4.1.7
Intuition Ek sentence mein ideaEk function continuous hota hai ek point par agar aap use us point se guzarte hue bina pen uthaye draw kar sako — jo value function approach karta hai woh us value ke barabar hoti hai jo woh actually leta hai .
Definition Ek point par ContinuityEk function f continuous at x = a hai agar teeno conditions poori hon:
f ( a ) defined hai (point exist karta hai),x → a lim f ( x ) exists karta hai (left limit = right limit, dono finite hain),x → a lim f ( x ) = f ( a ) barabar hai).
Compact form mein: x → a lim f ( x ) = f ( a )
TEEN conditions KYU? Kyunki har ek alag tarike se fail ho sakti hai:
Hole exist karta hai lekin koi value nahi → condition 1 fail.
Left aur right approaches alag-alag results deti hain → condition 2 fail.
Limit aur value dono exist karte hain lekin alag hain → condition 3 fail.
Ek function continuous on an interval hota hai agar woh us interval ke har point par continuous ho. Endpoints par hum sirf one-sided limit maangte hain (aap domain ke bahar se approach nahi kar sakte).
Intuition Teen tarike jab pen uthta hai
Removable : ek akela missing/galat dot — limit exist karta hai lekin value ke barabar nahi (ya value undefined hai). Ek point redefine karke theek ho sakta hai. Jump : left aur right pieces alag heights par land karte hain — dono one-sided limits finite hain lekin unequal. Theek nahi ho sakta. Infinite : kam se kam ek side ± ∞ Theek nahi ho sakta.
Definition Teeno classificationsMaano L − = lim x → a − f L + = lim x → a + f
Removable : L − = L + f ( a ) = Jump (discontinuity of the first kind) : L − , L + L − = L + Jump size hai ∣ L + − L − ∣ Infinite (essential / second kind) : L − , L + ± ∞
Worked example (1) Removable — famous holef ( x ) = x − 1 x 2 − 1 x = 1
Step 1: f ( 1 ) = 0 0 Kyun? Plug in karne par 0/0 Step 2: Factor karo: x − 1 ( x − 1 ) ( x + 1 ) = x + 1 x = 1 lim x → 1 f = 1 + 1 = 2 Factor kyun kiya? Limit point x = 1 Verdict: Limit exist karta hai (= 2 f ( 1 ) removable . Patch: f ( 1 ) = 2
Worked example (2) Jump — ek piecewise stepf ( x ) = { x + 1 , x 2 , x < 2 x ≥ 2 x = 2
Step 1: f ( 2 ) = 2 2 = 4 Doosra piece kyun use kiya? Kyunki x ≥ 2 x 2 Step 2: L − = lim x → 2 − ( x + 1 ) = 3 L + = lim x → 2 + x 2 = 4 Split kyun kiya? Left aur right alag formulas use karte hain.
Verdict: L − = 3 = 4 = L + jump , jump size ∣4 − 3∣ = 1
Worked example (3) Infinite — vertical asymptotef ( x ) = x − 3 1 x = 3
Step 1: f ( 3 ) = 0 1 Step 2: lim x → 3 − x − 3 1 = − ∞ lim x → 3 + x − 3 1 = + ∞ Opposite signs kyun? 3 se thoda neeche denominator ek chota negative number hota hai; thoda upar, ek chota positive.
Verdict: One-sided limits infinite hain → infinite discontinuity (x = 3
Worked example (4) Limit exist karta hai lekin value galat hai (phir bhi removable)g ( x ) = { x s i n x , 5 , x = 0 x = 0 x = 0
Step 1: g ( 0 ) = 5 Step 2: lim x → 0 x s i n x = 1 Verdict: Limit = 1 = 5 = g ( 0 ) removable (g ( 0 ) = 1 Jump kyun nahi hai? Limit exist karta hai; sirf dot galat jagah hai.
Recall Compute karne se pehle predict karoh ( x ) = x − 2 x 2 − 4 x = 2 type forecast karo, phir check karo.
Forecast: numerator = ( x − 2 ) ( x + 2 ) Verify: limit = x + 2 → 4 h ( 2 ) = 0/0 removable , patch h ( 2 ) = 4
f ( a ) Kyun sahi lagta hai: undefined often 0/0 Fix: x − 3 1 3 undefined hai lekin infinite hai. Undefined ≠ removable. Phir bhi aapko check karna hoga ki limit exist karti hai aur finite hai .
Common mistake "Limit exist karta hai, toh continuous hai."Kyun sahi lagta hai: limit ka exist karna poori kahani lagti hai.
Fix: continuity ke liye limit = actual value f ( a ) chahiye. Example (4): limit = 1 g ( 0 ) = 5
Common mistake "Jump aur removable basically same hain."Kyun sahi lagta hai: dono "chhote" defects lagte hain.
Fix: Removable ka matlab hai L − = L + exist karti hai) — ek redefinition se theek ho sakta hai. Jump ka matlab hai L − = L + koi bhi single value use fix nahi kar sakti.
f ( a ) Kyun sahi lagta hai: log boundary ke paas wala formula pakad lete hain.
Fix: exact inequality use karo. Agar definition kehti hai x ≥ 2 ⇒ x 2 f ( 2 ) = 4 x + 1 nahi .
Mnemonic 3 conditions yaad rakhne ke liye:"VLE"
V alue exist karta hai, L imit exist karta hai, dono E qual hain.
Types — "Hole, Step, Pole" = Removable (hole), Jump (step), Infinite (pole/asymptote).
Recall Feynman: ek 12-saal ke bachche ko samjhaoSocho tum crayon se ek line draw kar rahe ho. Agar tumhara crayon kabhi paper se nahi uthta, toh line continuous hai. Teen cheezein hain jo tumhe uthne par majboor kar sakti hain: (1) ek choti hole jahan ek dot missing hai — yeh removable hai, bas dot bharo. (2) Line achanak ek alag height par jump karti hai jaise ek stair step — yeh jump hai, aur koi akela dot poore step ko fix nahi karta. (3) Line ek jagah ke paas sky tak rocket karti hai (ek wall jise woh cross nahi kar sakti) — yeh infinite hai. Toh: hole, step, ya rocket — yehi teen tarike hain jab tumhara crayon uthta hai.
f x = a (1) f ( a ) lim x → a f ( x ) = f ( a )
Removable discontinuity tab hoti hai jab... two-sided limit exist karta hai (finite) lekin f ( a ) =
Jump discontinuity tab hoti hai jab... dono one-sided limits finite hain lekin L − = L +
Infinite discontinuity tab hoti hai jab... kam se kam ek one-sided limit ± ∞
Jump size formula ∣ L + − L − ∣ L ±
Kya x − 1 x 2 − 1 x = 1 Nahi — removable; limit = 2 f ( 1 )
x − 3 1 x = 3 Infinite (one-sided limits ∓ ∞
"Limit exist karta hai ⇒ continuous" — sach ya jhooth? Jhooth; limit = f ( a )
Kaun si discontinuity ek point redefine karke fix ho sakti hai? Removable.
Interval ke endpoint par continuity ke liye kya chahiye? Sirf relevant one-sided limit f ( endpoint )
Limits — definition and one-sided limits (continuity limits se bani hai)Standard limit sin(x)/x = 1 (Example 4 mein use hua)Differentiability implies continuity (derivatives ke liye continuity zaroori hai)Intermediate Value Theorem (ek closed interval par continuity chahiye)Vertical asymptotes and rational functions (infinite discontinuities ka source)Piecewise functions (jump discontinuities ka common source)
limit exists but wrong value
L- and L+ finite but unequal
