Intuition The one-sentence ideaA function is continuous at a point if you can draw it through that point without lifting your pen — the value the function approaches equals the value it actually takes .
Definition Continuity at a pointA function f f f continuous at x = a x = a x = a if all three of these hold:
f ( a ) f(a) f ( a ) defined (the point exists),lim x → a f ( x ) \displaystyle \lim_{x \to a} f(x) x → a lim f ( x ) exists (left limit = right limit, both finite),lim x → a f ( x ) = f ( a ) \displaystyle \lim_{x \to a} f(x) = f(a) x → a lim f ( x ) = f ( a ) equals the value).
Compactly: lim x → a f ( x ) = f ( a ) \quad \boxed{\displaystyle \lim_{x\to a} f(x) = f(a)} x → a lim f ( x ) = f ( a )
WHY three conditions? Because each can fail independently:
The hole exists but no value → condition 1 fails.
Left and right approaches disagree → condition 2 fails.
Limit and value both exist but differ → condition 3 fails.
A function is continuous on an interval if it is continuous at every point of it. At endpoints we only demand the one-sided limit (you can't approach from outside the domain).
Intuition Three ways the pen lifts
Removable : a single missing/misplaced dot — limit exists but ≠ value (or value undefined). Fixable by redefining one point. Jump : the left and right pieces land at different heights — both one-sided limits finite but unequal. Not fixable. Infinite : at least one side blows up to ± ∞ \pm\infty ± ∞ Not fixable.
Definition The three classificationsLet L − = lim x → a − f L^- = \lim_{x\to a^-}f L − = lim x → a − f L + = lim x → a + f L^+=\lim_{x\to a^+}f L + = lim x → a + f
Removable : L − = L + L^-=L^+ L − = L + f ( a ) f(a) f ( a ) ≠ \neq = Jump (discontinuity of the first kind) : L − , L + L^-, L^+ L − , L + L − ≠ L + L^-\neq L^+ L − = L + jump size is ∣ L + − L − ∣ |L^+ - L^-| ∣ L + − L − ∣ Infinite (essential / second kind) : at least one of L − , L + L^-,L^+ L − , L + ± ∞ \pm\infty ± ∞
Worked example (1) Removable — the famous holef ( x ) = x 2 − 1 x − 1 f(x) = \dfrac{x^2 - 1}{x - 1} f ( x ) = x − 1 x 2 − 1 x = 1 x=1 x = 1
Step 1: f ( 1 ) = 0 0 f(1) = \frac{0}{0} f ( 1 ) = 0 0 Why? Plugging in gives 0 / 0 0/0 0/0 Step 2: Factor: ( x − 1 ) ( x + 1 ) x − 1 = x + 1 \frac{(x-1)(x+1)}{x-1} = x+1 x − 1 ( x − 1 ) ( x + 1 ) = x + 1 x ≠ 1 x\neq 1 x = 1 lim x → 1 f = 1 + 1 = 2 \lim_{x\to1}f = 1+1 = 2 lim x → 1 f = 1 + 1 = 2 Why factor? The limit ignores the point x = 1 x=1 x = 1 Verdict: Limit exists (= 2 =2 = 2 f ( 1 ) f(1) f ( 1 ) removable . Patch: define f ( 1 ) = 2 f(1)=2 f ( 1 ) = 2
Worked example (2) Jump — a piecewise stepf ( x ) = { x + 1 , x < 2 x 2 , x ≥ 2 f(x) = \begin{cases} x+1, & x < 2 \\ x^2, & x \ge 2 \end{cases} f ( x ) = { x + 1 , x 2 , x < 2 x ≥ 2 x = 2 x=2 x = 2
Step 1: f ( 2 ) = 2 2 = 4 f(2) = 2^2 = 4 f ( 2 ) = 2 2 = 4 Why use the second piece? Because x ≥ 2 x\ge 2 x ≥ 2 x 2 x^2 x 2 Step 2: L − = lim x → 2 − ( x + 1 ) = 3 L^- = \lim_{x\to2^-}(x+1) = 3 L − = lim x → 2 − ( x + 1 ) = 3 L + = lim x → 2 + x 2 = 4 L^+ = \lim_{x\to2^+}x^2 = 4 L + = lim x → 2 + x 2 = 4 Why split? Left and right use different formulas.
