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

Definition Vocabulary you must have loaded first

Read these once — every trap below leans on them.

$f (a)$ = the value the function actually takes at $a$ (the dot on the graph).
$L^{-} = lim_{x \to a^{-}} f (x)$ = the height the graph approaches from the left.
$L^{+} = lim_{x \to a^{+}} f (x)$ = the height it approaches from the right.
Continuous at $a$ = all three agree: $L^{-} = L^{+} = f (a)$ , all finite.
The three continuity conditions (numbered — referenced by name throughout):
1. Condition 1: $f (a)$ is defined (the dot exists).
2. Condition 2: the two-sided limit exists and is finite ( $L^{-} = L^{+}$ , both finite).
3. Condition 3: limit equals value ( $L^{-} = L^{+} = f (a)$ ).
Hole = a removable discontinuity: the two-sided limit exists but the dot is missing or misplaced — one redefinition fixes it. Looks like a single missing point on an otherwise smooth curve.
Pole = an infinite discontinuity: at least one one-sided limit is $\pm \infty$ , so the graph shoots up a vertical asymptote. Cannot be fixed by any value.
Jump size = $∣ L^{+} - L^{-} ∣$ , the vertical gap between the two landing heights of a jump discontinuity. Only defined when both $L^{-}$ and $L^{+}$ are finite.

See the four trap-shapes side by side before you start:

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

True or false — justify

TF1. "If

lim_{x \to a} f (x)

exists, then

f

is continuous at

a

False. The limit existing only settles

L^{-} = L^{+}

(Condition 2); continuity also needs that shared limit to equal the actual dot

f (a)

(Condition 3). A misplaced dot (limit

= 1

but

f (0) = 5

) has a limit yet is discontinuous.

TF2. "If

f (a)

is undefined, the discontinuity must be removable."

False.

\frac{1}{x - 3}

is undefined at

3

but blows up to

\pm \infty

— an infinite discontinuity (a pole). Undefined only tells you Condition 1 failed; you must still check whether the limit is finite (Condition 2).

TF3. "A jump discontinuity can be fixed by redefining one point."

False. A jump means

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

: the left and right pieces land at different heights, so no single value can equal both. Only removable discontinuities (holes) are one-point fixable.

TF4. "If

f

is continuous at

a

, then

f (a)

equals the limit from the left."

True. Continuity forces

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

, so in particular the left limit equals the value. Both one-sided limits equal the dot.

TF5. "A function continuous on

(0, 1)

is automatically continuous on

[0, 1]

False. Continuity on the open interval says nothing about the endpoints

0

and

1

; the graph could rocket to infinity or jump right at an edge. Endpoints need their own one-sided check.

TF6. "Every polynomial is continuous everywhere."

True. Polynomials are built from adding and multiplying

x

and constants, all continuous operations, and never divide by a variable, so no holes or poles ever appear.

TF7. "If both one-sided limits are

+ \infty

, that is a jump discontinuity because both sides agree."

False. "Both finite but unequal" is the definition of a jump; here the limits are infinite, so it is an infinite discontinuity (pole) regardless of whether the two sides match.

TF8. "A function can be continuous at exactly one point."

True. For example

f (x) = x

x

is rational,

0

if irrational, is continuous only at

x = 0

. Continuity is a point-by-point property, not an all-or-nothing one.

TF9. "If

f

and

g

are both discontinuous at

a

, then

f + g

is discontinuous at

a

False. Two jumps can cancel: if

f

jumps up by

1

and

g

jumps down by

1

a

, their sum is continuous. Discontinuities can annihilate each other.

Spot the error

SE1. "

f (x) = \frac{x ^{2} - 4}{x - 2}

has a pole at

x = 2

because the denominator is zero."

The numerator is also zero there:

\frac{( x - 2 ) ( x + 2 )}{x - 2} = x + 2

cancels. A zero denominator only gives a pole (infinite discontinuity) when the numerator does not vanish; here it's a removable hole with limit

4

SE2. "For

f (x) = {x + 1, x^{2}, x < 2 x \geq 2

, we have

f (2) = 2 + 1 = 3

Wrong piece. The condition

x \geq 2

owns the boundary point, so

f (2) = 2^{2} = 4

. The

x + 1

formula only governs

x < 2

strictly, never

x = 2

itself.

SE3. "

\frac{s i n x}{x}

is discontinuous at

0

, so it has a jump there."

The limit is

1

from both sides, so it's not a jump (

L^{-} = L^{+}

). It is undefined at

0

only because of

0/0

; the two-sided limit exists, making it removable (a hole).

SE4. "

\frac{1}{x - 3}

is continuous everywhere except

x = 3

where the value is huge."

There is no value at

3

at all —

f (3)

is undefined, not "huge". The function grows without bound near

3

; the point itself simply doesn't exist on the graph.

