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

If limn->∞ an = a, an -> a <=> an - a -> 0 <=> |an - a| -> 0

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
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.

If -1 < q < 1 then limn->∞qn = 0.
limn->∞(1 - 1/n)n = limn->∞(1 + 1/n)n+1 = e
(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->∞(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 = ?

limn->0an = limn->05 . limn->04n = 5 . 40 = 5.1 = 5

Exercise 2:

If an =
 3n2 + 1 2n - n2
then limn->∞an = ?

limn->∞
 3n2 + 1 2n - n2
= limn->∞
 n2 n2
.
 3 + 1/n2 2/n - 1
= limn->∞
 3 + 0 0 - 1
= -3

Exercise 3:

If liman->1 =
 2an2 - an - 1 an - 1
= ?

liman->1 =
 2an2 - an - 1 an - 1
= liman->∞
 (an - 1)(2an + 1) an - 1
=
= liman->1(2an + 1) = 3

More about lim in the maths forum

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