One-sided limits — left-hand, right-hand: Worked examples

4.1.3 · D3 · Maths › Calculus I — Limits & Derivatives › One-sided limits — left-hand, right-hand

Scenario matrix

Har one-sided-limit question actually kuch gine-chune shapes mein se ek hoti hai. Yahan poori list hai, aur konsa example konse ko handle karta hai.

Cell	Case class	Mushkil kya hai	Example
A	Sides agree (piecewise)	dono rules check karne padte hain	Ex 1
B	Sides disagree, both finite (jump)	equal-existence trap	Ex 2
C	Domain edge — ek side undefined	sirf EK side exist karti hai	Ex 3
D	Pole / infinite — sides $\pm \infty$ jaate hain	denominator ka sign	Ex 4
E	Floor / step, tricky non-integer twist	floor sirf integers par jump karta hai	Ex 5
F	$0/0$ removable — pehle algebra zaroori	seedha plug in nahi kar sakte	Ex 6
G	Squeeze at a wiggle ( $sin \frac{1}{x}$ type)	bounded-times-small	Ex 7
H	Word problem (price/tax jump)	story ko piecewise mein translate karo	Ex 8
I	Exam twist — unknown constant ke liye solve karo	sides ko match karne par majboor karo	Ex 9

Hum pehle vocabulary zero se banate hain, phir A→I march karte hain.

Ground rules, seedhe shabdon mein dobara

Figure — One-sided limits — left-hand, right-hand

Upar ki figure hamaara map hai: ek single number line jisme $a$ marked hai. Coral arrow right se creep karta hai ( $a^{+}$ ), lavender arrow left se ( $a^{-}$ ). Neeche ka har example bas itna hai: "har arrow ki function kis value ki taraf ja rahi hai?"

Cell A — sides agree (piecewise)

Worked example Ex 1 · Do rules jo meet karte hain

$f (x) = {2 x + 1 x^{2} + 2 x \leq 1 x > 1$ $lim_{x \to 1^{-}} f$ , $lim_{x \to 1^{+}} f$ nikalo, aur two-sided limit decide karo.

Forecast: $x = 1$ par left rule $3$ deta hai; right rule deta hai $1 + 2 = 3$ . Guess: match karte hain, limit $= 3$ .

Left side: $x < 1$ ke liye hum $2 x + 1$ use karte hain. $x \to 1^{-}$ push karo: $2 (1) + 1 = 3$ . Ye step kyun? $1$ ke left mein, sirf rule $2 x + 1$ "active" hai, aur ye ek seedhi line hai — substitute karna safe hai.
Right side: $x > 1$ ke liye hum $x^{2} + 2$ use karte hain. $x \to 1^{+}$ push karo: $1^{2} + 2 = 3$ . Ye step kyun? Right mein, active rule $x^{2} + 2$ par switch ho jaata hai; ye bhi ek polynomial hai, isliye substitution legal hai.
Combine karo: dono sides $3$ deti hain, equal hain, toh law se $lim_{x \to 1} f = 3$ .

Verify: near-points test karo. $f (0.99) = 2 (0.99) + 1 = 2.98 \approx 3$ ✓. $f (1.01) = (1.01)^{2} + 2 = 3.0201 \approx 3$ ✓. Dono $3$ ke paas hain.

Cell B — sides disagree, both finite (ek jump)

Worked example Ex 2 · Classic sign-jump

$s (x) = \frac{∣ x - 4∣}{x - 4}, x \neq = 4. Find both one-sided limits at 4.$

Forecast: absolute value divided by wahi cheez — ye ek " $+ 1$ / $- 1$ " splitter hai. Guess: right $= + 1$ , left $= - 1$ , koi two-sided limit nahi.

Right side ( $x > 4$ ): tab $x - 4 > 0$ , toh $∣ x - 4∣ = x - 4$ . Ratio $= \frac{x - 4}{x - 4} = 1$ . Ye step kyun? Absolute value ki definition: ek positive quantity ke liye, $∣ u ∣ = u$ . Yahan $u = x - 4$ positive hai jab $x > 4$ .
Toh $lim_{x \to 4^{+}} s = 1$ .
Left side ( $x < 4$ ): tab $x - 4 < 0$ , toh $∣ x - 4∣ = - (x - 4)$ . Ratio $= \frac{- ( x - 4 )}{x - 4} = - 1$ . Ye step kyun? Ek negative quantity ke liye, $∣ u ∣ = - u$ (sign flip karo taaki positive ho jaye).
Toh $lim_{x \to 4^{-}} s = - 1$ . Kyunki $1 \neq = - 1$ , two-sided limit exist nahi karti — size $2$ ka ek jump hai.

