Exercises — Intuitive concept of a limit — table of values, graphical

KYA: Left column padho (jaise $x$ $2$ ki taraf badhta hai) aur right column padho (jaise $x$ $2$ ki taraf ghatta hai). KYUN: Definition kehti hai dono sides ko ek hi same number par aim karna chahiye. Left outputs: $4.9\to 4.99\to 4.999$ — $5$ ki taraf chadh rahe hain. Right outputs: $5.001\to 5.01\to 5.1$... ruko, inhe $2$ ki taraf padho: $5.001\to 5.01\to 5.1$ door ja raha hai; $2$ ki taraf yeh hain $5.001,5.0\dots$ yaani $5$ ki taraf shrink ho rahe hain. Dono sides $5$ par squeeze kar rahe hain. ${\lim }_{x\to 2}f(x)=5$

KYA: Dono one-sided limits compare karo, $f(3)$ ko ignore karo. KYUN: Limit approach ko dekhti hai, kabhi point khud ko nahi. Dono one-sided values $7$ ke barabar hain, toh yeh agree karte hain. $f(3)=100$ ki stray value ek akela dot hai jise trend kabhi touch nahi karta. ${\lim }_{x\to 3}f(x)=7$

KYA: Existence condition yaad karo. KYUN: $f(a)$ irrelevant hai (holes aur stray dots ke bhi limits hote hain). Dono one-sided limits exist bhi hone chahiye aur equal bhi hone chahiye. Sahi answers: (b), (c), (d). (a) nahi.

KYA: $1$ ke paas values plug in karo (kabhi $1$ khud nahi).

Dono sides $2$ par aim kar rahe hain. KYUN algebraically confirm karein: table sirf suggest karta hai. Sure hone ke liye factor karo — valid hai kyunki $x\to 1$ matlab $x\ne 1$, toh $x-1\ne 0$: $\frac{x^2-1}{x-1}=\frac{(x-1)(x+1)}{x-1}=x+1(x\ne 1).$ Jaise $x\to 1$, $x+1\to 2$. ${\lim }_{x\to 1}\frac{x^2-1}{x-1}=2.$

KYA: Plug in try karo: $\frac{16-16}{0}=\frac{0}{0}$ — ek $0/0$ trap, toh substitution fail ho jaati hai. KYUN factor karein: numerator ek difference of squares hai $x^2-16=(x-4)(x+4)$; $(x-4)$ cancel ho jaata hai (valid hai kyunki $x\ne 4$): $\frac{(x-4)(x+4)}{x-4}=x+4(x\ne 4).$ Jaise $x\to 4$: $x+4\to 8$. ${\lim }_{x\to 4}\frac{x^2-16}{x-4}=8.$

KYA: Substitution deta hai $\frac{3-3}{0}=\frac{0}{0}$. Table:

Dono sides $\approx 0.1667$ ki taraf squeeze ho rahe hain. KYUN algebra karein: conjugate $\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}$ se multiply karo root clear karne ke liye: $\frac{(\sqrt{x+9}-3)(\sqrt{x+9}+3)}{x(\sqrt{x+9}+3)}=\frac{(x+9)-9}{x(\sqrt{x+9}+3)}=\frac{x}{x(\sqrt{x+9}+3)}=\frac{1}{\sqrt{x+9}+3}.$ Jaise $x\to 0$: $\frac{1}{\sqrt{9}+3}=\frac{1}{6}$. ${\lim }_{x\to 0}\frac{\sqrt{x+9}-3}{x}=\frac{1}{6}\approx \mathrm{0.1667.}$

KYA / KAISA DIKHTA HAI: Figure mein, left piece (lavender) line $y=x+1$ hai; jaise $x\to 2^{-}$ uski height $2+1=3$ par chadh jaati hai. Right piece (coral) $y=2x-1$ hai; jaise $x\to 2^{+}$ uski height $2(2)-1=3$ par aa jaati hai. Akela butter dot $(2,5)$ par $g(2)$ hai.

• ${\lim }_{x\to 2^{-}}g=3$
• ${\lim }_{x\to 2^{+}}g=3$ KYUN two-sided limit exist karta hai: dono sides $3$ par agree karti hain, toh ${\lim }_{x\to 2}g=3.$ Lekin $g(2)=5$ — woh stray dot. Limit ($3$) aur value ($5$) alag hain. Yeh exactly ek removable-discontinuity picture hai; dekho Continuity at a point.

KYA: $x>0$ ke liye, $|x|=x$ toh $h(x)=\frac{x}{x}=1$. $x<0$ ke liye, $|x|=-x$ toh $h(x)=\frac{-x}{x}=-1$. KAISA DIKHTA HAI: do flat rays — right par mint ray height $+1$ par, left par coral ray height $-1$ par, $x=0$ ke paas ek jump ke saath.

