Worked examples — Intuitive concept of a limit — table of values, graphical

4.1.1 · D3 · Maths › Calculus I — Limits & Derivatives › Intuitive concept of a limit — table of values, graphical

Shuru karne se pehle, woh ek rule yaad karo jo yeh sab govern karta hai: ek limit dekhta hai ki $f(x)$ kahan ja raha hai jab $x$ dono sides se $a$ ki taraf creep karta hai, aur yeh ignore karta hai ki $a$ par khud kya hota hai. Ise pocket mein rakh lo.

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The scenario matrix

Har limit-at-a-point problem inhi cells mein se kisi ek mein aata hai. Har row ek behaviour class hai; last column us example ka naam hai jo use conquer karta hai.

#	Case class	Jab dekho toh kya dikhta hai	Trap	Example
A	Direct substitution kaam karta hai (continuous)	Curve smoothly point se guzarti hai	koi nahi	Ex 1
B	$0/0$ hole, factor & cancel	Line with a missing dot	plug in nahi kar sakte	Ex 2
C	$0/0$ jisme rationalising trick chahiye (roots)	Smooth curve, hidden hole	factoring kaam nahi karega	Ex 3
D	Two-sided limit fail hoti hai — a jump	Do pieces alag-alag heights par	ek side kaafi nahi	Ex 4
E	Blow-up to infinity ($k/0$, $k\ne 0$)	Curve rockets up/down	"$=\infty$" koi number nahi hai	Ex 5
F	Sign matters — left aur right opposite infinities par jaate hain	Ek branch neeche dives, doosri upar soars	denominator ke signs	Ex 5
G	Limit value $\ne$ function value (stray dot)	Curve ek jagah ja rahi hai, dot kahin aur rakha hai	$f(a)$ par trust karna	Ex 6
H	Wild oscillation — koi destination nahi	Curve $a$ ke paas forever wobble karta rahta hai	yeh maanana ki limit exist karni chahiye	Ex 7
I	Real-world word problem	Average speed instantaneous ki taraf ja rahi hai	"approaches" ko interpret karna	Ex 8
J	Exam twist — limit at infinity	Curve ek shelf ki taraf flatten ho rahi hai	large-$x$ dominance	Ex 9

Ab hum board clear karte hain, ek cell at a time.

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Ex 1 — Cell A: substitution bas kaam kar jaata hai

Forecast: yeh ek plain polynomial hai jisme koi fraction nahi, koi roots nahi — kuch bhi toot nahi sakta. Aage padhne se pehle value guess karo: agar simply $x=2$ rakh do toh kya milta hai?

Steps.

1. $x=2$ expression mein daalo: $3(2{)}^2-2+4$. Yeh step kyun? Ek polynomial ke liye koi division by zero nahi hai aur koi undefined operation nahi hai, toh curve ink ki ek unbroken line hai — uski destination uski actual value ke barabar hoti hai. (Yahi idea hai Continuity at a point ki: aise functions ke liye, limit = plug-in.)
2. Compute karo: $3\cdot 4-2+4=12-2+4=14$.

${\lim }_{x\to 2}(3x^2-x+4)=14$

Verify: $2$ se thodi kum value test karo, maano $x=1.99$: $3(1.99{)}^2-1.99+4=3(3.9601)-1.99+4=11.8803+2.01=13.8903$, jo $14$ ke paas aa raha hai. ✓

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Ex 2 — Cell B: $0/0$ hole, factoring se theek hota hai

Forecast: $x=3$ daalo aur milta hai $\frac{9-9}{3-3}=\frac{0}{0}$ — ek trap hai, answer nahi. Lekin top clearly factor karna chahta hai. Tumhare khayal mein curve kahan ja rahi hai?

Steps.

1. Notice karo ki $x^2-9=(x-3)(x+3)$ (difference of two squares). Yeh step kyun? Hume $0/0$ isliye mila kyunki top aur bottom dono mein secretly $(x-3)$ factor hai; factoring use expose karta hai taaki hum common cause deal kar sakein.
2. Shared $(x-3)$ cancel karo: $\frac{(x-3)(x+3)}{x-3}=x+3(x\ne 3).$ Yeh step kyun? Cancelling legal hai kyunki $x\to 3$ ka matlab hai $x$ paas hai lekin kabhi equal nahi $3$ ke, toh $x-3\ne 0$ aur isse divide karna allowed hai. Single point $x=3$ se door, function sirf tame line $x+3$ hai.
3. Ab substitution trap khatam ho gayi — $x+3$ ka limit $3$ par lo: $3+3=6$.

