One-sided limits — left-hand, right-hand: Question bank

4.1.3 · D5 · Maths › Calculus I — Limits & Derivatives › One-sided limits — left-hand, right-hand

Dimaag mein rakhne wali pictures

Neeche di gayi do figures woh mental images hain jinpar is page ki har item tiki hui hai: ek number-line "dono sides se approach" ki picture, aur $ε$ – $δ$ band picture jo dikhati hai kyun ek inequality secretly do sides karti hai.

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

Doosri figure dekho: $L$ ke around half-width $ε$ ka horizontal band woh jagah hai jahan hum chahte hain $f (x)$ land kare; $a$ ke around half-width $δ$ ka vertical strip woh jagah hai jahan $x$ roam kar sakta hai. Woh strip $a$ ko straddle karti hai — uske left ke points aur uske right ke points dono $0 < ∣ x - a ∣ < δ$ ke andar hain. Yahi geometric reason hai ki ek single inequality dono sides se agreement maangti hai.

True or false — justify karo

Ek claimed one-sided limit $lim_{x \to a^{+}} f (x) = L$ ke liye: " $+$ " ka matlab hai hum $a$ ko sirf $x > a$ use karke approach karte hain.

True or false:

x \to 5^{-}

ka matlab hai

x

negative values leta hai.

False. Iska matlab hai

x

5

se chhote numbers (jaise

4.9, 4.99

) se approach karta hai, jo sab positive hain; minus ek direction hai, sign nahi.

True or false: Agar

lim_{x \to a^{-}} f = lim_{x \to a^{+}} f

, toh two-sided limit exist karti hai aur us common value ke barabar hoti hai.

True. Equal one-sided limits exactly woh condition hai jo boxed rule require karta hai; shared value hi two-sided limit hai.

True or false: Agar dono one-sided limits exist karti hain, toh two-sided limit zaroor exist karegi.

False. Unhe equal bhi hona chahiye.

∣ x ∣/ x

ke liye

0

par dono sides exist karti hain (

+ 1

aur

- 1

) lekin disagree karti hain, isliye koi two-sided limit nahi.

True or false:

lim_{x \to a^{+}} f (x)

compute karne ke liye

f (a)

ki value zaroori hai.

False. Ek limit sirf neighbourhood padhti hai; point

a

khud kabhi plug in nahi hota aur undefined bhi ho sakta hai.

True or false:

lim_{x \to 0^{+}} \frac{1}{x} = + \infty

ek existing (real-valued) limit hai.

False.

+ \infty

ek real number nahi hai; hum kehte hain limit diverges to

+ \infty

— yeh ek behaviour ka naam hai, existing limit nahi.

True or false: Agar

lim_{x \to a^{-}} f = + \infty

aur

lim_{x \to a^{+}} f = + \infty

, toh

lim_{x \to a} f = + \infty

True is sense mein ki dono sides

+ \infty

ki taraf diverge karne par "agree" karti hain (jaise

1/ x^{2}

0

); hum two-sided divergence ko

+ \infty

likhte hain, halaanki yeh abhi bhi ek non-existent finite limit hai.

True or false: Ek continuous function apne domain ke kisi point par unequal one-sided limits rakh sakti hai.

False.

a

par continuity demand karti hai ki dono one-sided limits

f (a)

ke barabar hon; unequal sides exactly woh cheez hai jo continuity ko break karti hai (ek jump).

True or false: Agar

f

sirf

x \geq a

ke liye defined hai, toh

lim_{x \to a} f (x)

phir bhi two-sided exist kar sakti hai.

False. Left mein koi domain na hone se left-hand limit undefined hai, isliye two-sided limit exist nahi kar sakti; sirf right-hand limit meaningful hai.

True or false:

lim_{x \to 3^{-}} ⌊ x ⌋ = ⌊ 3 ⌋ = 3

False.

3

se thoda neeche,

2.999

jaise values floor hokar

2

banti hain, isliye left-hand limit

2

hai,

3

nahi; floor integer par jump karta hai.

Error dhundho

Har line mein ek flawed reasoning hai. Batao kya galat hua.

g (2) = 7

, aur

g

wahan defined hai, isliye

lim_{x \to 2} g (x) = 7

." (parent note se piecewise

g

)

Error: limit

g (2)

ko ignore karti hai. One-sided rules

x + 1

aur

3 x - 3

dono

3

ki taraf approach karte hain, isliye limit

3

hai; point value

g (2) = 7

irrelevant hai.

∣ x ∣/ x

0

: kyunki

∣ x ∣ = x

hamesha hota hai,

f (x) = 1

dono sides par hai, isliye limit

1

hai."

Error:

∣ x ∣ = x

sirf

x > 0

ke liye hota hai.

x < 0

ke liye,

∣ x ∣ = - x

, jisse

f (x) = - 1

milta hai, isliye sides disagree karti hain aur koi two-sided limit exist nahi karti.

1/ x

ke dono sides

0

par blow up karte hain, isliye two-sided limit

\infty

hai."

Error: sides opposite infinities ki taraf jaati hain (right se

+ \infty

, left se

- \infty

), isliye woh disagree karti hain; limit exist nahi karti aur use ek single

\infty

label nahi diya ja sakta.

x

0

: left-hand limit

- 0.001

hai jo imaginary hai, isliye limit imaginary hai."

Error: real numbers mein

x

ka left domain bilkul nahi hai, isliye left-hand limit undefined hai (imaginary nahi); sirf

lim_{x \to 0^{+}} x = 0

exist karta hai.

x \to a^{-}

f

ki chhoti root/branch pick karta hai."