• ${\lim }_{x\to 0^{-}}h=-1$
• ${\lim }_{x\to 0^{+}}h=+1$ KYUN limit nahi hai: left $(-1)\ne$ right $(+1)$. Dono paths alag heights par aim kar rahe hain, toh ${\lim }_{x\to 0}h\text{ does not exist.}$

KYA: $3$ ke paas values daalo. Right se ($x=3.1,3.01,3.001$): $x-3$ ek tiny positive number hai, toh $\frac{1}{x-3}$ bahut bada aur positive hai — yeh $+\infty$ ki taraf blow up karta hai. Left se ($x=2.9,2.99,2.999$): $x-3$ ek tiny negative number hai, toh $\frac{1}{x-3}$ bahut bada aur negative hai — yeh $-\infty$ ki taraf dive karta hai. KYUN finite limit nahi hai: outputs kisi bhi number par settle nahi hote; woh infinity ki taraf bhaag jaate hain, aur dono sides par opposite infinities ki taraf. ${\lim }_{x\to 3}p\text{ does not exist (finite); }{\lim }_{x\to 3^{+}}p=+\infty ,{\lim }_{x\to 3^{-}}p=-\infty .$ Yeh "blow-up" behaviour One-sided limits and limits at infinity ka topic hai.

KYA: Dono one-sided limits compute karo. Left: $x^2+1\to 1^2+1=2$. Right: $3x-1\to 3(1)-1=2$. KYUN limit $k$ se independent hai: dono sides pehle se hi $2$ par agree karti hain, aur limit $q(1)=k$ ko completely ignore karti hai. Toh: ${\lim }_{x\to 1}q=2\text{for **any** }k.$ $k=2$ choose karna $q$ ko additionally continuous bana deta hai; lekin limit $2$ hi rehti hai chahe $k$ kuch bhi ho. (Contrast: continuity demand karti $k=2$; limit nahi karti — dekho Continuity at a point.)

KYA: Substitution deta hai $\frac{\sin 0}{0}=\frac{0}{0}$ — indeterminate. Cancel karne ke liye koi algebraic factor nahi hai ($\sin x$ koi polynomial nahi hai jisme se $x$-factor cleanly nikala ja sake), toh hum table banate hain:

KYUN dono sides matter karti hain: left aur right dono $1$ ki taraf chadh rahe hain jaise $x\to 0$. ${\lim }_{x\to 0}\frac{\sin x}{x}=1.$ KYUN substitution/cancelling fail hote hain: substitution $0/0$ pe atak jaati hai; aur $\sin x$ ko $x^2-1$ ki tarah factor nahi kiya ja sakta, toh hum cancel karke bahar nahi nikal sakte. Yahi precisely woh case hai jiske liye Squeeze (Sandwich) Theorem banaya gaya tha rigorously, aur yeh The derivative as a limit of difference quotients ka seed hai.

KYA / KAISA DIKHTA HAI: Figure mein, coral line ek secant hai — yeh parabola ko $(3,9)$ aur $(3+h,(3+h{)}^2)$ par kaatti hai. Jaise $h$ shrink hota hai, doosra point pehle ki taraf slide karta hai aur secant lavender tangent ki taraf pivot karti hai. Difference quotient secant ka slope hai. Substitution $\frac{0}{0}$ deta hai, toh table banate hain:

Dono sides $6$ par aim kar rahe hain. KYUN / algebra confirm karne ke liye: expand karo (valid hai kyunki limit mein $h\ne 0$, toh hum $h$ cancel kar sakte hain): $\frac{(3+h{)}^2-9}{h}=\frac{9+6h+h^2-9}{h}=\frac{6h+h^2}{h}=6+h(h\ne 0).$ Jaise $h\to 0$: $6+h\to 6$. ${\lim }_{h\to 0}\frac{f(3+h)-f(3)}{h}=6.$ MATLAB: $6$ exactly $y=x^2$ ki steepness (slope) hai single point $x=3$ par — lavender tangent line ka slope. Yahi derivative hai; dekho The derivative as a limit of difference quotients.

KYA: Jaise $x\to 0$, andar ka $\frac{1}{x}$ $\pm \infty$ ki taraf race karta hai, toh $\cos (1/x)$ apni poori range $[-1,1]$ mein over and over swing karta hai — zyada se zyada tezi se.

KYUN claim fail hota hai: outputs kabhi settle nahi hote — chahe $x$ $0$ ke kitna bhi paas aa jaaye, woh $-1$ aur $1$ ke beech alternate karte rehte hain. Koi single approached value nahi hai. ${\lim }_{x\to 0}\cos \left(\frac{1}{x}\right)\text{ does not exist.}$ (Contrast: ${\lim }_{x\to 0}x\cos (1/x)=0$, kyunki $x$ factor wobble ko crush kar deta hai — yeh kaam hai Squeeze (Sandwich) Theorem ka.)

(1) Jump — left aur right one-sided limits alag hain (Ex 3.2). (2) Blow-up — outputs $\pm \infty$ ki taraf bhaag jaate hain (Ex 3.3). (3) Oscillation — outputs kabhi ek value par settle nahi hote (Ex 5.2).

$6$ sirf ek number kyun nahi balki ek slope kyun hai? Kyunki difference quotient $y=x^2$ par do points se guzarne wali secant line ka slope hai; jaise $h\to 0$ secant $x=3$ par tangent ban jaati hai, jiska slope $6$ hai.

$0/0$ ka matlab "no limit" nahi tha lekin Ex 3.3 mein blow-up ka matlab no finite limit tha? $0/0$ indeterminate hai — ek shared vanishing factor cancel ho sakta hai aur ek finite value reveal kar sakta hai. Ex 3.3 mein sirf denominator $0$ ki taraf gaya (numerator $1$ tha), toh output genuinely explode ho gaya.