${\lim }_{x\to 3}\frac{x^2-9}{x-3}=6$

Verify: $x=2.999$ par table check: $\frac{2.999^2-9}{2.999-3}=\frac{8.994001-9}{-0.001}=\frac{-0.005999}{-0.001}=5.999$, jo $6$ ki taraf aa raha hai. ✓

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Ex 3 — Cell C: root ke saath $0/0$, rationalising se theek hota hai

Forecast: $x=0$ daalo: $\frac{\sqrt{4}-2}{0}=\frac{0}{0}$. Factoring kaam nahi karega — culprit square root hai. Guess karo: kya answer small lagta hai (fraction jaisa) ya bada?

Steps.

1. Top aur bottom ko conjugate $\sqrt{x+4}+2$ se multiply karo: $\frac{\sqrt{x+4}-2}{x}\cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}.$ Yeh step kyun? Conjugate kyun, factoring kyun nahi? Kyunki $(\sqrt{u}-2)(\sqrt{u}+2)=u-4$ top par se root mita deta hai. Hum yeh specific tool isliye use karte hain kyunki square roots ka difference exactly wahi hai jise "difference of squares" flatten karne ke liye banaya gaya tha.
2. Numerator ban jaata hai $(x+4)-4=x$: $\frac{x}{x\left(\sqrt{x+4}+2\right)}.$ Yeh step kyun? Yahi hamaara reward hai — woh offensive $x$ jo $0/0$ cause kar raha tha, ab ek clean common factor ki tarah appear ho raha hai.
3. $x$ cancel karo (legal hai kyunki $x\to 0$ ka matlab $x\ne 0$): $\frac{1}{\sqrt{x+4}+2}.$
4. $x=0$ substitute karo: $\frac{1}{\sqrt{4}+2}=\frac{1}{2+2}=\frac{1}{4}$.

${\lim }_{x\to 0}\frac{\sqrt{x+4}-2}{x}=\frac{1}{4}$

Verify: $x=0.001$ par: $\frac{\sqrt{4.001}-2}{0.001}=\frac{2.00024998-2}{0.001}=\frac{0.00024998}{0.001}=0.24998\approx 0.25$. ✓ (Yeh trick Indeterminate forms and algebraic simplification ka dil hai.)

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Ex 4 — Cell D: ek jump, isliye two-sided limit fail ho jaati hai

Forecast: $|x|$ hai "$x$ ko positive banao." $x$ se divide karna… negatives ke liye kya hoga versus positives ke liye? Guess karo ki koi ek clean number aata hai ya nahi.

Steps.

1. $x>0$ ke liye, $|x|=x$, toh $f(x)=\frac{x}{x}=1$. Right-hand limit $\underset{x\to 0^{+}}{\lim }f(x)=1$. Yeh step kyun? Right branch par (figure mein red horizontal segment) function height $1$ par flat line hai.
2. $x<0$ ke liye, $|x|=-x$, toh $f(x)=\frac{-x}{x}=-1$. Left-hand limit $\underset{x\to 0^{-}}{\lim }f(x)=-1$. Yeh step kyun? Left branch par (black segment) function $-1$ par flat baitha hai. Figure dekho: dono segments $x=0$ ki taraf alag-alag heights se aa rahe hain.
3. Compare karo: left $-1$ deta hai, right $+1$ deta hai. Dono disagree kar rahe hain.

${\lim }_{x\to 0}\frac{|x|}{x}\text{does not exist.}$

Verify: $x=-0.001\Rightarrow \frac{0.001}{-0.001}=-1$; $x=+0.001\Rightarrow \frac{0.001}{0.001}=1$. Do alag targets ⇒ koi single limit nahi. ✓ (Aur detail One-sided limits and limits at infinity mein.)

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Ex 5 — Cells E & F: blow-up, aur dono sides opposite infinities par jaana

Forecast: $x=2$ par dono denominators $0$ hain lekin tops $1$ hain, $0$ nahi — toh yeh not $0/0$ hai. Yeh hai "ek nonzero number ek aisi cheez par jo zero ki taraf shrink ho rahi hai," jo explode karta hai. Lekin kya upar explode hoga ya neeche? Yeh denominator ki sign par depend karta hai.

Part I — $\frac{1}{x-2}$ (Cell F, opposite signs).

