Continuity — definition, types of discontinuity (removable, jump, infinite): Question bank

4.1.7 · D5 · Maths › Calculus I — Limits & Derivatives › Continuity — definition, types of discontinuity (removable,

Definition Vocabulary jo pehle load honi chahiye

Inhe ek baar padh lo — neeche ke har trap mein inhi ka sahara hai.

$f (a)$ = woh value jo function actually leta hai $a$ par (graph par dot).
$L^{-} = lim_{x \to a^{-}} f (x)$ = woh height jis par graph left se approach karta hai.
$L^{+} = lim_{x \to a^{+}} f (x)$ = woh height jis par graph right se approach karta hai.
$a$ par Continuous = teeno agree karte hain: $L^{-} = L^{+} = f (a)$ , sab finite.
Teen continuity conditions (numbered — poore note mein naam se reference hongi):
1. Condition 1: $f (a)$ defined hai (dot exist karta hai).
2. Condition 2: two-sided limit exist karta hai aur finite hai ( $L^{-} = L^{+}$ , dono finite).
3. Condition 3: limit value ke barabar hai ( $L^{-} = L^{+} = f (a)$ ).
Hole = ek removable discontinuity: two-sided limit exist karta hai lekin dot missing ya misplaced hai — ek redefinition se theek ho jaata hai. Ek otherwise smooth curve par ek single missing point jaisa dikhta hai.
Pole = ek infinite discontinuity: kam se kam ek one-sided limit $\pm \infty$ hai, toh graph ek vertical asymptote ki taraf shoot karta hai. Kisi bhi value se fix nahi ho sakta.
Jump size = $∣ L^{+} - L^{-} ∣$ , ek jump discontinuity ki do landing heights ke beech ka vertical gap. Sirf tab defined hai jab $L^{-}$ aur $L^{+}$ dono finite hon.

Shuru karne se pehle charon trap-shapes ek saath dekho:

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

True ya false — justify karo

TF1. "Agar

lim_{x \to a} f (x)

exist karta hai, toh

f

a

par continuous hai."

False. Limit ka exist karna sirf

L^{-} = L^{+}

(Condition 2) settle karta hai; continuity ke liye yeh bhi zaroori hai ki woh shared limit actual dot

f (a)

ke barabar ho (Condition 3). Ek misplaced dot (limit

= 1

lekin

f (0) = 5

) mein limit hota hai phir bhi discontinuous hai.

TF2. "Agar

f (a)

undefined hai, toh discontinuity zaroor removable hogi."

False.

\frac{1}{x - 3}

3

par undefined hai lekin

\pm \infty

tak blow up karta hai — ek infinite discontinuity (pole). Undefined hona sirf batata hai ki Condition 1 fail hui; tumhe phir bhi check karna hoga ki limit finite hai ya nahi (Condition 2).

TF3. "Ek jump discontinuity ko ek point redefine karke fix kiya ja sakta hai."

False. Jump ka matlab hai

L^{-} \neq = L^{+}

: left aur right pieces alag-alag heights par land karti hain, toh koi single value dono ke barabar nahi ho sakti. Sirf removable discontinuities (holes) ek-point fixable hoti hain.

TF4. "Agar

f

a

par continuous hai, toh

f (a)

left ki taraf se limit ke barabar hai."

True. Continuity force karti hai

L^{-} = L^{+} = f (a)

, toh khaskar left limit value ke barabar hai. Dono one-sided limits dot ke barabar hote hain.

TF5. "

(0, 1)

par continuous ek function automatically

[0, 1]

par continuous hoga."

False. Open interval par continuity endpoints

0

aur

1

ke baare mein kuch nahi kehti; graph kisi edge par infinity tak ja sakta hai ya jump kar sakta hai. Endpoints ko apna alag one-sided check chahiye.

TF6. "Har polynomial har jagah continuous hoti hai."

True. Polynomials

x

aur constants ko add aur multiply karke bani hoti hain, ye sab continuous operations hain, aur kabhi kisi variable se divide nahi karti, toh kabhi koi hole ya pole nahi aata.

TF7. "Agar dono one-sided limits

+ \infty

hain, toh yeh ek jump discontinuity hai kyunki dono sides agree karti hain."

False. "Dono finite lekin unequal" jump ki definition hai; yahan limits infinite hain, toh yeh ek infinite discontinuity (pole) hai chahe dono sides match karein ya na karein.

TF8. "Ek function exactly ek point par continuous ho sakta hai."

True. For example

f (x) = x

agar

x

rational hai,

0

agar irrational, sirf

x = 0

par continuous hai. Continuity ek point-by-point property hai, sab kuch ya kuch nahi wali nahi.

TF9. "Agar

f

aur

g

dono

a

par discontinuous hain, toh

f + g

bhi

a

par discontinuous hai."

False. Do jumps cancel ho sakte hain: agar

f

a

par

1

se upar jump karta hai aur

g

1

se neeche, toh unka sum continuous hai. Discontinuities ek doosre ko annihilate kar sakti hain.

Why questions

WHY1. "Continuity ko 'limit value ke barabar' ki jagah teen conditions kyun chahiye?"

Kyunki tum limit ko value se compare nahi kar sakte agar koi bhi missing ho. Conditions 1 aur 2 guarantee karte hain ki equation

lim = f (a)

ke dono sides actually exist karte hain Condition 3 ke unhe agree check karne se pehle.

WHY2. "Ek removable discontinuity ko 'removable' kyun kaha jaata hai?"

Kyunki graph ka ek two-sided limit

L

hai, toh sirf ek dot galat hai —

f (a) = L

redefine karna flaw ko poori tarah remove karta hai aur ek unbroken curve restore karta hai.

