SME
Grade 12 Math · Problem type
One-sided limits and continuity
Use this for “f(x) = x + 1 for x < 2 and 2x − 3 for x ≥ 2. Is f continuous at 2?”
Chart
As x → 2 from the left, f(x) → 3; from the right, f(x) → 1.
The two sides differ, so the limit does not exist there.
Limit from the left: 3
Limit from the right, f(c): 1
Jump, L⁺ − L⁻: −2
This is an example. Type your own values to start a new problem.
m₁
Slope left of c
example
b₁
Intercept left of c
example
m₂
Slope from c on
example
b₂
Intercept from c on
example
c
Where the rule changes
example
L⁻
Limit from the left
calculated
L⁺
Limit from the right, f(c)
calculated
J
Jump, L⁺ − L⁻
worked out (not typed)
Formulas
L⁻
=
m₁
×
c
+
b₁
3 = 1 × 2 + 1
L⁺
=
m₂
×
c
+
b₂
1 = 2 × 2 + (−3)
J
=
L⁺
−
L⁻
−2 = 1 − 3
Assumptions
•
Left of c the rule is m₁x + b₁; from c on it is m₂x + b₂, so f(c) = m₂c + b₂.
•
The limit at c exists only when the left and right limits are equal (J = 0).
•
f is continuous at c when the limit exists and equals f(c).
Step-by-step
WE KNOW
m₁ = 1, b₁ = 1, m₂ = 2, b₂ = −3, c = 2
FIND
limit from the left (L⁻), limit from the right, f(c) (L⁺), jump, L⁺ − L⁻ (J)
Step 1 · Find limit from the left (L⁻)
Use L⁻ = m₁ × c + b₁
From
the
left
the
left
rule
applies:
put
in
x
=
c
.
L⁻ = 1 × 2 + 1
L⁻ = 3
Step 2 · Find limit from the right, f(c): L⁺
Use L⁺ = m₂ × c + b₂
From
the
right
(and
at
c
)
the
right
rule
applies:
put
in
x
=
c
.
L⁺ = 2 × 2 − 3
L⁺ = 4 − 3
L⁺ = 1
Step 3 · Find jump, L⁺ − L⁻ (J)
Use J = L⁺ − L⁻
How
far
the
graph
jumps
at
c
:
0
means
the
two
sides
meet.
J
=
1
−
3
J
=
−2
→
the
sides
differ:
no
limit
at
c
,
so
not
continuous
Check
Put the numbers back in. Both sides should match.
3 = 1 × 2 + 1 ✓
1 = 2 × 2 − 3 ✓
−2 = 1 − 3 ✓
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Limits, continuity and an introduction to derivatives
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Limits, continuity and an introduction to derivatives
Main lesson
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