1000$ . ✓ Opposite signs, unbounded — figure mein vertical asymptote dekho.

C6 — Product jahan ek factor ka koi limit nahi

Worked example Example 6 (cell C6)

$x \to 0 lim x sin \frac{1}{x}$ find karo.

Forecast: piece $sin \frac{1}{x}$ $x \to 0$ ke paas tezi se oscillate karta rehta hai — iska koi limit nahi hai. Kya product phir bhi settle hoga?

Tum product law use nahi kar sakte. Ise dono limits exist karne chahiye, aur $lim_{x \to 0} sin \frac{1}{x}$ exist nahi karta. Yeh step kyun? Mistake B ulta — ek missing piece limit law ko void kar deta hai.
Wild factor ko bound karo: har input ke liye, $- 1 \leq sin \frac{1}{x} \leq 1$ . $∣ x ∣$ se multiply karo: $- ∣ x ∣ \leq x sin \frac{1}{x} \leq ∣ x ∣.$ Yeh step kyun? Hum wobble track nahi kar sakte, lekin hum ise do tame walls ke beech trap kar sakte hain jo dono $0$ tak jaati hain. Yahi Squeeze theorem hai.
Dono walls $lim_{x \to 0} (- ∣ x ∣) = 0$ aur $lim_{x \to 0} ∣ x ∣ = 0$ . Squeezed: $lim_{x \to 0} x sin \frac{1}{x} = 0.$

Verify: $x = 0.001$ : value $= 0.001 \cdot sin (1000)$ , magnitude $\leq 0.001$ . ✓ Ek shrinking corridor ke andar trapped — figure mein do straight walls $\pm ∣ x ∣$ oscillation ko zero tak pinch karte hue dikhti hain.

C7 — Sum jahan har piece diverge kare lekin total theek ho

C8 — Real-world word problem

Worked example Example 8 (cell C8)

Ek tank drain hota hai jisse pani ki depth $D (t) = \frac{60 t}{t + 4}$ centimetres hai $t$ minutes baad. Jab pump $t = 6$ minutes par apni steady running speed pakadta hai, level kaunsi depth ki taraf ja raha hai?

Forecast: compute karne se pehle cm mein ek depth guess karo.

Hume chahiye $t \to 6 lim \frac{60 t}{t + 4}$ . Bottom check karo: $lim_{t \to 6} (t + 4) = 10 \neq = 0$ . Yeh step kyun? Quotient-law bouncer — $10 \neq = 0$ hai, toh law fire karta hai aur model wahan continuous hai.
Top: $lim_{t \to 6} 60 t = 360$ ( $lim t = 6$ ka constant-multiple).
Quotient law: $\frac{360}{10} = 36$ .

Verify (units ke saath): $\frac{60 \cdot 6}{6 + 4} = \frac{360}{10} = 36$ cm. Units: cm-numerator over dimensionless $(t -in-min)$ ratio cm rehta hai. ✓ Physically sensible: depth positive hai aur $60$ cm ceiling se neeche jo formula $t \to \infty$ mein approach karta hai.

C9 — Exam twist: law ko fire karne dena

Worked example Example 9 (cell C9)

Constant $k$ find karo taaki $x \to 5 lim \frac{x ^{2} + k x - 5}{x - 5}$ exist kare (finite ho).

Forecast: bottom $0$ ki taraf jaata hai. Finite limit ke liye, top ko kya karna chahiye?

Bottom $\to 0$ . Finite limit tabhi possible hai jab top bhi $\to 0$ ho — warna yeh cell C5 ka $\frac{nonzero}{0}$ blow-up hai. Yeh step kyun? Hum jaan-bujhkar ek $\frac{0}{0}$ engineer karte hain taaki ek common factor cancel ho sake aur limit bach jaye.
Top ko $x = 5$ par vanish karne ke liye force karo: $x = 5$ ko $x^{2} + k x - 5$ mein substitute karo aur $0$ set karo: $25 + 5 k - 5 = 0 \Rightarrow 5 k = - 20 \Rightarrow k = - 4.$ Yeh step kyun? Agar $(x - 5)$ top ka factor banna hai, toh top ko $x = 5$ par $0$ equal karna chahiye (factor theorem).
Check karo ki cancel hota hai. $k = - 4$ ke saath: top $= x^{2} - 4 x - 5 = (x - 5) (x + 1)$ . Toh $\frac{( x - 5 ) ( x + 1 )}{x - 5} = x + 1, x \neq = 5.$
Ab limit exist karta hai: $lim_{x \to 5} (x + 1) = 6$ .

Verify: $k = - 4$ , $x = 5.001$ : $\frac{( 5.001 ) ^{2} - 4 ( 5.001 ) - 5}{0.001} \approx 6.001$ . ✓ Limit $6$ ke barabar hai, aur yeh exactly isliye exist karta hai kyunki humne $k$ tune kiya.

Connections

Limit laws — sum, product, quotient, constant multiple — parent rules jo yeh examples exercise karte hain.
Indeterminate forms 0 over 0 — cells C3, C4, C9 sab is form ko resolve karte hain.
One-sided limits — cell C5 ka engine.
Squeeze theorem — cell C6 mein rescue jab product law void ho.
Continuity — isliye C2 aur C8 plug-in tak collapse karte hain.
Epsilon-Delta definition of a limit — har "cancel then substitute" ke peeche ultimate justification.
Derivative as a limit — cell C4 ka rationalising trick exactly wahi hai jo $x$ differentiate karne mein use hota hai.