Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e: Exercises

4.1.6 · D4 · Maths › Calculus I — Limits & Derivatives › Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e

Level 1 — Recognition

L1 ka poora game: kya upar ka argument wahi hai jo neeche hai? Agar haan, toh ratio $1$ hai; agar nahi, toh fix karo.

Exercise 1.1

$x \to 0 lim \frac{sin 7 x}{7 x}$ evaluate karo.

Recall Solution 1.1

KYA: upar ka argument $7 x$ hai, neeche bhi $7 x$ hai. YEH KYUN MATTER KARTA HAI: atomic limit $\frac{s i n ( stuff )}{same stuff} \to 1$ ke liye identical arguments chahiye — yahan woh already match kar rahe hain. $u = 7 x$ lo. Jab $x \to 0$ , $u \to 0$ , aur expression exactly $\frac{s i n u}{u} \to 1$ hai. $1$

Exercise 1.2

$x \to 0 lim \frac{tan x}{x}$ evaluate karo.

Recall Solution 1.2

KYA: $tan x = \frac{sin x}{cos x}$ split karo. KYUN: hum chahte hain ki ek $\frac{s i n x}{x}$ factor appear ho. $\frac{t a n x}{x} = \frac{s i n x}{x} \cdot \frac{1}{c o s x} ⟶ 1 \cdot \frac{1}{1} = 1.$ Jab $x \to 0$ , $cos x \to 1$ , toh doosra factor $\to 1$ . $1$

Level 2 — Application

Ab top aur bottom match nahi karte. Tumhara kaam: $1$ ki ek clever form se multiply karo taaki matching argument force ho, phir leftover constant padh lo.

Exercise 2.1

$x \to 0 lim \frac{sin 4 x}{9 x}$ evaluate karo.

Recall Solution 2.1

KYA: $4 x$ ko sin ke neeche force karo. KYUN: atomic limit ko $\frac{s i n 4 x}{4 x}$ chahiye, isliye hum $\frac{4 x}{4 x} = 1$ insert karte hain. $\frac{s i n 4 x}{9 x} = \frac{s i n 4 x}{4 x} \cdot \frac{4 x}{9 x} = \to 1 \frac{sin 4 x}{4 x} \cdot \frac{4}{9} .$ $\frac{4}{9}$

Exercise 2.2

$x \to 0 lim \frac{sin 3 x}{sin 5 x}$ evaluate karo.

Recall Solution 2.2

KYA: do sines — har ek ke liye $\frac{s i n}{arg}$ banao. KYUN: top aur bottom dono ko $x$ se divide karo aur arguments match karo. $\frac{s i n 3 x}{s i n 5 x} = \frac{\frac{sin 3 x}{3 x} \cdot 3 x}{\frac{sin 5 x}{5 x} \cdot 5 x} = \frac{\frac{sin 3 x}{3 x}}{\frac{sin 5 x}{5 x}} \cdot \frac{3 x}{5 x} ⟶ \frac{1}{1} \cdot \frac{3}{5} .$ $\frac{3}{5}$

Exercise 2.3

$n \to \infty lim (1 + \frac{5}{n})^{n}$ evaluate karo.

Recall Solution 2.3

KYA: $a = 5$ ke saath $e$ -pattern recognize karo. KYUN: parent se master rule, $(1 + \frac{a}{n})^{n} \to e^{a}$ . Log se check karo: $ln [(1 + \frac{5}{n})^{n}] = n ln (1 + \frac{5}{n}) = \frac{ln ( 1 + 5/ n )}{1/ n} \to 5$ , helper limit $lim_{h \to 0} \frac{l n ( 1 + h )}{h} = 1$ (upar stated, Exercise 4.1 mein prove kiya) use karke, $h = 5/ n \to 0$ ke saath. Kyunki expression ka $ln$ $\to 5$ hai, expression khud $\to e^{5}$ . $e^{5}$

Level 3 — Analysis

Yahan tumhe ek mushkil limit ko atomic pieces ke product mein split karna hai, har ek ko exactly woh power deni hai jo use chahiye.

