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Limits of Sequences, Lim

We already know what are arithmetic and geometric progression - a sequences of values. Let us take the sequence a_n = 1/n, if k and m are natural numbers then for every k < m is true a_k > a_m, so as big as it gets n as smaller is becoming a_n and it's always positive, but it never reaches null. In this case we say that 0 is
the lim a_n->∞ if n->∞, or the other way to write it down lim_n->∞ a_n = 0.

Limit Definition

The number a is called limit of a sequence, if for every ε > 0 it can be found a number n_ε, so that for all the members of the sequence a_n with index n > n_ε is true that a - ε < a_n < 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 a_n and b_n both have real limits then, the sequences
a_n + b_n, a_n - b_n, a_n.b_n and a_n / b_n also have real limit and:

If a_n < b_n for every natural n and lim_n->∞a_n = a,
lim_n->∞b_n = b then a ≤ b

If a_n ≤ b_n ≤ c_n or every real n and if lim_n->∞a_n = lim_n->∞c_n = A
then lim_n->∞b_n = A.

If a_n ≥ 0 and lim_n->∞a_n = a, then the sequence b_n = √a_n also has a limit and lim_n->∞√a_n = √a_n.

If a_n = ¹/_nk and k ≥ 1 then lim_n->∞a_n = 0.

e is the number of Neper.

If sequence a_n has a unreal limit( -∞ or +∞ ) then the sequence 1/a_n has a limit and lim_n->∞¹/_an = 0

If sequences a_n and b_n have unreal limits and lim_n->∞a_n=+∞, lim_n->∞b_n=+∞ then:

Lim problems

Exercise 1:
If aⁿ = 5.4ⁿ, lim_n->0a_n = ?

Answer:
lim_n->0a_n = lim_n->05 . lim_n->04ⁿ = 5 . 4⁰ = 5.1 = 5

Exercise 2:

If an =

3n2 + 1 2n - n2

then limn->∞an = ?

Answer:

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

= ?

Answer:

liman->1 =

2an2 - an - 1 an - 1

=

liman->∞

(an - 1)(2an + 1) an - 1

=

= lim_an->1(2a_n + 1) = 3

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