- #1
Korisnik
- 62
- 1
I kind of know what limits are, or at least believe I do: I think that a limit of a sequence is just an approximation/intuitive way to finding a number (if it exists) to which a sequence tends. For example, 1, 2, 3, 4... tends to +∞, while 1/10, 1/100, 1/1000... tends, "obviously/intuitively", to 0.
Now formal definition of a limit of a sequence in my book is:
A sequence of real numbers [itex]a_n[/itex] converges (tends) to a real number [itex]L[/itex] if for all [itex]\varepsilon>0[/itex] exists [itex]n_0\in\mathbb{N}[/itex] such that for all [itex]n\geq n_0[/itex] it's true [itex]|a_n-L|<\varepsilon[/itex]. Then [itex]L[/itex] is limit of a sequence [itex]a_n[/itex],
[tex]L=\lim_{n\to \infty} a_n.[/tex]
Or symbolically: [tex]\lim_{n\to\infty}a_n=L\Longleftrightarrow(\forall\varepsilon>0)(\exists n_0\in\mathbb{N})(\forall n\geq n_0)(|a_n-L|<\varepsilon).[/tex]
I understand the first part of the definition: for every [itex]\varepsilon>0[/itex] there is [itex]n_0\in\mathbb{N}[/itex] such that for every [itex]n\geq n_0[/itex] it's true...
This means that we first pick a epsilon bigger than 0, then there must be (if the limit exists) some [itex]n_0[/itex] from which (to infinity) all members will be of a greater ordinal number -- and they ([itex]\varepsilon[/itex] we picked and [itex]n_0[/itex] that corresponds) comply with the condition [itex]|a_n-L|<\varepsilon.[/itex]
Now this condition, the inequality with the absolute value represents a interval:
[tex]-\varepsilon<a_n-L<\varepsilon\Longleftrightarrow a_n\in(L-\varepsilon, L+\varepsilon).[/tex]
This, I think, means that [itex]a_n[/itex] (as [itex]n\to\infty[/itex]) is near the limit [itex]L[/itex] value, but in the range of left and right boundaries of [itex]\varepsilon[/itex].
Represented (poorly) by a number line:
-∞ · · · ------(-ε) a_n-L (ε)------ · · · +∞
From the equations I now think that as [itex]n[/itex] approaches infinity, the [itex]a_n[/itex] approaches limit (obviously), and that makes sense -- and it agrees with the equation (because if [itex]a_n[/itex] reaches [itex]L[/itex], they will subtract to a 0, and since epsilon basically tends to a 0, that's the definition of a zero [itex](-0.0000...1 < 0 < 0.0000...1)[/itex]).
Finally, limit of a sequence is just an approximation of the "infinitieth" member of a sequence.
Tell me if I have understood anything wrongly. Thanks in advance.
Now formal definition of a limit of a sequence in my book is:
A sequence of real numbers [itex]a_n[/itex] converges (tends) to a real number [itex]L[/itex] if for all [itex]\varepsilon>0[/itex] exists [itex]n_0\in\mathbb{N}[/itex] such that for all [itex]n\geq n_0[/itex] it's true [itex]|a_n-L|<\varepsilon[/itex]. Then [itex]L[/itex] is limit of a sequence [itex]a_n[/itex],
[tex]L=\lim_{n\to \infty} a_n.[/tex]
Or symbolically: [tex]\lim_{n\to\infty}a_n=L\Longleftrightarrow(\forall\varepsilon>0)(\exists n_0\in\mathbb{N})(\forall n\geq n_0)(|a_n-L|<\varepsilon).[/tex]
I understand the first part of the definition: for every [itex]\varepsilon>0[/itex] there is [itex]n_0\in\mathbb{N}[/itex] such that for every [itex]n\geq n_0[/itex] it's true...
This means that we first pick a epsilon bigger than 0, then there must be (if the limit exists) some [itex]n_0[/itex] from which (to infinity) all members will be of a greater ordinal number -- and they ([itex]\varepsilon[/itex] we picked and [itex]n_0[/itex] that corresponds) comply with the condition [itex]|a_n-L|<\varepsilon.[/itex]
Now this condition, the inequality with the absolute value represents a interval:
[tex]-\varepsilon<a_n-L<\varepsilon\Longleftrightarrow a_n\in(L-\varepsilon, L+\varepsilon).[/tex]
This, I think, means that [itex]a_n[/itex] (as [itex]n\to\infty[/itex]) is near the limit [itex]L[/itex] value, but in the range of left and right boundaries of [itex]\varepsilon[/itex].
Represented (poorly) by a number line:
-∞ · · · ------(-ε) a_n-L (ε)------ · · · +∞
From the equations I now think that as [itex]n[/itex] approaches infinity, the [itex]a_n[/itex] approaches limit (obviously), and that makes sense -- and it agrees with the equation (because if [itex]a_n[/itex] reaches [itex]L[/itex], they will subtract to a 0, and since epsilon basically tends to a 0, that's the definition of a zero [itex](-0.0000...1 < 0 < 0.0000...1)[/itex]).
Finally, limit of a sequence is just an approximation of the "infinitieth" member of a sequence.
Tell me if I have understood anything wrongly. Thanks in advance.