Worked examples — Continuity — definition, types of discontinuity (removable, jump, infinite)

4.1.7 · D3 · Maths › Calculus I — Limits & Derivatives › Continuity — definition, types of discontinuity (removable,

Yeh sab kuch parent Continuity note aur Limits — definition and one-sided limits pe based hai. Agar neeche koi word unfamiliar lage, toh woh pehli baar aane par define kiya gaya hai.

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The scenario matrix

Table ko is tarah padho: "Yahan ek shape hai jo world tumhe de sakti hai, yahan bataya gaya hai ki teen VLE conditions mein se kaunsi tooti, aur yahan woh example hai jo ise drill karta hai." (VLE = Value, Limit, Equal — parent se mnemonic.)

#	Case class	Kya aata hai	Kaunsi condition tooti	Expected verdict	Example
A	$0/0$ jo cancel hota hai	rational, factor zero ko khatam karta hai	V fails (undefined)	removable	Ex 1
B	$0/0$ jo cancel NAHI hota	rational, denom zyada strong hai	V + L fail	infinite	Ex 2
C	Piecewise, pieces agree karte hain	do formulas cleanly milte hain	koi nahi — continuous!	continuous	Ex 3
D	Piecewise, pieces disagree karte hain	stair-step	E fails, $L^{-}\ne L^{+}$	jump	Ex 4
E	Wall jahan dono sides same direction mein jaayein	$1/(x-a{)}^2$ type	V + L fail (both $+\infty$)	infinite (same sign)	Ex 5
F	Oscillation — kabhi settle nahi hota	$\sin (1/x)$ type	L fails (no limit)	essential (not removable)	Ex 6
G	Real-world word problem	parking-cost step function	E fails at breakpoints	jump	Ex 7
H	Exam twist — $k$ dhundho	piecewise with unknown constant	force E to hold	solve for $k$	Ex 8

"Signs/quadrants" wale columns denominator ka left-vs-right sign ban jaate hain (cases B, E), "zero/degenerate input" $0/0$ column hai (A, B), "limiting values" oscillation hai (F), "word problem" G hai, aur "exam twist" H hai. Neeche har cell cover hoga.

Neeche wali figure is matrix ka tumhara visual index hai — chaar shapes ki ek gallery jo yeh cells produce karti hain. Examples padhne se pehle ise dhyan se dekho:

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Example 1 — Cell A: woh $0/0$ jo cancel hota hai (removable)

Forecast: $x=3$ plug karne par top $=9-3-6=0$ aur bottom $=0$ milta hai. Ek $0/0$. Abhi guess karo: kya top factor $(x-3)$ share karta hai? Agar haan → removable hole. Agar nahi → infinite wall.

1. $f(3)$ compute karo. Top $=3^2-3-6=0$, bottom $=3-3=0$, toh $f(3)=0/0$ — undefined. Condition V fails. Yeh step kyun? Limits touch karne se pehle hum hamesha test karte hain ki point exist bhi karta hai ya nahi; yahan karta nahi.
2. Top ko factor karo. $x^2-x-6=(x-3)(x+2)$. Toh $x\ne 3$ ke liye, $f(x)=\frac{(x-3)(x+2)}{x-3}=x+2.$ Yeh step kyun? Limit single point $x=3$ ko ignore karti hai; common $(x-3)$ cancel karna woh value reveal karta hai jis par graph aim kar raha hai. Yeh bilkul legal hai kyunki near the point $x\ne 3$ hai.
3. Limit lo. $\underset{x\to 3}{\lim }(x+2)=5$, finite, aur left $=$ right (yeh ek polynomial hai). Condition L holds. Yeh step kyun? Removable-vs-infinite ka decision is baat se hota hai ki limit finite hai ya nahi; ek polynomial ka ek obvious two-sided value hota hai, toh hum seedha read kar lete hain.
4. Compare karo. Limit $=5$ exist karta hai lekin $f(3)$ undefined hai → removable. Patch: $f(3)=5$ define karo. Yeh step kyun? Classification exactly L aur V ka comparison hai; yahan L exist karta hai (finite) lekin V missing hai, jo removable ki definition hai.