Exercise 3.1

$x \to 0 lim \frac{1 - cos x}{x sin x}$ evaluate karo.

Recall Solution 3.1

KYA: hum helper limit $\frac{1 - c o s x}{x ^{2}} \to \frac{1}{2}$ use karte hain (upar listed). $x^{2}$ expose karne ke liye denominator rewrite karo. KYUN: $x sin x = x^{2} \cdot \frac{s i n x}{x}$ split karo, taaki har factor ek known limit ho. $\frac{1 - cos x}{x sin x} = \frac{1 - cos x}{x ^{2}} \cdot \frac{x ^{2}}{x sin x} = \frac{1 - cos x}{x ^{2}} \cdot \frac{x}{sin x} .$ Jab $x \to 0$ : $\frac{1 - cos x}{x ^{2}} \to \frac{1}{2}$ aur $\frac{x}{sin x} \to 1$ . $\frac{1}{2}$

Exercise 3.2

Taylor expansion $sin x = x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots$ use karke $x \to 0 lim \frac{x - sin x}{x ^{3}}$ evaluate karo.

Recall Solution 3.2

KYA: series ko $x$ se subtract karo. KYUN: leading terms cancel ho jaate hain, $x^{3}$ behaviour reveal hota hai. (Taylor Series dekho.) $x - sin x = x - (x - \frac{x ^{3}}{6} + \frac{x ^{5}}{120} - \dots) = \frac{x ^{3}}{6} - \frac{x ^{5}}{120} + \dots$ $x^{3}$ se divide karo: $\frac{x - s i n x}{x ^{3}} = \frac{1}{6} - \frac{x ^{2}}{120} + \dots ⟶ \frac{1}{6} .$ $\frac{1}{6}$

Exercise 3.3

$n \to \infty lim (\frac{2 n + 3}{2 n - 1})^{n}$ evaluate karo.

Recall Solution 3.3

KYA: base ko $1 + \frac{kuch}{2 n - 1}$ ki tarah rewrite karo. KYUN: $e$ -pattern match karne ke liye hume $1 + \frac{u}{denominator}$ chahiye. $\frac{2 n + 3}{2 n - 1} = \frac{( 2 n - 1 ) + 4}{2 n - 1} = 1 + \frac{4}{2 n - 1} .$ Ab exponent ko denominator se match karo: $(1 + \frac{4}{2 n - 1})^{n} = [(1 + \frac{4}{2 n - 1})^{2 n - 1}]^{\frac{n}{2 n - 1}} .$ Inner bracket $\to e^{4}$ ; exponent $\frac{n}{2 n - 1} \to \frac{1}{2}$ . $\to (e^{4})^{1/2} = e^{2} .$ $e^{2}$

Level 4 — Synthesis

Tools ko rebuild karo aur ek hi problem mein dono families ko combine karo.

Exercise 4.1

Sirf $e = lim_{n \to \infty} (1 + \frac{1}{n})^{n}$ ki definition aur $ln$ ki continuity use karke, L'Hôpital ke bina $h \to 0 lim \frac{ln ( 1 + h )}{h} = 1$ prove karo.

Recall Solution 4.1

KYA: $e$ -limit ko $ln$ ke baare mein ek statement mein convert karo. KYUN: $ln$ continuous hai, isliye yeh limits ke saath commute karta hai.