Verdict: L − = 3 ≠ 4 = L + L^- = 3 \ne 4 = L^+ L − = 3 = 4 = L + jump , jump size ∣ 4 − 3 ∣ = 1 |4-3| = 1 ∣4 − 3∣ = 1
Worked example (3) Infinite — vertical asymptotef ( x ) = 1 x − 3 f(x) = \dfrac{1}{x-3} f ( x ) = x − 3 1 x = 3 x=3 x = 3
Step 1: f ( 3 ) = 1 0 f(3) = \frac{1}{0} f ( 3 ) = 0 1 Step 2: lim x → 3 − 1 x − 3 = − ∞ \lim_{x\to3^-}\frac{1}{x-3} = -\infty lim x → 3 − x − 3 1 = − ∞ lim x → 3 + 1 x − 3 = + ∞ \lim_{x\to3^+}\frac{1}{x-3} = +\infty lim x → 3 + x − 3 1 = + ∞ Why opposite signs? Just below 3 the denominator is a tiny negative; just above, a tiny positive.
Verdict: One-sided limits are infinite → infinite discontinuity (vertical asymptote at x = 3 x=3 x = 3
Worked example (4) Limit exists but value wrong (still removable)g ( x ) = { sin x x , x ≠ 0 5 , x = 0 g(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 5, & x = 0 \end{cases} g ( x ) = { x s i n x , 5 , x = 0 x = 0 x = 0 x=0 x = 0
Step 1: g ( 0 ) = 5 g(0) = 5 g ( 0 ) = 5 Step 2: lim x → 0 sin x x = 1 \lim_{x\to0}\frac{\sin x}{x} = 1 lim x → 0 x s i n x = 1 Verdict: Limit = 1 ≠ 5 = g ( 0 ) =1 \ne 5 = g(0) = 1 = 5 = g ( 0 ) removable (redefine g ( 0 ) = 1 g(0)=1 g ( 0 ) = 1 Why removable not jump? The limit exists ; only the dot is misplaced.
Recall Predict before you computeFor h ( x ) = x 2 − 4 x − 2 h(x) = \dfrac{x^2-4}{x-2} h ( x ) = x − 2 x 2 − 4 x = 2 x=2 x = 2 forecast the type, then check.
Forecast: numerator = ( x − 2 ) ( x + 2 ) =(x-2)(x+2) = ( x − 2 ) ( x + 2 ) Verify: limit = x + 2 → 4 =x+2\to 4 = x + 2 → 4 h ( 2 ) = 0 / 0 h(2)=0/0 h ( 2 ) = 0/0 removable , patch h ( 2 ) = 4 h(2)=4 h ( 2 ) = 4
f ( a ) f(a) f ( a ) Why it feels right: undefined often comes from 0 / 0 0/0 0/0 The fix: 1 x − 3 \frac{1}{x-3} x − 3 1 also undefined at 3 3 3 infinite . Undefined ≠ removable. You must still check whether the limit exists and is finite .
Common mistake "Limit exists, so it's continuous."Why it feels right: the limit existing feels like the whole story.
The fix: continuity needs limit = the actual value f ( a ) f(a) f ( a ) . Example (4): limit = 1 =1 = 1 g ( 0 ) = 5 g(0)=5 g ( 0 ) = 5
Common mistake "Jump and removable are basically the same."Why it feels right: both are "small" defects.
The fix: Removable means L − = L + L^-=L^+ L − = L + exists ) — fixable with one redefinition. Jump means L − ≠ L + L^-\neq L^+ L − = L + no single value can fix it.
Common mistake Picking the wrong piece forf ( a ) f(a) f ( a ) Why it feels right: people grab the formula nearest the boundary.
The fix: use the exact inequality. If the definition says x ≥ 2 ⇒ x 2 x\ge 2 \Rightarrow x^2 x ≥ 2 ⇒ x 2 f ( 2 ) = 4 f(2)=4 f ( 2 ) = 4 not the x + 1 x+1 x + 1
Mnemonic Remember the 3 conditions:"VLE"
V alue exists, L imit exists, they're E qual.