1. Right se approach karo ($x\to 2^{+}$): tab $x-2$ ek tiny positive number hai, toh $\frac{1}{x-2}$ ek huge positive number hai. $\underset{x\to 2^{+}}{\lim }\frac{1}{x-2}=+\infty$. Yeh step kyun? $1$ ko $0.001$ jaisi cheez se divide karne par $1000$ milta hai; $0.0001$ se divide karne par $10000$ — koi ceiling nahi. Figure mein red branch upar shoot karta hai.
2. Left se approach karo ($x\to 2^{-}$): tab $x-2$ ek tiny negative number hai, toh $\frac{1}{x-2}$ ek huge negative number hai. $\underset{x\to 2^{-}}{\lim }\frac{1}{x-2}=-\infty$. Yeh step kyun? Black branch neeche dive karta hai. Kyunki dono sides opposite infinities ki taraf jaate hain, two-sided limit exist nahi karti (aur "$\infty$" bhi nahi hai).

Part II — $\frac{1}{(x-2{)}^2}$ (Cell E, same infinity).

3. Ab denominator $(x-2{)}^2$ ek square hai, toh yeh dono sides se positive hai. Dono one-sided limits $+\infty$ par jaate hain. Yeh step kyun? Squaring sign issue khatam kar deta hai; kisi bhi side se hum $1$ ko ek tiny positive number se divide karte hain. Yahaan hum likh sakte hain $\underset{x\to 2}{\lim }\frac{1}{(x-2{)}^2}=+\infty$ kyunki dono sides same runaway direction par agree karte hain.

Verify: $\frac{1}{2.001-2}=1000$ (upar) lekin $\frac{1}{1.999-2}=-1000$ (neeche) → sides disagree. Square ke liye: $\frac{1}{(2.001-2{)}^2}=10^6$ aur $\frac{1}{(1.999-2{)}^2}=10^6$ → dono huge positive. ✓

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Ex 6 — Cell G: stray dot (limit ≠ value)

Forecast: definition ek random dot height $7$ par park karti hai jab $x=1$. Lekin limit sirf approach dekhti hai — kya woh dot matter karta hai?

Steps.

1. Har $x\ne 1$ ke liye rule hai $x^2+1$, aur limit kabhi $x$ ko $1$ equal nahi hone deti. Yeh step kyun? Kyunki limit single point $x=1$ ko ignore karti hai, hum simply woh formula use karte hain jo saare nearby points govern karta hai.
2. $x^2+1$ ka limit $1$ par lo: $1^2+1=2$.
3. Lekin defined value hai $g(1)=7$.

${\lim }_{x\to 1}g(x)=2,g(1)=7,2\ne 7.$

Curve height $2$ ki taraf ja rahi hai; $7$ par akela dot limit ke liye irrelevant hai.

Verify: $x=0.999$ par: $0.999^2+1=1.998001$, aur $x=1.001$ par: $1.001^2+1=2.002001$ — dono $2$ ki taraf squeeze ho rahe hain, $7$ ki taraf nahi. ✓

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Ex 7 — Cell H: endless oscillation, koi destination nahi

Forecast: jab $x\to 0$, fraction $\frac{1}{x}$ enormous ho jaata hai, toh sine ke andar ka angle tezi se spin karta hai. Kya output kabhi settle hota hai?

Steps.

1. $0$ ke paas $\frac{1}{x}$ ko samjho: $x=0.1$ ke liye yeh $10$ hai; $x=0.01$ ke liye $100$; $x=0.001$ ke liye $1000$. Yeh infinity ki taraf race karta hai. Yeh step kyun? $\sin$ ka behaviour poori tarah uske input angle par depend karta hai, toh pehle dekhna zaroori hai ki woh angle kitni tezi se badh raha hai.
2. Ek ever-growing angle ka $\sin$ full range $[-1,1]$ ko sweep karta rehta hai, baar baar, $0$ ke aas-paas kisi bhi tiny window mein infinitely many times. Yeh step kyun? Figure mein red curve dekho: $x=0$ ke paas yeh oscillation ka ek solid band hai — koi single height nahi hai jiske paas yeh jaata ho.
3. Kyunki hum $0$ ke arbitrarily close aisi points dhundh sakte hain jahan value $+1$ hai aur utne hi close doosre points hain jahan value $-1$ hai, koi single number $L$ approach nahi ho raha.

${\lim }_{x\to 0}\sin \left(\frac{1}{x}\right)\text{does not exist.}$

Verify: $x=\frac{2}{\pi }$-type inputs par value $1$ hai, doosron par $-1$. Lo $x_1=\frac{1}{5000\pi }$ (value $0$) aur $x_2=\frac{2}{(4k+1)\pi }$ (value $1$), dono $0$ ke paas — alag heights ⇒ no limit. ✓ (Squeeze (Sandwich) Theorem se compare karo: ise $x$ se multiply karna wobble ko clamp kar deta, isliye ek limit force ho jaati.)