Right side, $h \to 0^{+}$ . $h = \frac{1}{n}$ rakho, toh $h \to 0^{+}$ ka matlab $n \to \infty$ hai. Phir $\frac{l n ( 1 + h )}{h} = n ln (1 + \frac{1}{n}) = ln (1 + \frac{1}{n})^{n} .$ Kyunki $ln$ continuous hai, hum limit andar pass kar sakte hain: $lim_{n \to \infty} ln (1 + \frac{1}{n})^{n} = ln (lim_{n \to \infty} (1 + \frac{1}{n})^{n}) = ln e = 1.$

Left side, $h \to 0^{-}$ . Ab $h < 0$ hai, isliye $h = \frac{1}{n}$ (positive) use nahi ho sakta — hum $h = - \frac{1}{n}$ substitute karte hain $n \to \infty$ ke saath. Phir $\frac{l n ( 1 + h )}{h} = \frac{l n ( 1 - \frac{1}{n} )}{- \frac{1}{n}} = n ln (\frac{1}{1 - \frac{1}{n}}) = n ln (\frac{n}{n - 1}) = n ln (1 + \frac{1}{n - 1}) .$ $n = (n - 1) + 1$ likho taaki exponent denominator se match kare: $ln [(1 + \frac{1}{n - 1})^{n - 1} \cdot (1 + \frac{1}{n - 1})^{1}] \to ln (e \cdot 1) = 1.$ Dono one-sided limits $1$ ke barabar hain, isliye two-sided limit $1$ hai. $■$ (Yeh exactly $ln x$ ka $x = 1$ par derivative hai — The Number e and ln dekho.)

Exercise 4.2

$x \to 0 lim \frac{e ^{2 x} - 1}{sin 3 x}$ evaluate karo. (Page ke top par listed helper $lim_{u \to 0} \frac{e ^{u} - 1}{u} = 1$ use karo.)

Recall Solution 4.2

KYA: do atomic limits milte hain — exponential ratio aur sine ratio. KYUN: matching arguments insert karke har ratio build karo. $\frac{e ^{2 x} - 1}{s i n 3 x} = \to 1 \frac{e ^{2 x} - 1}{2 x} \cdot \to 1 \frac{3 x}{sin 3 x} \cdot \frac{2 x}{3 x} .$ Last factor $\frac{2}{3}$ hai. $\frac{2}{3}$

Level 5 — Mastery

Edge cases, degenerate inputs, aur poora geometric proof.

Exercise 5.1 (edge case)

$n \to \infty lim (1 + \frac{1}{n ^{2}})^{n}$ kya hai? Aur $n \to \infty lim (1 + \frac{1}{n})^{n}$ kya hai?

Recall Solution 5.1

Principle: $a_{n} \to 0$ ke saath $(1 + a_{n})^{b_{n}}$ ke liye, log deta hai $b_{n} ln (1 + a_{n}) \approx b_{n} a_{n}$ ( $\frac{l n ( 1 + h )}{h} \to 1$ use karke). Limit $e^{l i m b_{n} a_{n}}$ hai.

Pehla: $a_{n} = \frac{1}{n ^{2}}$ , $b_{n} = n$ , product $= n \cdot \frac{1}{n ^{2}} = \frac{1}{n} \to 0$ . Toh limit $= e^{0} = 1$ . $1$ KYUN: base $1$ ki taraf exponent ke badhne se tez ja raha hai, isliye compounding kabhi accumulate nahi hota — yeh $1$ par aa ke khatam ho jaata hai.

Doosra: $a_{n} = \frac{1}{n}$ , $b_{n} = n$ , product $= \frac{n}{n} = n \to \infty$ . Toh limit $= e^{+ \infty} = + \infty$ . $+ \infty$ KYUN: ab base exponent ke relative vanish hone ke liye bahut slowly badh raha hai — product blow up ho jaata hai. Sirf exact balance $a_{n} = \frac{c}{n}$ (product $\to c$ ) finite $e^{c}$ par land karta hai.

Exercise 5.2 (sign / two-sided case)

Kya $x \to 0 lim \frac{sin ∣ x ∣}{x}$ defined hai? $x \to 0^{+}$ aur $x \to 0^{-}$ alag alag analyse karo.