Types — "Hole, Step, Pole" = Removable (hole), Jump (step), Infinite (pole/asymptote).
Recall Feynman: explain to a 12-year-oldImagine drawing a line with a crayon. If your crayon never leaves the paper, the line is continuous . Three things can make you lift it: (1) there's a tiny hole where one dot is missing — that's removable , just fill the dot. (2) The line suddenly jumps to a different height like a stair step — that's a jump , and no single dot fixes a whole step. (3) The line rockets up to the sky near a spot (a wall it can't cross) — that's infinite . So: hole, step, or rocket — those are the three ways your crayon jumps.
What are the 3 conditions for f f f x = a x=a x = a (1)
f ( a ) f(a) f ( a ) defined, (2)
lim x → a f ( x ) \lim_{x\to a}f(x) lim x → a f ( x ) exists & finite, (3) limit
= f ( a ) = f(a) = f ( a ) .
A removable discontinuity occurs when... the two-sided limit exists (finite) but
f ( a ) f(a) f ( a ) is undefined or
≠ \neq = the limit.
A jump discontinuity occurs when... both one-sided limits are finite but
L − ≠ L + L^-\neq L^+ L − = L + .
An infinite discontinuity occurs when... at least one one-sided limit is
± ∞ \pm\infty ± ∞ (vertical asymptote).
Jump size formula ∣ L + − L − ∣ |L^+ - L^-| ∣ L + − L − ∣ , where
L ± L^\pm L ± are the one-sided limits.
Is x 2 − 1 x − 1 \frac{x^2-1}{x-1} x − 1 x 2 − 1 x = 1 x=1 x = 1 No — removable; limit
= 2 =2 = 2 but
f ( 1 ) f(1) f ( 1 ) undefined.
Type of discontinuity of 1 x − 3 \frac1{x-3} x − 3 1 x = 3 x=3 x = 3 Infinite (one-sided limits
∓ ∞ \mp\infty ∓ ∞ ).
"Limit exists ⇒ continuous" — true or false? False; also need limit
= f ( a ) =f(a) = f ( a ) .
Which discontinuity is fixable by redefining one point? Removable.
Continuity at an interval endpoint requires? Only the relevant one-sided limit equals
f ( endpoint ) f(\text{endpoint}) f ( endpoint ) .
Limits — definition and one-sided limits (continuity is built from limits)Standard limit sin(x)/x = 1 (used in Example 4)Differentiability implies continuity (continuity is necessary for derivatives)Intermediate Value Theorem (requires continuity on a closed interval)Vertical asymptotes and rational functions (source of infinite discontinuities)Piecewise functions (common source of jump discontinuities)
limit exists but wrong value
L- and L+ finite but unequal
Intuition Hinglish mein samjhoContinuity ka matlab simple hai: agar tum graph ko bina pen uthaaye ek point ke through draw kar sako, toh wahaan function continuous hai. Formally, teen cheezein chahiye — f ( a ) f(a) f ( a ) lim x → a f ( x ) \lim_{x\to a}f(x) lim x → a f ( x )
Discontinuity teen type ki hoti hai. Removable (hole) — limit exist karti hai par value missing ya galat hai, jaise x 2 − 1 x − 1 \frac{x^2-1}{x-1} x − 1 x 2 − 1 x = 1 x=1 x = 1 = 2 =2 = 2 Jump (step) — left limit aur right limit dono finite hain par alag-alag, jaise piecewise function jisme ek piece 3 3 3 4 4 4 Infinite (pole) — koi ek side ± ∞ \pm\infty ± ∞ 1 x − 3 \frac{1}{x-3} x − 3 1 x = 3 x=3 x = 3
Sabse common galti: log sochte hain "limit exist kar gayi matlab continuous" — galat! Limit ko actual value f ( a ) f(a) f ( a ) f ( a ) f(a) f ( a ) 1 x − 3 \frac{1}{x-3} x − 3 1