Verify: $x=2.99$ se approach karo: $x+2=4.99$; $x=3.01$ se: $5.01$. Dono $5$ pe squeeze karte hain. Single missing dot → removable. ✓

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Example 2 — Cell B: woh $0/0$ jo cancel NAHI hota (infinite)

Forecast: Phir se top aur bottom dono $3$ par vanish karte hain. Lekin bottom dekho: $x^2-6x+9=(x-3{)}^2$ — ek squared factor. Ek cancel ke baad bhi, ek $(x-3)$ denominator mein chhupa hai. Guess: wall, hole nahi.

1. $f(3)$ compute karo. Top $=0$, bottom $=(3-3{)}^2=0$, toh $0/0$ — undefined. V fails. Kyun? Hamesha wahi pehla check.
2. Factor karo aur ek baar cancel karo. $f(x)=\frac{x-3}{(x-3{)}^2}=\frac{1}{x-3},x\ne 3.$ Yeh step kyun? Hum exactly utne $(x-3)$'s cancel karte hain jitne dono sides share karte hain. Yahan bottom mein do the aur top mein ek, toh ek neeche survive karta hai — ek surviving wall ka tell-tale sign.
3. One-sided limits lo (dekho Vertical asymptotes and rational functions).
   • Left se, $x=2.999$: $x-3=-0.001$, toh $1/(x-3)=-1000\to -\infty$.
   • Right se, $x=3.001$: $x-3=+0.001$, toh $1/(x-3)=+1000\to +\infty$. Sides mein kyun split kiya? Kyunki tiny denominator ka sign $3$ cross karte waqt flip hota hai; ek two-sided statement yeh chhupaata.
4. Compare karo. Ek side infinite hai → infinite discontinuity. Yeh parent-note mistake in action hai: undefined ≠ removable. Yeh step kyun? Classification hamesha parent-note definitions se apni findings match karke khatam hoti hai; ek infinite one-sided limit exactly "infinite" case hai (parent note ka "pole" ya vertical-asymptote type).

Verify: $f(2.999)=-1000$, $f(3.001)=+1000$ — magnitudes explode karti hain, signs oppose karte hain → genuine pole (infinite discontinuity). ✓

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Example 3 — Cell C: piecewise jo actually continuous hai

Forecast: Do alag formulas $x=1$ par milte hain. Guess: kya dono heights wahan match karti hain? Agar haan, toh pen kabhi nahi uthta.

1. Value. $x=1$ uses $x\ge 1\Rightarrow x^2+2$, toh $f(1)=1^2+2=3$. V holds. Doosra piece kyun? Inequality $x\ge 1$ hai, aur $1$ "$\ge 1$" satisfy karta hai. (Parent mistake: exact inequality se choose karo, proximity se nahi.)
2. Left limit. $L^{-}={\lim }_{x\to 1^{-}}(2x+1)=2(1)+1=3$. Yeh piece kyun? Neeche se aate waqt, $x<1$, jo $2x+1$ use karta hai.
3. Right limit. $L^{+}={\lim }_{x\to 1^{+}}(x^2+2)=1+2=3$. Yeh piece kyun? Upar se aate waqt, $x\ge 1$, jo $x^2+2$ use karta hai.
4. Compare karo. $L^{-}=L^{+}=f(1)=3$ — teeno equal hain → continuous. Koi discontinuity nahi. Yeh step kyun? Continuity exactly VLE triple ka agree karna hai; hum "continuous" tabhi declare karte hain jab literally teeno numbers match hone check kar lein.

Verify: $L^{-}=3$, $L^{+}=3$, $f(1)=3$ — VLE triple sab agree karte hain. ✓ Dekho Piecewise functions ke liye ki seam par heights ka match karna exactly continuity condition kyun hai.

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Example 4 — Cell D: piecewise jo jump karta hai

Forecast: Almost Example 3, lekin doosra piece ab $x^2+5$ hai. Guess: right side zyada upar land karega — ek step.

1. Value. $f(1)=1^2+5=6$. V holds. Yeh piece kyun? $x=1$ satisfies $x\ge 1$, toh hum $x^2+5$ use karte hain.
2. Left limit. $L^{-}=2(1)+1=3$. Yeh piece kyun? Neeche se approach karne par $x<1\Rightarrow 2x+1$ use hota hai.
3. Right limit. $L^{+}=1^2+5=6$. Yeh piece kyun? Upar se approach karne par $x\ge 1\Rightarrow x^2+5$ use hota hai.
4. Compare karo. $L^{-}=3\ne 6=L^{+}$, dono finite hain → jump. Jump size $=|L^{+}-L^{-}|=|6-3|=3$. "Jump" kyun, "removable" kyun nahi? Removable ke liye $L^{-}=L^{+}$ chahiye (limit exist kare). Yahan dono sides disagree karte hain, toh koi single dot gap fill nahi kar sakta.