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Ex 8 — Cell I: real-world word problem (average → instantaneous)

Forecast: jab time window $h$ zero ki taraf shrink hota hai, average speed $t=2$ ke instant par "speedometer reading" par home in karni chahiye. Guess karo: $20$ se bada ya chhota?

Steps.

1. Pieces compute karo: $s(2+h)=5(2+h{)}^2=5(4+4h+h^2)=20+20h+5h^2$, aur $s(2)=5(4)=20$. Yeh step kyun? Average karne se pehle hume window mein change in distance chahiye.
2. Average speed banao: $v_{\text{avg}}(h)=\frac{(20+20h+5h^2)-20}{h}=\frac{20h+5h^2}{h}.$ Yeh step kyun? Direct substitution $h=0$ se $\frac{0}{0}$ milta hai — wahi trap jaise pehle, kyunki window ki width zero hai. Toh hum simplify karte hain.
3. $h$ cancel karo (legal hai kyunki $h\to 0$ ka matlab $h\ne 0$): $v_{\text{avg}}(h)=20+5h$.
4. Limit lo: $\underset{h\to 0}{\lim }(20+5h)=20$.

$\text{Instantaneous speed at }t=2\text{ s}=20\text{m/s}.$

Verify (units + number): $5t^2$ mein metres hain; metres ko seconds se divide karne par m/s milta hai ✓. $h=0.001$ par: $20+5(0.001)=20.005$ m/s, jo $20$ ke paas hai. ✓ (Yahi The derivative as a limit of difference quotients ka seed hai.)

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Ex 9 — Cell J: exam twist — limit at infinity

Forecast: ab $x$ kisi point ki jagah infinity ki taraf jaata hai. Top aur bottom dono explode hote hain, toh yeh ek $\infty /\infty$ tug-of-war hai. Winner decide hota hai highest powers se. Shelf height guess karo.

Steps.

1. Har term ko, top aur bottom mein, highest power $x^2$ se divide karo: $\frac{3+\frac{5}{x}}{2-\frac{7}{x^2}}.$ Yeh step kyun? Hum yeh dividing trick isliye use karte hain kyunki jab $x\to \infty$, $\frac{\text{constant}}{x^k}$ form ka koi bhi term $0$ ki taraf shrink ho jaata hai — toh yeh rewrite isolate karta hai jo survive karta hai aur jo vanish hota hai.
2. Jab $x\to \infty$: $\frac{5}{x}\to 0$ aur $\frac{7}{x^2}\to 0$. Yeh step kyun? Ek fixed number ko ever-larger $x$ se divide karna use zero ki taraf le jaata hai — wahi "$k/\text{big}$" idea Ex 5 se, ab terms ko khatam karne ke liye use ho raha hai.
3. Jo bachta hai: $\frac{3+0}{2-0}=\frac{3}{2}$.

${\lim }_{x\to \infty }\frac{3x^2+5x}{2x^2-7}=\frac{3}{2}$

Verify: $x=1000$ par: $\frac{3\cdot 10^6+5000}{2\cdot 10^6-7}=\frac{3005000}{1999993}=1.50250\dots \approx 1.5$. ✓ (Poori treatment One-sided limits and limits at infinity mein.)

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Recall

Recall Kaun se do cells "does not exist" dete hain, aur woh alag kyun hain?

Cell D (jump): dono sides ek finite value par disagree karte hain. Cell H (oscillation): koi bhi value approach nahi hoti. Cell F ($\frac{1}{x-2}$) bhi two-sided fail karta hai kyunki sides opposite infinities ki taraf jaate hain.

Recall Jab

$0/0$ mile, tumhare do main tools kya hain aur choose kaise karte ho? Factor & cancel (Ex 2) jab polynomial ho; conjugate se multiply karo (Ex 3) jab square root involved ho.

Recall

${\lim }_{x\to 2}\frac{1}{(x-2{)}^2}=+\infty$ answer ki taur par kyun allowed hai lekin ${\lim }_{x\to 2}\frac{1}{x-2}$ nahi? Square denominator ko dono sides par positive rakhta hai, toh dono $+\infty$ par agree karte hain; plain version right par $+\infty$ deta hai lekin left par $-\infty$ — sides disagree karte hain.

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Connections

• Intuitive concept of a limit — table of values, graphical
• One-sided limits and limits at infinity
• Indeterminate forms and algebraic simplification
• Continuity at a point
• The derivative as a limit of difference quotients
• Squeeze (Sandwich) Theorem
• Formal epsilon-delta definition of a limit