SME
Grade 12 Math · Problem type
Limits at infinity by degree
Use this for “Find the limit of (3n + 1) ÷ (2n − 1) as n → ∞” or “of (2n² + 1) ÷ (n² − 3), or say there is none.”
Table
Term number (n)
Term n (aₙ)
This is an example. Type your own values to start a new problem.
p
Leading number on top
example
j
Power of n on top
example
q
Constant on top
example
r
Leading number in the bottom
example
k
Power of n in the bottom
example
s
Constant in the bottom
example
n
Term number
example
aₙ
Term n
calculated
L
Limit
worked out (not typed)
Formulas
aₙ
=
p
×
n
j
+
q
r
×
n
k
+
s
2.0722
=
2 ×
10
2
+ 1
1 ×
10
2
+ (−3)
L
=
p
r
if
j
=
k
,
0
if
j
<
k
2
=
2
1
if
2
=
2,
0
if
2
<
2
Assumptions
•
Divide the top and bottom by the bottom’s highest power of n; every term with n left under it goes to 0.
•
Same power on top and bottom: the limit is the ratio of the leading numbers. A higher power in the bottom: 0.
•
A higher power on top: the terms grow without bound, so there is no limit.
Step-by-step
WE KNOW
p = 2, j = 2, q = 1, r = 1, k = 2, s = −3, n = 10
FIND
term n (aₙ), limit (L)
Step 1 · Find term n (aₙ)
Use aₙ = (p × nʲ + q) ÷ (r × nᵏ + s)
Put
n
into
the
rule.
aₙ
=
2 ×
10
2
+ 1
1 ×
10
2
− 3
aₙ
=
2 × 100 + 1
100 − 3
aₙ
=
200 + 1
97
aₙ
=
201
97
aₙ = 2.0722
Step 2 · Find limit (L)
Use L = p ÷ r if j = k, 0 if j < k
Compare the powers: equal gives the leading numbers’ ratio, a higher power in the bottom gives 0.
L
=
p
r
L
=
2
1
L
=
2
→
equal
powers:
divide
the
top
and
bottom
by
n
2
,
and
only
2
1
is
left
Check
Put the numbers back in. Both sides should match.
2.0722
≈
2 ×
10
2
+ 1
1 ×
10
2
− 3
✓
2
=
2
1
✓
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