Limits of Sequences, Lim
We already know what are arithmetic and geometric progression -
a sequences of values. Let us take the sequence an = 1/n, if k
and m are
natural numbers then for every k < m is true ak > am, so as
big as it gets n as smaller is becoming an
and it's always positive, but it never reaches null. In this case we say that 0 is
the lim an->∞
if n->∞, or the other
way to write it down limn->∞ an = 0.
Limit Definition
The number a is called limit of a sequence, if for every ε > 0 it can be found a number nε, so that for all the members of the sequence an with index n > nε is true that a - ε < an < a + ε.
Basic rule
A sequence not always has limit, and sometimes has unreal limit( -∞ or +∞ ). The limits +∞ and -∞ are called unreal limits.
If the sequences an and bn both have real limits then,
the sequences
an + bn,
an - bn, an.bn and an / bn also have real limit and:
limn -> ∞(an - bn) = limn -> ∞an - limn -> ∞bn
limn -> ∞(an . bn) = limn -> ∞an . limn -> ∞bn
limn -> ∞(an/ bn) = limn -> ∞an / limn -> ∞bn
if bn ≠ 0 and limn->∞bn ≠ 0
If an < bn for every natural n
and limn->∞an = a,
limn->∞bn = b
then a ≤ b
If an ≤ bn ≤ cn or every real
n and if limn->∞an = limn->∞cn = A
then limn->∞bn = A.
If an ≥ 0 and limn->∞an = a, then the sequence bn = √an also has a limit and limn->∞√an = √an.
If an = 1/nk and k ≥ 1 then limn->∞an = 0.
(1+1/n)n < e < (1 + 1/n)n-1
e is the number of Neper.
If sequence an has a unreal limit( -∞ or +∞ ) then the sequence 1/an has a limit and limn->∞1/an = 0
If sequences an and bn have unreal limits and limn->∞an=+∞, limn->∞bn=+∞ then:
limn->∞(an . bn) = +∞
limn->∞ank = +∞ if k > 0
limn->∞ank = 0; if k < 0
limn->∞-an = -∞
Lim problems
Exercise 1:
If an = 5.4n, limn->0an = ?
Answer:
limn->0an = limn->05 . limn->04n = 5 . 40 = 5.1 = 5
Exercise 2:
If an = |
|
then limn->∞an = ? |
Answer:
limn->∞ |
|
= limn->∞ |
|
. |
|
= limn->∞ |
|
= -3 |
Exercise 3:
If liman->1 = |
|
= ? |
Answer:
liman->1 = |
|
= | liman->∞ |
|
= |
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