Verify: $s (4.5) = \frac{∣0.5∣}{0.5} = 1$ ✓ aur $s (3.5) = \frac{∣ - 0.5∣}{- 0.5} = \frac{0.5}{- 0.5} = - 1$ ✓.

Cell C — domain edge (sirf ek side zinda hai)

Worked example Ex 3 · Ek square-root ki deewar

$r (x) = x - 2$ ki one-sided limits $x = 2$ par nikalo.

Forecast: tum negative real ka square root nahi le sakte. Toh left side ke paas khadne ki jagah nahi hai. Guess: sirf right limit exist karti hai, aur wo $0$ hai.

Pehle domain check karo. $x - 2$ real tabhi hai jab $x - 2 \geq 0$ , yaani $x \geq 2$ . Ye step kyun? Ek limit tabhi "approach" kar sakti hai jab function defined ho. Left par koi domain nahi = koi left-hand limit nahi.
Right side ( $x > 2$ ): jaise $x \to 2^{+}$ , $x - 2 \to 0^{+}$ , toh $x - 2 \to 0$ . Is liye $lim_{x \to 2^{+}} r = 0$ .
Left side ( $x < 2$ ): function undefined hai; $lim_{x \to 2^{-}} r$ exist nahi karta (koi domain nahi).
Two-sided: exist nahi karta, kyunki ek side missing hai — lekin ek khaas wajah se: ye ek domain edge hai, mismatch nahi.

Verify: $r (2.01) = 0.01 = 0.1$ , aur $r (2.0001) = 0.0001 = 0.01$ — $0$ ki taraf shrink ho raha hai ✓. $r (1.9) = - 0.1$ real number nahi hai ✓.

Cell D — pole / infinite behaviour

Worked example Ex 4 · Sides opposite directions mein blow up hoti hain

$p (x) = \frac{1}{x - 5}$ ki one-sided limits $x = 5$ par nikalo.

Forecast: tiny denominator = huge output; denominator ka sign decide karta hai upar ya neeche. Guess: right $\to + \infty$ , left $\to - \infty$ .

Right side ( $x > 5$ ): $x - 5$ ek tiny positive number hai, jaise $0.001$ . Tab $\frac{1}{0.001} = 1000$ , bina bound ke badhta hai. Toh $lim_{x \to 5^{+}} p = + \infty$ . Ye step kyun? $1$ ko ek shrinking positive number se divide karna arbitrarily large aur positive ho jaata hai.
Left side ( $x < 5$ ): $x - 5$ ek tiny negative number hai, jaise $- 0.001$ . Tab $\frac{1}{- 0.001} = - 1000$ . Toh $lim_{x \to 5^{-}} p = - \infty$ .
Two-sided: exist nahi karta — sides opposite infinities ki taraf diverge karti hain. Ye $x = 5$ par ek vertical asymptote hai.

Verify: $p (5.001) = \frac{1}{0.001} = 1000 > 0$ ✓, $p (4.999) = \frac{1}{- 0.001} = - 1000 < 0$ ✓.

Cell E — floor function, non-integer twist

Worked example Ex 5 · Floor jahan approach point integer NAHI hai

$⌊ 2 x ⌋$ ki dono one-sided limits $x = 1.5$ par nikalo.

Forecast: floor tabhi jump karta hai jab uska input integer hit kare. Input hai $2 x$ ; $x = 1.5$ par wo hai $2 (1.5) = 3$ , ek integer — toh yahan jump hoga. Guess: left $= 2$ , right $= 3$ .

Pata karo kahan andar integer hai. $2 x = 3$ exactly $x = 1.5$ par hai. Toh floor ka "step" bilkul hamare point par land karta hai. Ye step kyun? $⌊ \cdot ⌋$ integers ke beech constant hota hai aur unpar jump karta hai; hume check karna hoga ki $2 x$ , $x = 1.5$ par integer cross karta hai ya nahi. Karta hai.
Left side ( $x < 1.5$ ): maano $x = 1.49$ , tab $2 x = 2.98$ , aur $⌊ 2.98 ⌋ = 2$ . Toh $lim_{x \to 1. 5^{-}} ⌊ 2 x ⌋ = 2$ . Ye step kyun? $1.5$ ke thoda neeche, $2 x$ $3$ se thoda neeche hai, toh wo $2$ tak floor down karta hai.
Right side ( $x > 1.5$ ): maano $x = 1.51$ , tab $2 x = 3.02$ , aur $⌊ 3.02 ⌋ = 3$ . Toh $lim_{x \to 1. 5^{+}} ⌊ 2 x ⌋ = 3$ .
Two-sided: $2 \neq = 3$ , toh exist nahi karta — $1$ ka ek jump.

