Single-Variable-Calculus · Unit 2 · Video 4 · Interactive Practice

Continuity Is a Comparison of Two Numbers: Jumps and Removable Holes

IKey Formulas

Formula	Name	What you need
$\lim\limits_{x \to x_0} f(x) = f(x_0)$	Continuity at $x_0$	A limit and a value that agree
$\lim\limits_{x \to x_0^-} f(x) \ne \lim\limits_{x \to x_0^+} f(x)$	Jump discontinuity	Two one-sided limits that disagree
$\lim\limits_{x \to x_0^-} f(x) = \lim\limits_{x \to x_0^+} f(x) = L \ne f(x_0)$	Removable discontinuity	$f(x_0)$ missing or wrong; repair with $f(x_0) := L$
$\lim\limits_{x \to 0} \dfrac{\sin x}{x} = 1 \qquad \lim\limits_{x \to 0} \dfrac{1 - \cos x}{x} = 0$	The two standard holes	$x$ measured in radians

Key Insight: The two sides are computed by procedures that never overlap — the limit uses every point except $x_0$ , the value uses $x_0$ alone. Let the limit peek at $f(x_0)$ and the definition collapses into a tautology.

IIThree Ingredients, Three Ways to Fail

Continuity needs a limit, a value, and their agreement; which ingredient breaks names the discontinuity.

Left limit 2 Right limit 1 f(0)

1 · the limit exists

2 · f(0) is defined

3 · the two agree

IIIWhy a Jump Is Permanent

One number can match at most one of two different one-sided limits.

f(0) 1.00

💡 With time on the horizontal axis the left limit is the past and the right limit the future, so a price model must choose the value at a jump — the question behind Merton's option-pricing work.

IVFilling a Removable Hole

Each curve is undefined at $0$ yet has a limit there; commit to the number that fills the hole.

Function Value at 0 0.55

💡 Both limits are asserted here, not derived; their proofs come in the next lecture.

VQuiz Questions

Problem 1 · Classify and Repair

Given: $f(x) = x + 2$ for $x \lt 3$ , $f(3) = 4$ , and $f(x) = 2x - 1$ for $x \gt 3$ — find the status of $f$ at $x = 3$ .

✅ Correct! Both branches head for

5

x = 3

, so the limit exists and equals

5

; only the value

f(3) = 4

is out of line, and one number repairs it.

❌ Different formulas, same destination.

x + 2 \to 5

and

2x - 1 \to 5

x \to 3

. A jump needs the two one-sided limits to differ, not the two formulas.

❌ Ingredient 2 is not enough.

f(3) = 4

is defined and the limit is

5

; continuity also demands that the two numbers agree.

❌ Averaging is not repairing. The repair value is the common limit

5

itself, not a compromise between the limit and the old value.

❌ Not quite. Compute

\lim\limits_{x \to 3^-} f(x)

and

\lim\limits_{x \to 3^+} f(x)

first, then compare the result with

f(3)

Show solution

Ingredient 1 — does the limit exist?

\lim_{x \to 3^-} f(x) = 3 + 2 = 5, \qquad \lim_{x \to 3^+} f(x) = 2(3) - 1 = 5

The two sides agree, so $\lim\limits_{x \to 3} f(x) = 5$ . ✅

Ingredient 2 — is $f(3)$ defined? Yes, $f(3) = 4$ . ✅

Ingredient 3 — do they agree? $5 \ne 4$ . ❌

Only the third check fails, so the discontinuity is removable. Setting $f(3) = 5$ makes all three hold, and the repaired graph is a single unbroken bend.

Problem 2 · One Formula, Two Limits

Given: $f(x) = \dfrac{|x|}{x}$ for $x \ne 0$ and $f(0) = 0$ — find the status of $f$ at $x = 0$ .

✅ Correct! For

x \gt 0

the quotient is

1

and for

x \lt 0

it is

-1

. Ingredient 1 fails, and nothing about

f(0)

can rescue it.

❌ The midpoint is not a limit.

0

agrees with neither

-1

nor

1

, and a removable hole requires the two one-sided limits to be equal in the first place.

❌ Being defined is only ingredient 2. The limit does not exist at

0

, so the test fails before

f(0)

is even consulted.

❌ Continuous from the right only.

f(0) = 1

matches

\lim\limits_{x \to 0^+} f(x) = 1

but not

\lim\limits_{x \to 0^-} f(x) = -1

❌ Not quite. Evaluate

|x|/x

separately for

x \gt 0

and for

x \lt 0

before deciding.