WHY3. "Hum ek interval endpoint par dono sides se approach kyun nahi kar sakte?"

Function interval ke bahar defined nahi hai, toh approach ka ek side empty space mein land karta hai. Hum sirf woh one-sided limit maangte hain jo domain ke andar rahe ki woh

f (endpoint)

ke barabar ho.

WHY4. "

\frac{1}{x - 3}

left se

- \infty

aur right se

+ \infty

kyun deta hai?"

3

ke theek neeche,

x - 3

ek chhota negative number hai, toh uska reciprocal ek bada negative hai;

3

ke theek upar yeh ek chhota positive hai, ek bada positive deta hai. Shrinking denominator ka sign

3

ke aas-paas flip hota hai.

WHY5. "Ek removable point par

0/0

form ke liye factoring pehla move kyun hai?"

Limit

x = a

ko khud ignore karta hai, toh hum

0/0

cause karne wala common factor cancel kar sakte hain. Cancelled expression woh height reveal karta hai jis par graph approach karta hai.

WHY6. "Differentiability ke liye pehle continuity kyun chahiye?"

Derivative slopes ka limit hai; agar graph jump kare ya hole ho, toh slope-limit ek single number par settle nahi ho sakta. Dekho Differentiability implies continuity — continuity admission ki kimat hai.

WHY7. "Piecewise functions mein itni baar jumps kyun aate hain?"

Alag-alag formulas boundary ke har side ko govern karte hain, aur koi rule nahi hai jo unhe same height par milne ke liye force kare. Jab tak pieces deliberately match na ki jayein,

L^{-} \neq = L^{+}

hoga. Dekho Piecewise functions.

WHY8. "

x = 0

par

sin (\frac{1}{x})

removable, jump, ya infinite kyun nahi hai?"

Jaise

x \to 0

, input

\frac{1}{x}

infinitely many cycles se race karta hai, toh graph hamesha ke liye

- 1

aur

1

ke beech wiggle karta rehta hai bina settle hue — na

L^{-}

exist karta hai na

L^{+}

, phir bhi kuch

\infty

tak blow up nahi hota. Yeh ek chautha kind hai: ek oscillatory discontinuity, non-removable aur non-fixable.

Edge cases

EC1. "Kya

f (x) = \frac{1}{x}

x = 5

par discontinuous hai?"

Nahi.

f

sirf wahan toot-ti hai jahan denominator zero hota hai —

x = 0

par.

x = 5

par yeh ek bilkul ordinary continuous point hai jahan

f (5) = \frac{1}{5}

hai.

EC2. "Agar

L^{-} = L^{+} = 2

lekin

f (a) = 2

bhi, toh discontinuity kis type ki hai?"

Koi nahi — teeno conditions hold karti hain, toh

f

a

par continuous hai. Matching limits plus matching value exactly continuity ki definition hai, koi defect nahi.

EC3. "Kya ek function

a

par continuous ho sakta hai even if

a

uske domain ka ek isolated point hai (aas-paas koi aur domain points nahi)?"

Haan — vacuously.

ε

–

δ

definition sirf

a

δ

andar domain points par range karti hai; agar sirf wahi point

a

hai, toh condition

∣ f (x) - f (a) ∣ < ε

trivially hold karti hai. Toh isolated point par continuity automatic hai, undefined nahi.

EC4. "

tan x

ka graph

x = \frac{π}{2}

par infinity tak shoot karta hai. Wahan continuous hai ya nahi?"

Infinite discontinuity (pole). Jaise

x \to \frac{π}{2}^{-}

tan x \to + \infty

aur right se

\to - \infty

— ek vertical asymptote, bilkul Vertical asymptotes and rational functions jaisa.

EC5. "Ek step function

x < 0

ke liye

0

aur

x \geq 0

ke liye

1

ke barabar hai. Kya yeh

x = 0.5

par continuous hai?"

Haan —

0.5

ke paas function constantly

1

hai, toh

L^{-} = L^{+} = f (0.5) = 1

. Sirf discontinuity

x = 0

par jump hai; step se dur yeh flat aur continuous hai.

EC6. "

g (x) = \frac{s i n x}{x}

g (0) = 1

ke roop mein redefine ki gayi hai, kya

g

0

par continuous hai?"

Haan. Limit

1

hai (dekho Standard limit sin(x)/x = 1) aur ab

g (0) = 1

bhi, toh teeno conditions hold karti hain — hole patch ho gaya.

EC7. "Intermediate Value Theorem ko chahiye ki

f

[a, b]

par continuous ho. Kya andar ek jump use ruin kar deta hai?"

Haan — ek jump

f

ko intermediate values skip karne deta hai, toh guaranteed crossing fail ho sakti hai. IVT ko truly puri closed interval mein unbroken continuity chahiye; dekho Intermediate Value Theorem.

EC8. "Kya

f (x) = sin (\frac{1}{x})

(jahan

f (0) = 0

)

x = 0

par removable hai?"

Nahi. Ek hole tabhi removable hai jab two-sided limit exist kare; yahan graph hamesha ke liye oscillate karta hai, toh koi limit exist nahi karta — yeh ek oscillatory discontinuity hai,

f (0)

ki kisi bhi choice se fix nahi hoti.

Connections

Continuity — Definition & Types of Discontinuity (parent)
Limits — definition and one-sided limits
Standard limit sin(x)/x = 1
Differentiability implies continuity
Intermediate Value Theorem
Vertical asymptotes and rational functions
Piecewise functions

Question bank — Continuity — definition, types of discontinuity (removable, jump, infinite)

True ya false — justify karo

Error dhundho

Why questions

Edge cases

Connections