Verify: left $3$ approach karta hai, right $6$; gap $=3$. Figure dekho: pink left-piece open dot height $3$ par baitha hai, blue right-piece filled dot $6$ par. ✓

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Example 5 — Cell E: ek wall jahan DONO sides same direction mein jaati hain

Forecast: Example 2 mein dono sides $-\infty$ aur $+\infty$ par shoot hue the. Yahan denominator squared hai, toh woh hamesha positive hai. Guess: dono sides same direction mein rocket karti hain (upar).

1. Value. $f(2)=1/0$ — undefined. V fails. Yeh step kyun? Hum hamesha value check se shuru karte hain; ek $1/0$ batata hai ki point exist nahi karta aur possible wall ki warning deta hai.
2. Denominator ka sign analyse karo. $(x-2{)}^2\ge 0$ har $x$ ke liye, aur yeh $2$ ke paas ek tiny positive number hai (kabhi negative nahi). Sign kyun matter karta hai? Ek positive tiny denominator $1/\text{tiny}$ ko ek bada positive number banata hai, chahe hum kisi bhi side se aayein — squared power woh sign flip remove kar deta hai jo humne Ex 2 mein dekha tha.
3. One-sided limits.
   • Left, $x=1.99$: $(x-2{)}^2=0.0001$, toh $f=10000\to +\infty$.
   • Right, $x=2.01$: $(x-2{)}^2=0.0001$, toh $f=10000\to +\infty$. Dono sides kyun compute karein? Ek infinite discontinuity ke liye sirf ek side ka blow up karna kaafi hai; dono check karne se confirm hota hai ki yeh ek two-sided upward wall hai, na ki one-sided oddity.
4. Compare karo. Dono sides $=+\infty$ → infinite discontinuity (ek two-sided upward asymptote / pole). Yeh step kyun? Apni findings ko parent definitions se match karna: koi bhi side $=\pm \infty$ infinite case hai.

Verify: $f(1.99)=10000=f(2.01)$; dono positively blow up karte hain — Ex 2 se contrast karo jahan odd power ne sign flip kiya tha. ✓

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Example 6 — Cell F: oscillation jo kabhi settle nahi hoti

Forecast: Jaise-jaise $x\to 0$, andar ka $1/x$ $\pm \infty$ ki taraf race karta hai, toh $\sin$ $-1$ aur $1$ ke beech tezi se zyada swing karta hai. Guess: graph kabhi ek number approach nahi karta — limit simply exist nahi karti, aur infinity ki wajah se nahi.

1. Value. $f(0)=\sin (1/0)$ — undefined. V fails. Yeh step kyun? Usual pehla check: andar $1/0$ matlab point exist nahi karta.
2. Test karo ki limit exist karti hai ya nahi. Do shrinking sequences lo jo $0$ ki taraf jaati hain:
   • $x_n=\frac{1}{2\pi n}$ deta hai $\sin (2\pi n)=0$.
   • $x_n=\frac{1}{2\pi n+\pi /2}$ deta hai $\sin (2\pi n+\pi /2)=1$. Do sequences kyun? Agar limit exist hoti, toh $0$ approach karne ka har tarika same value deta. Humne do approaches find ki jो $0$ aur $1$ deti hain — toh koi single limit exist nahi karti.
3. Classify karo. One-sided limits na finite-and-equal hain (not removable) na $\pm \infty$ (not simple pole) — function oscillate karta hai. Yeh step kyun? Hum har named type ko ek-ek karke rule out karte hain: not removable (koi finite limit nahi), not jump (koi do finite one-sided values nahi), not infinite (yeh blow up nahi karta) — jo bachta hai woh oscillating case hai.

Verify: sequence $x_n=1/(2\pi n)$ par value $0$ hai; $x_n=1/(2\pi n+\pi /2)$ par value $1$ hai. Do limits ≠ ek → koi limit nahi. ✓

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Example 7 — Cell G: real-world word problem (parking meter)

Forecast: Cost constant hai, phir har pore ghante par ₹20 jump karta hai. Guess: $t=1$ par jump discontinuity, size ₹20.