SE5. "Since

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

equals

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

in size, the jump size is

∣ + \infty - (- \infty) ∣

Jump size

∣ L^{+} - L^{-} ∣

is only defined for finite one-sided limits (see the vocabulary). When a side is infinite it's an infinite discontinuity (pole), and "jump size" simply doesn't apply.

SE6. "To patch

\frac{x ^{2} - 1}{x - 1}

x = 1

, redefine

f (1) = 0

since the original gave

0/0

You patch with the limit, not the indeterminate form. The limit is

x + 1 \to 2

, so

f (1) = 2

makes it continuous;

0

would leave a hole (misplaced dot).

Why questions

WHY1. "Why does continuity need three conditions instead of just 'limit equals value'?"

Because you can't compare a limit to a value if either is missing. Conditions 1 and 2 guarantee both sides of the equation

lim = f (a)

actually exist before Condition 3 checks they agree.

WHY2. "Why is a removable discontinuity called 'removable'?"

Because the graph has a two-sided limit

L

, so the only thing wrong is one dot — redefining

f (a) = L

removes the flaw entirely and restores an unbroken curve.

WHY3. "Why can't we approach an interval endpoint from both sides?"

The function isn't defined outside the interval, so one side of the approach lands in empty space. We only demand the one-sided limit that stays inside the domain to equal

f (endpoint)

WHY4. "Why does

\frac{1}{x - 3}

give

- \infty

from the left but

+ \infty

from the right?"

Just below

3

x - 3

is a tiny negative number, so its reciprocal is a huge negative; just above

3

it's a tiny positive, giving a huge positive. The sign of the shrinking denominator flips across

3

WHY5. "Why is factoring the first move for a

0/0

form at a removable point?"

A limit ignores the single point

x = a

itself, so we may cancel the common factor that caused

0/0

. The cancelled expression reveals the height the graph approaches.

WHY6. "Why does differentiability require continuity first?"

A derivative is a limit of slopes; if the graph jumps or has a hole, the slope-limit can't settle to a single number. See Differentiability implies continuity — continuity is the price of admission.

WHY7. "Why do piecewise functions so often produce jumps?"

Different formulas govern each side of the boundary, and there's no rule forcing them to meet at the same height. Unless the pieces are deliberately matched,

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

. See Piecewise functions.

WHY8. "Why isn't

sin (\frac{1}{x})

x = 0

removable, a jump, or infinite?"

x \to 0

the input

\frac{1}{x}

races through infinitely many cycles, so the graph wiggles between

- 1

and

1

forever without settling — neither

L^{-}

nor

L^{+}

exists, yet nothing blows up to

\infty

. This is a fourth kind: an oscillatory discontinuity, non-removable and non-fixable.

Edge cases

EC1. "Is

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

discontinuous at

x = 5

No.

f

is only broken where the denominator is zero — at

x = 0

. At

x = 5

it's a perfectly ordinary continuous point with

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

EC2. "If

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

but

f (a) = 2

as well, what type of discontinuity is it?"

None — all three conditions hold, so

f

is continuous at

a

. Matching limits plus a matching value is exactly the definition of continuity, not a defect.

EC3. "Can a function be continuous at

a

even if

a

is an isolated point of its domain (no other domain points nearby)?"

Yes — vacuously. The

ε

–

δ

definition ranges only over domain points within

δ

a

; if the only such point is

a

itself, the condition

∣ f (x) - f (a) ∣ < ε

holds trivially. So continuity at an isolated point is automatic, not undefined.

EC4. "The graph of

tan x

shoots to infinity at

x = \frac{π}{2}

. Continuous or not there?"

Infinite discontinuity (pole). As

x \to \frac{π}{2}^{-}

tan x \to + \infty

and from the right

\to - \infty

— a vertical asymptote, exactly like Vertical asymptotes and rational functions.

EC5. "A step function equals

0

for

x < 0

and

1

for

x \geq 0

. Is it continuous at

x = 0.5

Yes — near

0.5

the function is constantly

1

, so

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

. The only discontinuity is the jump at

x = 0

; away from the step it's flat and continuous.

EC6. "For

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

redefined as

g (0) = 1

, is

g

now continuous at

0

Yes. The limit is

1

(see Standard limit sin(x)/x = 1) and now

g (0) = 1

too, so all three conditions hold — the hole is patched.

EC7. "The Intermediate Value Theorem needs

f

continuous on

[a, b]

. Does one jump inside ruin it?"

Yes — a jump lets

f

skip over intermediate values, so the guaranteed crossing can fail. IVT truly requires unbroken continuity across the whole closed interval; see Intermediate Value Theorem.

EC8. "Is

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

(with

f (0) = 0

) removable at

x = 0

No. A hole is removable only if the two-sided limit exists; here the graph oscillates forever, so no limit exists — it's an oscillatory discontinuity, unfixable by any choice of

f (0)

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