SME
Grade 12 Math · Precalculus
Limits, continuity and an introduction to derivatives
Use this for “Find the limit of (x² − 9) ÷ (x − 3) as x → 3,” where the function has a hole.
Chart
f(2) = 5
As x → 3 from the left, f(x) → 6; from the right, f(x) → 6.
So the limit is 6, though f(3) has no value.
This is an example. Type your own values to start a new problem.
a
x approaches
example
b
The other zero
example
L
The limit
worked out (not typed)
x
Another x
example
y
f(x) there
calculated
Formulas
L
=
a
−
b
6 = 3 − (−3)
y
=
x
−
b
5 = 2 − (−3)
Assumptions
•
f(x) = (x − a)(x − b) ÷ (x − a): factor the top first, as x² − 9 = (x − 3)(x + 3).
•
The limit is the value f(x) approaches as x gets close to a, not the value f(a).
•
Here f has no value at x = a (a hole in the graph), but the limit exists.
Step-by-step
WE KNOW
a = 3, b = −3, x = 2
FIND
the limit (L), f(x) there (y)
Step 1 · Find the limit (L)
Use L = a − b
Cancel
x
−
a
:
f
(
x
)
=
x
−
b
away
from
a
,
and
x
−
b
heads
to
a
−
b
as
x
→
a
.
L
=
3
−
(−3)
x
=
2.9:
f
(
x
)
=
2.9
−
(−3)
=
5.9
x
=
2.99:
f
(
x
)
=
2.99
−
(−3)
=
5.99
x
=
3.01:
f
(
x
)
=
3.01
−
(−3)
=
6.01
L
=
6
Step 2 · Find f(x) there (y)
Use y = x − b
For
x
≠
a
,
(
x
−
a
)
x
−
b
x
−
a
is
just
x
−
b
.
y
=
2
−
(−3)
y
=
5
Check
Put the numbers back in. Both sides should match.
6 = 3 − (−3) ✓
5 = 2 − (−3) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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More problem types
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?”
From a secant slope to the derivative
Use this for “Find the slope of the secant of x² from 3 to 3.1, then the derivative at 3.”
Limits at infinity
Use this for “Find the limit of (2x + 1) ÷ (x − 3) as x → ∞.” Other powers of x are on “Limits at infinity by degree.”