Show solution

For $x \gt 0$ , $|x| = x$ ; for $x \lt 0$ , $|x| = -x$ . So

\lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1, \qquad \lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1

Both one-sided limits exist, but $1 \ne -1$ , so ingredient 1 fails: this is a jump discontinuity.

One number can equal at most one of $-1$ and $1$ , so no choice of $f(0)$ makes $f$ continuous at $0$ ; the best any choice achieves is continuity from one side.

Problem 3 · Locate the Failure, Then Repair It

Given: $f(x) = \dfrac{x^2 - 5x + 6}{x - 2}$ , defined for every $x \ne 2$ — find which ingredient of continuity fails at $x = 2$ and the value that repairs it.

Which ingredient fails?

What value repairs it?

✅ Correct!

\dfrac{(x-2)(x-3)}{x-2} = x - 3

for every

x \ne 2

, so the limit is

-1

while

f(2)

was never assigned. Setting

f(2) = -1

makes all three ingredients hold.

❌ Check the ingredients in order. Cancel the common factor first: the limit turns out to exist, so ask next whether

f(2)

is defined.

❌ The limit does exist. After cancelling

x - 2

the function agrees with

x - 3

at every point except

2

, and

x - 3 \to -1

from both sides.

❌ There is nothing to disagree with. Ingredient 3 compares two numbers, and

f(2)

is not one of them — the denominator

x - 2

vanishes there, so no value was ever assigned.

❌ Check the repair value. It is the common limit: cancel the factor

x - 2

, then evaluate what is left at

x = 2

❌ $\tfrac{0}{0}$ is not $0$ . Substituting

x = 2

gives

\tfrac{0}{0}

, which carries no information; the factor

x - 2

has to cancel before you substitute.

❌ Check the factorisation.

x^2 - 5x + 6 = (x-2)(x-3)

, not

(x-2)(x+3)

— the roots are

2

and

3

, so the surviving factor is

x - 3

❌ $\tfrac{0}{0}$ is not $1$ . The indeterminate form has to be removed by cancellation, and

x - 3

evaluated at

x = 2

-1

Show solution

Step 1 — factor and cancel.

\frac{x^2 - 5x + 6}{x - 2} = \frac{(x-2)(x-3)}{x-2} = x - 3 \quad (x \ne 2)

Step 2 — the limit. The limit never looks at $x = 2$ , and everywhere else $f$ agrees with $x - 3$ :

\lim_{x \to 2} f(x) = \lim_{x \to 2} (x - 3) = -1

Step 3 — run the checklist. The limit exists ✅; $f(2)$ is undefined ❌; ingredient 3 has nothing to compare. Exactly one ingredient fails, and it is the repairable one.

Step 4 — repair. Setting $f(2) := -1$ fills the hole, and the repaired function is the line $y = x - 3$ .

Problem 4 · Two Holes at Once

Given: $F(x) = \dfrac{\sin x}{x} + \dfrac{1 - \cos x}{x}$ for $x \ne 0$ , with $x$ in radians — find the status of $F$ at $x = 0$ .

✅ Correct!

\dfrac{\sin x}{x} \to 1

and

\dfrac{1 - \cos x}{x} \to 0

, so

F(x) \to 1 + 0 = 1

from both sides while

F(0)

is undefined — exactly one missing number.

❌ Two summands, one limit. Each fraction has the same limit from the left as from the right, so

F

has a genuine two-sided limit; a jump needs the one-sided limits of

F

itself to differ.

❌ Only the second limit was used.

\lim\limits_{x \to 0} \dfrac{1 - \cos x}{x} = 0

, but

\lim\limits_{x \to 0} \dfrac{\sin x}{x} = 1

, and the limit of a sum is the sum of the limits.

❌ Undefined is exactly the repairable case. A removable discontinuity is a point where the limit exists and the value is missing;

\tfrac{0}{0}

says the limit must be found some other way, not that it fails to exist.

❌ Not quite. Take the limit of each fraction separately, add the two results, and compare with

F(0)

Show solution

At $x = 0$ both fractions read $\tfrac{0}{0}$ , so $F(0)$ is undefined — ingredient 2 fails.

The limit, term by term:

\lim_{x \to 0} \frac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0

\lim_{x \to 0} F(x) = 1 + 0 = 1

Both one-sided limits equal $1$ , so ingredient 1 holds. A limit that exists with a value that is missing is precisely a removable discontinuity, and the repair is the common limit:

F(0) := 1

Sanity check: near $0$ the first fraction is a hair under $1$ and the second is a small number carrying the sign of $x$ , so $F$ hugs the height $1$ from both sides.

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