SME
Grade 12 Math · Problem type
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.”
Chart
This is an example. Type your own values to start a new problem.
p
Number before x on top
example
q
Constant on top
example
r
Number before x below
example
s
Constant below
example
v
Vertical asymptote, −s ÷ r
worked out (not typed)
L
Limit as x → ∞
worked out (not typed)
x
A large x
example
y
f(x) there
calculated
Formulas
v
=
−
s
r
3
=
−
−3
1
L
=
p
r
2
=
2
1
y
=
p
×
x
+
q
r
×
x
+
s
2.007
=
2 × 1,000 + 1
1 × 1,000 + (−3)
Assumptions
•
For (px + q) ÷ (rx + s), far out the x terms swamp the constants: f(x) approaches p ÷ r.
•
So y = p ÷ r is a horizontal asymptote; x = −s ÷ r is a vertical one.
•
Try x = 1000: f(1000) is already close to the limit.
Step-by-step
WE KNOW
p = 2, q = 1, r = 1, s = −3, x = 1,000
FIND
vertical asymptote, −s ÷ r (v), limit as x → ∞ (L), f(x) there (y)
Step 1 · Find vertical asymptote, −s ÷ r (v)
Use v = −s ÷ r
The
bottom
is
0
there:
rx
+
s
=
0.
v
=
−
−3
1
v
=
3
Step 2 · Find limit as x → ∞ (L)
Use L = p ÷ r
Divide
top
and
bottom
by
x
:
q
x
and
s
x
go
to
0,
leaving
p
r
.
L
=
2
1
L
=
2
Step 3 · Find f(x) there (y)
Use y = (p × x + q) ÷ (r × x + s)
Put
x
into
the
function.
y
=
2 × 1,000 + 1
1 × 1,000 − 3
y
=
2,000 + 1
997
y
=
2,001
997
y
=
2.007
Check
Put the numbers back in. Both sides should match.
3
=
−
−3
1
✓
2
=
2
1
✓
2.007
≈
2 × 1,000 + 1
1 × 1,000 − 3
✓
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