Recall Solution 5.2

KYA: $∣ x ∣$ numerator mein sign ignore karta hai, lekin denominator sign rakhta hai. KYUN split karo: dono one-sided limits alag ho sakte hain.

$x \to 0^{+}$ : yahan $∣ x ∣ = x$ , toh $\frac{s i n x}{x} \to 1$ .
$x \to 0^{-}$ : yahan $∣ x ∣ = - x$ , toh $\frac{s i n ( - x )}{x} = \frac{- s i n x}{x} \to - 1$ ( $sin (- x) = - sin x$ aur $\frac{s i n x}{x} \to 1$ use karke).

Left limit $= - 1$ , right limit $= + 1$ . Woh alag hain, isliye: $the two-sided limit does not exist.$ Plain $\frac{s i n x}{x}$ se compare karo, jo even hai ( $\frac{s i n ( - x )}{- x} = \frac{s i n x}{x}$ ) aur isliye dono sides se $1$ limit rakhta hai.

Exercise 5.3 (poora geometric proof, sabhi cases)

Areas use karke $x \to 0^{+}$ ke liye $lim_{x \to 0} \frac{s i n x}{x} = 1$ prove karo, phir $x \to 0^{-}$ justify karo.

Recall Solution 5.3

Setup (neeche figure dekho). Unit circle par $0 < x < \frac{π}{2}$ ka central angle $x$ lo. Teen nested regions dekho.

Figure — Important limits — lim(sin x - x) = 1, lim((1+1 - n)ⁿ) = e

Figure — Radius $1$ ka unit circle jisme central angle $x$ hai. White inner triangle (centre, point $(1, 0)$ , aur circle point) ki height $sin x$ hai, amber mein bani hai, aur area $\frac{s i n x}{2}$ hai. Cyan shaded wedge circular sector hai, area $\frac{x}{2}$ . Amber outer triangle tangent line tak height $tan x$ (cyan segment) tak jaata hai, area $\frac{t a n x}{2}$ . Bahar ki taraf padhne par shapes strictly nest karti hain: white $\subset$ cyan $\subset$ amber, jo exactly neeche di gayi inequality chain hai.

Kyunki inner triangle $\subset$ sector $\subset$ outer triangle: $\frac{s i n x}{2} \leq \frac{x}{2} \leq \frac{t a n x}{2} .$ $2$ se multiply karo, $sin x > 0$ se divide karo (direction preserved kyunki positive hai): $1 \leq \frac{x}{s i n x} \leq \frac{1}{c o s x} .$ Reciprocals lo (yeh inequalities flip kar deta hai): $cos x \leq \frac{s i n x}{x} \leq 1.$ Jab $x \to 0^{+}$ , $cos x \to 1$ ; right bound $1$ hai. Squeeze (Sandwich) Theorem se middle trap ho jaata hai, toh $\frac{s i n x}{x} \to 1$ . Left side ( $x \to 0^{-}$ ): $\frac{s i n x}{x}$ even hai — $x$ ki jagah $- x$ rakhne se yeh unchanged rehta hai kyunki $sin$ aur denominator dono ka sign ek saath flip hota hai. Toh left limit right limit ke barabar hai, aur poora two-sided limit $1$ hai. $■$

Connections

Squeeze (Sandwich) Theorem — yahan har $\frac{s i n x}{x}$ exercise ka engine.
Radian Measure — kyun L5 proofs degrees mein collapse ho jaate hain.
The Number e and ln — L2/L3/L4 $e$ -limits aur $\frac{l n ( 1 + h )}{h}$ tool.
Taylor Series — 3.2 mein shortcut.
Indeterminate Forms — is page par har jagah $0/0$ aur $1^{\infty}$ .
Derivative of sin and cos — yeh limits jo unlock karte hain uska payoff.
L'Hôpital's Rule — 3.2 / 4.2 ka ek alternative (lekin atomic limits ke liye circular hai).