Verify: $⌊ 2 (1.499)⌋ = ⌊ 2.998 ⌋ = 2$ ✓ aur $⌊ 2 (1.501)⌋ = ⌊ 3.002 ⌋ = 3$ ✓.

Cell F — removable $0/0$ (pehle algebra)

Worked example Ex 6 · Seedha plug in nahi kar sakte

$q (x) = \frac{x ^{2} - 9}{x - 3}$ ka $lim_{x \to 3^{-}}$ aur $lim_{x \to 3^{+}}$ nikalo.

Forecast: $x = 3$ par hume $\frac{0}{0}$ milega — undefined, toh plug karna illegal hai. Lekin ye factor hota hai. Guess: dono sides $6$ deti hain; limit exist karti hai $= 6$ .

Diagnose karo. Direct substitution deta hai $\frac{3 ^{2} - 9}{3 - 3} = \frac{0}{0}$ , ek indeterminate form — abhi koi information nahi. Ye step kyun? $\frac{0}{0}$ ka matlab hai "read nahi kar sakte"; limit lene se pehle simplify karna zaroori hai.
Top factor karo. $x^{2} - 9 = (x - 3) (x + 3)$ (difference of squares). Toh $q (x) = \frac{( x - 3 ) ( x + 3 )}{x - 3} = x + 3$ for $x \neq = 3$ . Ye step kyun? Common $(x - 3)$ cancel karna mushkil hatata hai; valid hai kyunki limit ke paas $x \neq = 3$ hota hai (hum $x = 3$ ko kabhi touch nahi karte).
Ab dono sides $x + 3$ use karti hain. Left: $x \to 3^{-}$ , $x + 3 \to 6$ . Right: $x \to 3^{+}$ , $x + 3 \to 6$ .
Dono $6$ deti hain, equal hain, toh $lim_{x \to 3} q = 6$ . ( $x = 3$ par hole removable hai.)

Verify: $q (2.99) = \frac{2.9 9 ^{2} - 9}{2.99 - 3} = \frac{- 0.0599}{- 0.01} = 5.99$ ✓ aur $q (3.01) = \frac{3.0 1 ^{2} - 9}{0.01} = 6.01$ ✓.

Cell G — squeeze at a wiggle

Worked example Ex 7 · Infinite oscillation, kabu mein

$w (x) = x sin (\frac{1}{x})$ ka $lim_{x \to 0^{+}}$ aur $lim_{x \to 0^{-}}$ nikalo.

Forecast: $sin (1/ x)$ , $0$ ke paas pagal tarah oscillate karta hai, lekin $x$ shrink ho kar $0$ hota hai. Ek chota number times ek bounded number → $0$ . Guess: dono sides $= 0$ .

Wiggle ko bound karo. Har $x \neq = 0$ ke liye, $- 1 \leq sin (1/ x) \leq 1$ . Ye step kyun? $sin$ kisi bhi cheez ka $[- 1, 1]$ ke andar rehta hai; hume nahi pata kahan hai, bas itna ki bounded hai.
$∣ x ∣$ se multiply karo. Tab $- ∣ x ∣ \leq x sin (1/ x) \leq ∣ x ∣$ . Ye step kyun? Bound ko $∣ x ∣$ se multiply karna $w (x)$ ko do cheezon ke beech trap karta hai jo dono $0$ ki taraf jaati hain — squeeze.
Dono bounds → 0. Jaise $x \to 0$ kisi bhi side se, $∣ x ∣ \to 0$ aur $- ∣ x ∣ \to 0$ , toh $w (x)$ squeeze hokar $0$ ho jaata hai.
Is liye $lim_{x \to 0^{-}} w = lim_{x \to 0^{+}} w = 0$ , aur two-sided limit $0$ hai — chahe $w$ infinitely baar oscillate kare.

Verify: $w (0.01) = 0.01 sin (100) \approx 0.01 (- 0.506) = - 0.00506$ , tiny ✓. $w (- 0.01) = - 0.01 sin (- 100) \approx - 0.00506$ , bhi tiny ✓. Dono $0$ ke paas.

Cell H — word problem (story translate karo)

Worked example Ex 8 · Parking-garage price jump

Ek garage $3 per hour charge karta hai, lekin jaise hi tum $2$ hours cross karte ho, poore session ke liye flat $8 bill hota hai jo $4$ hours tak valid hai. Cost $C (t)$ model karo aur $t = 2$ hours par one-sided limits nikalo.