Error: minus purely ek direction of approach hai (

x < a

); yeh nahi batata ki

f

ki kaunsi branch ya value choose karni hai.

"Kyunki

lim_{x \to 3^{+}} ⌊ x ⌋ = 3 = ⌊ 3 ⌋

, floor right-continuous hai aur isliye

3

par continuous hai."

Error: right-continuity akeli continuity nahi hoti. Left-hand limit

2 \neq = 3

hai, isliye

⌊ x ⌋

3

par discontinuous hai, chahe right se match karta ho.

Why questions

Kyun ek two-sided limit ke liye sides ka sirf exist karna kaafi nahi, unhe equal bhi hona chahiye?

Ek limit ek single destination

L

ka promise hai jo har approach se reachable ho; agar left

2

ki taraf jaaye aur right

5

ki taraf, koi single

L

woh promise fulfill nahi kar sakta, isliye yeh fail ho jaata hai.

Kyun

ε

–

δ

condition

0 < ∣ x - a ∣ < δ

automatically "dono sides" encode karta hai?

a

ke around

δ

-strip ki picture lo (doosri figure): yeh

a

ke left mein

δ

tak aur

a

ke right mein

δ

tak extend karti hai, isliye

∣ x - a ∣ < δ

mein

- δ < x - a < 0

aur

0 < x - a < δ

dono cover hote hain; two-sided limit inhi do halvon ka logical AND hai.

Kyun

f (a)

ko deliberately exclude kiya jaata hai (

0 < ∣ x - a ∣

ke zariye,

0 \leq

nahi)?

Strict

0 < ∣ x - a ∣

exactly

x = a

par ek hole punch karta hai, isliye limit describe karta hai ki

f

a

ke paas kahan ja raha hai aur tab bhi meaningful rehta hai jab

f (a)

undefined ho ya ek outlier ho.

Kyun ek limit "

= + \infty

" likhi ja sakti hai phir bhi non-existent kahi jaati hai?

Symbol

+ \infty

faithfully record karta hai ki outputs har bound se aage badhte hain, lekin kyunki

+ \infty

ek real number nahi hai, koi real value

L

nahi hai jiske barabar ho — isliye koi finite limit exist nahi karti; "diverges to

+ \infty

" honest phrasing hai.

Kyun floor function har integer par two-sided limit nahi rakhta lekin baaki jagah rakhta hai?

Ek integer par floor value

1

se jump karta hai, isliye left aur right limits differ karti hain; strictly integers ke beech

⌊ x ⌋

constant hota hai, isliye dono sides agree karti hain aur limit exist karti hai.

Kyun

1/ x

ke vertical asymptote ko left se approach karne par

- \infty

milta hai lekin right se

+ \infty

Tiny negative denominators

1/ x

ko hugely negative banate hain, tiny positive ones use hugely positive banate hain; shrinking denominator ka sign output ke sign ko flip kar deta hai.

Edge cases

f (x) = x

x = 0

: kaunsi one-sided limit meaningful bhi hai, aur two-sided verdict kya hai?

Sirf right-hand limit meaningful hai,

lim_{x \to 0^{+}} x = 0

; left ka koi real domain nahi, isliye two-sided limit exist nahi karti.

Ek function sirf

[a, b]

par defined hai:

lim_{x \to a} f

aur

lim_{x \to b} f

ka status kya hai?

Sirf

lim_{x \to a^{+}} f

aur

lim_{x \to b^{-}} f

exist kar sakte hain (domain ke andar se approach); endpoints par two-sided limits exist nahi karti kyunki ek side par koi points nahi hain.

f (x) = 1/ x^{2}

x = 0

: kya sides agree karti hain, aur kya two-sided limit "exist" karti hai?

Dono sides

+ \infty

ki taraf diverge karti hain, isliye behaviour mein agree karti hain aur hum likhte hain

lim_{x \to 0} 1/ x^{2} = + \infty

; yeh abhi bhi ek non-existent finite limit hai lekin ek genuine two-sided divergence hai.

Ek removable discontinuity jahan dono sides

4

ke barabar hain lekin

f (a) = 9

: kya

lim_{x \to a} f

exist karta hai?

Haan, two-sided limit

4

hai kyunki dono sides agree karti hain; mismatched

f (a) = 9

ka sirf matlab hai

f

discontinuous hai, limit fail nahi hoti.

f (x) = sin (1/ x)

x \to 0^{+}

: kya right-hand limit exist karti hai?

Nahi —

0

ke paas input

1/ x

infinitely many cycles se race karta hai, isliye

f

- 1

aur

1

ke beech oscillate karta hai bina settle kiye; koi single approach value nahi, isliye koi one-sided limit nahi.

Domain ka ek isolated point (maano

f

sirf

a

par defined hai lekin kisi punctured neighbourhood mein nahi): kya koi one-sided limit exist kar sakti hai?

Nahi. Kisi bhi side se

a

approach karne wale koi domain points na hone se, na one-sided na two-sided limit defined hai; sirf

f (a)

ek plain value ke roop mein exist karta hai.

Dono one-sided limits

2

ke barabar hain aur

f (a) = 2

bhi hai — kya

f

discontinuous hai, aur koi jump hai?

Dono mein nahi. Teeno agree karte hain, isliye

f

a

par continuous hai; koi jump nahi — ek jump discontinuity ke liye dono one-sided limits ka differ karna zaroori hai, jo yahan nahi hota.

Connections

One-sided Limits — Left-hand & Right-hand
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