A function is continuous where its graph draws without lifting the pencil. Three checks make that precise — and three kinds of failure cover every discontinuity.
The problem
f(x) = x² − 4x − 2 — what happens at x = 2?
The limit exists (it is 4 — factor and cancel) but f(2) itself does not, so the graph is the line y = x + 2 with one missing point: a hole, the removable kind. D is the trap: "undefined at 2" does not mean the function blows up — here it quietly skips one point.
1. f(a) exists
the point has a value
2. the limit as x → a exists
both sides head to the same height
3. they are equal
the value sits where the limit points
Fail any one and the pencil lifts. The three kinds of discontinuity are just the three interesting ways of failing.
| Type | What fails | Classic source |
|---|---|---|
| removable (hole) | limit exists, value missing or mismatched | a cancelled common factor |
| jump | left and right limits disagree | piecewise definitions |
| infinite | the function blows up | a denominator zero that does NOT cancel |
x² − 4x − 2: the (x − 2) cancels
hole at x = 2 — removable
1x − 3: nothing cancels
vertical asymptote at x = 3 — infinite
Same test that sorts asymptotes from holes on rational-function graphs: factor everything first, then judge the zero of the denominator.
And a denominator with no real zero gives no break at all: x² + 9 is never 0, so x + 5x² + 9 is continuous for every real x. Polynomials are continuous everywhere too. A bottom like x² − 9 or x³ is zero somewhere, so it breaks there.
Set x − 2 = 0 in x² − 4x − 2, get x = 2, and write “vertical asymptote at x = 2” — that is the wrong answer the opener's D is built on. Test it: f(1.99) = 3.99 and f(2.01) = 4.01. An asymptote would send those values towards ±∞; these sit calmly next to 4. The (x − 2) cancels, so it is a hole.
define f(2) = 4
fill the hole with the limit's value
now all three checks pass
the repaired function is continuous
Jumps and asymptotes cannot be repaired with one value — no single number bridges a gap or tames a blow-up. Only holes are removable, hence the name.
Let f(x) = 3x − 1 for x < 2 and f(x) = x² + k for x ≥ 2. Find k so that f is continuous.
left piece at 2: 3(2) − 1 = 5
where the left side is heading
right piece at 2: 2² + k = 4 + k
this one also gives f(2)
4 + k = 5 → k = 1
no jump allowed
Put the boundary x into both rules, whichever one owns the “=”, and set them equal. With two unknowns — say ax + 2 for x < 1, f(1) = 5, x² + b for x > 1 — each side must equal the given value: a + 2 = 5 and 1 + b = 5, so a = 3, b = 4.
5 is where the pieces meet; k = 1 is what was asked. Both will be in the choices.
The value exists but sits at the wrong height. f(x) = x² + x − 2x − 1 for x ≠ 1, and f(1) = 2. Is f continuous at 1?
check 1: f(1) = 2 exists ✓
given separately
check 2: (x + 2)x − 1x − 1 = x + 2 → 3 ✓
the limit exists
check 3: 3 ≠ 2 ✗ → not continuous
removable: redefine f(1) = 3
What is different: nothing is missing. The first two checks pass, and only the third fails — the dot is off the curve.
k on both sides. f(x) = kx + 3 for x < 2 and f(x) = k²x − 1 for x ≥ 2. Find every k that makes f continuous.
2k + 3 = 2k² − 1
both rules at x = 2
2k² − 2k − 4 = 0 → k² − k − 2 = 0
divide by 2
(k − 2)(k + 1) = 0 → k = 2 or −1
k = 2: 7 and 7 ✓; k = −1: 1 and 1 ✓
What is different: the matching equation is a quadratic, so two values of k both work. Stopping at one loses half the answer.
No k can fix it. f(x) = x + 1x − 2 for x ≠ 2, and f(2) = k. Which k makes f continuous?
at x = 2: 2 + 12 − 2 = 30
not 00
nothing cancels → the limit does not exist
an infinite break
no value of k
check 2 fails whatever f(2) is
What is different: the question invites a number, but an asymptote cannot be repaired. Only a hole is removable.
What happens to f(x) = |x + 1| ÷ (x + 1) at x = −1?
x > −1: |x + 1| = x + 1 → f = 1
the inside is positive
x < −1: |x + 1| = −(x + 1) → f = −1
the inside is negative
right limit 1, left limit −1 → jump
the sides disagree
Not an asymptote: top and bottom shrink together, so the values stay at ±1 instead of blowing up. Not removable either: no single f(−1) can sit at both heights.
Expect a k that joins two pieces, a graph to read (limit vs the filled dot), or a rational function to sort into hole and asymptote. Factor first; for pieces, plug the boundary into both rules. A 10-second check for a graph: cover the dot — where the curve is heading is the limit; the dot is the value.
The problem
g(x) = 1x − 3 — what kind of discontinuity sits at x = 3?
Nothing cancels the (x − 3), so near 3 the outputs explode — 10.01 = 100, 10.001 = 1000 — an infinite discontinuity with a vertical asymptote. A would need the limit to exist; here it runs off to ±∞ instead of settling on a height.
