text: Limit of a sequence (explanation)
  
  1. Aug 25, 2014
  
  Korisnik
  
  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 $a_n$ converges (tends) to a real number $L$ if for all $\varepsilon>0$ exists $n_0\in\mathbb{N}$ such that for all $n\geq n_0$ it's true $|a_n-L|<\varepsilon$. Then $L$ is limit of a sequence $a_n$,
  $$L=\lim_{n\to \infty} a_n.$$
  Or symbolically: $$\lim_{n\to\infty}a_n=L\Longleftrightarrow(\forall\varepsilon>0)(\exists n_0\in\mathbb{N})(\forall n\geq n_0)(|a_n-L|<\varepsilon).$$
  
  I understand the first part of the definition: for every $\varepsilon>0$ there is $n_0\in\mathbb{N}$ such that for every $n\geq n_0$ it's true...
  
  This means that we first pick a epsilon bigger than 0, then there must be (if the limit exists) some $n_0$ from which (to infinity) all members will be of a greater ordinal number -- and they ($\varepsilon$ we picked and $n_0$ that corresponds) comply with the condition $|a_n-L|<\varepsilon.$
  
  Now this condition, the inequality with the absolute value represents a interval:
  $$-\varepsilon<a_n-L<\varepsilon\Longleftrightarrow a_n\in(L-\varepsilon, L+\varepsilon).$$
  
  This, I think, means that $a_n$ (as $n\to\infty$) is near the limit $L$ value, but in the range of left and right boundaries of $\varepsilon$.
  
  Represented (poorly) by a number line:
  
  -∞ · · · ------(-ε) a_n-L (ε)------ · · · +∞
  
  From the equations I now think that as $n$ approaches infinity, the $a_n$ approaches limit (obviously), and that makes sense -- and it agrees with the equation (because if $a_n$ reaches $L$, they will subtract to a 0, and since epsilon basically tends to a 0, that's the definition of a zero $(-0.0000...1 < 0 < 0.0000...1)$).
  
  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.
  
  2. Aug 25, 2014
  
  xAxis
  
  Yes, seems to me you nailed it. Someone might make problem about your last sentence about "infinitieth member", but don't pay much attention to them :).
  There are few other corollaries involving "all but finitely many members of sequence being out of interval" which will help solidify your understanding.
  
  3. Aug 25, 2014
  
  jbriggs444
  
  Intuitively, this is a decent way to think about limits. However the term "approximation" is not apt. A limit is not an approximation. It is an exact number.
  
  The "infinitieth" member, if it were to exist and if the sequence were to be continuous at infinity would have to be exactly equal to the limit.
  
  4. Aug 25, 2014
  
  mal4mac
  
  .. unless it's your professor marking you down for it :)
  
  5. Aug 25, 2014
  
  Stephen Tashi
  
  It's tempting to introduce some idea of motion into the intuitive idea of a limit since the formal definition uses the word "approaches". But in the formal definition, there isn't any motion or dynamic process. Epsilon isn't going anywhere. It isn't "tending" to anything. The freedom of choice guarnteed by the phrase "for each epsilon greater than 0" mean's you could pick a sequence of epsilons that "tended" to zero, but doing this is not specified in the definition.
  
  Understanding the formal definition of limit involves comprehending how it avoids using concepts of dynamic motion ("tending", "approaching") by clever use of the logical quantifiers "for each" and "there exists".
  
  6. Aug 25, 2014
  
  economicsnerd
  
  OP, everything you put looks basically right to me.
  
  A bit of terminology that might be helpful for phrasing this stuff with less notation: the word "eventually". Given a property "blahblahblah", say a sequence $(x_n)_{n=1}^\infty$ is eventually blahblahblah if there exists a number $n_0\in\mathbb N$ such that for every $n\in\mathbb N$ with $n\geq n_0$, the number $x_n$ is blahblahblah.
  
  For example, the sequence with $x_n = n - 72304551$ is eventually positive, as we can see by setting $n_0 = 72304552$. The sequence of integers with $x_n= \begin{cases} 12 & \text{ if n is a power of } 10, \\ 7 & \text{ otherwise} \end{cases}$ isn't eventually odd, because there are arbitrarily late members of the sequence that fail to be odd.
  
  Now limits become even easier to describe:
  - $(x_n)_n$ converges to $x\in\mathbb R$ if for every $\epsilon>0,$ the sequence is eventually in $(x-\epsilon, x+\epsilon)$.
  - $(x_n)_n$ converges to $\infty$ if for every $K\in\mathbb R$ the sequence is eventually bigger than $K$.
  
  7. Aug 25, 2014
  
  Korisnik
  
  Yes, I expressed myself badly (english isn't my mother tongue), what I meant was -- a focus of a better and better approximation?
  
  economicsnerd, thank You for the tips.
  
  Thank You very much for this post, you have a really good point!
  
  When I wrote the sentence you quoted, I myself didn't agree with it. Somehow I realized I could've written that at the beginning and then everything would be easier to explain; but I didn't, because I wasn't sure in it, it wasn't "my original idea"... Though it made more sense, it was just so simple... like an oversimplification of something and abstraction to a broader concept so it basically lost it's meaning (my meaning, my thinking of the problem). Your post made me at the beginning again. Thank You for that.
  
  I think I'll have to sleep on the idea you mentioned and to think through it. But I'll write down my first thoughts of the definition for now: first it defines $L$ as a number, so one number, not an interval. This will be very important... Now they could've just made this condition $|a_n-L|=0$ but then that could be any number, what exactly are we looking for? We are looking for one number. By clever use of: for all $\varepsilon$ exist all $n$ that satisfy the condition -- mentally plugging in all possible numbers for $\varepsilon$ and $a_n$ -- it's obvious that all choices of $\varepsilon$ result in a range of $a_n$ members (there is always a $n_0$, such that $n\geq n_0$), except for the smallest $\varepsilon$ possible (a number "next to" $0$), for which exists only one member of $a_n$: the last one. This last member is just $n_0$, one number, because if there were a next number, it would be smaller than $0$ and would add to $L$, that would be bigger than $0$ which is impossible by the condition. Now that we've eliminated all other members of the sequence, we are left with a number that is in this range $-0.0...1 < a_n-L < 0.0...1$. And what number is between these infinitely small numbers? It's $0$. This is where the definition ends. Because now we can say that $a_{n=last}=L$, because if $a-b=0$ then $a=b$.
  
  Last edited: Aug 26, 2014
metadata: {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}