1. $t=1$ par value. Clause $0<t\le 1$ includes $t=1$, toh $C(1)=30$. V holds. Yeh clause kyun? $t=1$ satisfies "$\le 1$", toh ₹30 rate exactly ek ghante par bhi apply hota hai.
2. Left limit. $t$ just below $1$ ke liye, abhi bhi $0<t\le 1$ mein: $L^{-}=30$. Yeh piece kyun? Ek ghante se thodi kam values flat ₹30 rate par bill hoti hain.
3. Right limit. $t$ just above $1$ ke liye, hum $1<t\le 2$ mein enter karte hain: $L^{+}=50$. Yeh piece kyun? Jaise hi hum ek ghanta cross karte hain, doosre ghante ka band ₹50 apply hota hai.
4. Compare karo. $L^{-}=30\ne 50=L^{+}$, dono finite hain → jump. Jump size $=|50-30|=20$, yaani ₹20 — exactly extra ghante ka charge. Yeh step kyun? Do finite lekin unequal one-sided values exactly jump definition hai; gap size wahi hai jo customer boundary cross karne par charge hota hai.

Verify (with units): left-approach cost ₹30, right-approach cost ₹50, difference ₹20 = stated second-hour fee. Model ko jump karna chahiye — real fees step karti hain, ramp nahi karti. ✓

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Example 8 — Cell H: exam twist ($k$ dhundho)

Forecast: Top piece mein ek removable hole hai (Ex-1 style). Guess: $k$ ko wahi equal hona chahiye jis par pieces limit karti hain — sahi dot se hole band karo.

1. Value. By definition $f(3)=k$ (yahi toh poora point hai — $k$ humara choose karne ka hai). Kisi bhi $k$ ke liye V holds. Yeh step kyun? Hume actual value at $3$ pata honi chahiye limit ke equal force karne se pehle; definition ise $k$ ke roop mein hamare haath mein deta hai.
2. Doosre piece ki limit dhundho. $x\ne 3$ ke liye, $\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3,$ toh $\underset{x\to 3}{\lim }f(x)=3+3=6$. Yeh step kyun? Factor–cancel woh value reveal karta hai jis par curve $3$ par aim kar raha hai (hole); yeh woh target hai jise $k$ ko hit karna hai.
3. Condition E force karo. Continuity ke liye limit $=f(3)$ chahiye, yaani $6=k$. Toh $\boxed{k=6}$. Yeh equation solve kyun karein? Hum yahan test nahi kar rahe — hum design kar rahe hain. $k$ ko limit ke equal set karna pen-lift band kar deta hai.

Verify: $k=6$ ke saath: $L^{-}=L^{+}=6$ (polynomial $x+3$) aur $f(3)=6$ — teeno agree karte hain, continuous. Koi bhi doosra $k$ (maano $k=5$) ise removable-again banata misplaced dot ke saath. ✓

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Recall Move on se pehle self-test karo

Ek rational function $0/0$ deta hai lekin denominator mein squared factor tha. Type? ::: Infinite (ek pole) — ek $(x-a)$ downstairs survive karta hai (Cell B/E). $f$ $x=1$ par $3$ se $6$ tak jump karta hai. Jump size? ::: $|6-3|=3$. $\sin (1/x)$ at $0$ removable kyun nahi hai? ::: Limit exist nahi karti (yeh oscillate karta hai), aur removable ek finite limit maangta hai — toh yeh essential/second-kind hai. $\frac{x^2-9}{x-3}$ ko $3$ par continuous banane ke liye, $k=$ ? ::: $6$. Cell E mein, dono sides $+\infty$ par kyun jaati hain? ::: Denominator squared hai, toh woh point ke paas hamesha positive hota hai.

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Connections

• Continuity — definition, types of discontinuity (removable, jump, infinite) (parent — VLE test aur type definitions)
• Limits — definition and one-sided limits (har example $L^{-}$ aur $L^{+}$ use karta hai)
• Vertical asymptotes and rational functions (Cells B, E — surviving-factor walls / poles)
• Piecewise functions (Cells C, D, G, H — seams aur unknown constants)