Forecast: $2$ hours se neeche cost $3\times2=\$ 6 $k i t a r a f ba d h t ihai;$ 2 $hours par/baad ye \$ 8 par jump karti hai. Guess: left $= 6$ , right $= 8$ , koi two-sided limit nahi — customer ko ek price cliff dikhti hai.

Piecewise rule mein translate karo. $C (t) = {3 t 8 0 \leq t < 2 2 \leq t \leq 4$ Ye step kyun? "Per-hour up to 2h" line $3 t$ hai; "2h cross karne ke baad flat 8" ek constant hai.
Left side ( $t \to 2^{-}$ ): $3 t$ use karo: $3 (2) = 6$ . Toh $\lim_{t\to2^-}C=\$ 6 $. * Y es t e p k y u n ? *$ 2$ hours se thoda neeche tum abhi bhi per-hour meter par ho.
Right side ( $t \to 2^{+}$ ): $8$ use karo: constant hai, toh $\lim_{t\to2^+}C=\$ 8$.
Interpret karo: $6 \neq = 8$ , koi two-sided limit nahi — ek asal jump discontinuity $2 ki. Wo extra $2 hi 2-hour line cross karne ki "penalty" hai.

Verify (units $): $C (1.99) = 3 (1.99) = 5.97 \approx 6$ ✓; $C (2.01) = 8$ ✓. Prices forecast ke anusaar behave karti hain.

Cell I — exam twist: unknown ke liye solve karo

Worked example Ex 9 · Pieces ko milaao

$f (x) = {k x + 2 x^{2} + k x < 1 x \geq 1$ Kis value of $k$ ke liye $lim_{x \to 1} f (x)$ exist karta hai? (Ye Continuity at a point se connect hota hai.)

Forecast: two-sided limit tabhi exist karti hai jab sides match karein; unhe equal set karo aur $k$ ke liye solve karo. Guess: $k$ ki ek clean value.

Left side ( $x \to 1^{-}$ ): rule $k x + 2$ deta hai $k (1) + 2 = k + 2$ . Ye step kyun? $1$ ke neeche active rule line $k x + 2$ hai; approached value $1$ substitute karo.
Right side ( $x \to 1^{+}$ ): rule $x^{2} + k$ deta hai $1 + k$ .
Equality force karo (law). Two-sided limit exist karne ke liye: $k + 2 = 1 + k ⟹ 2 = 1.$ Ye step kyun? Humne demand ki ki sides match karein; lekin $2 = 1$ har k ke liye galat hai.
Conclusion: koi bhi aisi $k$ ki value nahi hai jo limit ko exist karaye — constant terms chahe kuch bhi ho, clash karte hain. (Ek sneaky exam trap: jis unknown ke liye solve kar rahe ho wo cancel ho jaata hai.)

Verify: dono expressions subtract karo: $(k + 2) - (1 + k) = 1 \neq = 0$ , $k$ se independent hai ✓. Sides hamesha exactly $1$ se alag rehti hain.

Wrap-up: matrix, filled

Recall Konsa example kis shape ke liye tha?

A agree ::: Ex 1 (limit $= 3$ ) B jump ::: Ex 2 ( $+ 1$ vs $- 1$ ) C domain edge ::: Ex 3 (right $= 0$ , left undefined) D pole ::: Ex 4 ( $+ \infty$ vs $- \infty$ ) E floor twist ::: Ex 5 ( $2$ vs $3$ ) F removable $0/0$ ::: Ex 6 (dono $= 6$ ) G squeeze wiggle ::: Ex 7 (dono $= 0$ ) H word problem ::: Ex 8 ($6 vs $8) I unknown constant ::: Ex 9 (koi $k$ kaam nahi karta)

Connections

One-sided Limits — Left-hand & Right-hand (parent)
Limit of a function — intuitive & ε-δ definition
Continuity at a point
Jump, removable & infinite discontinuities
Vertical asymptotes
Greatest integer / floor function
Differentiability — left & right derivatives

Worked examples — One-sided limits — left-hand, right-hand

Scenario matrix

Ground rules, seedhe shabdon mein dobara

Cell A — sides agree (piecewise)

Cell B — sides disagree, both finite (ek jump)

Cell C — domain edge (sirf ek side zinda hai)

Cell D — pole / infinite behaviour

Cell E — floor function, non-integer twist

Cell F — removable $0/0$ (pehle algebra)

Cell G — squeeze at a wiggle

Cell H — word problem (story translate karo)

Cell I — exam twist: unknown ke liye solve karo

Wrap-up: matrix, filled

Connections

Concept Map