Difference between revisions of "Limit (sequence)"

Here we have two definitions

1. [math]\lim_{n\rightarrow\infty}(x_n)=x\iff\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}\left[n> N\implies d(x_n,x)<\epsilon\right][/math] (given at the top of the page)
2. [math]\lim_{n\rightarrow\infty}(x_n)=x\iff\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}\left[n\ge N\implies d(x_n,x)<\epsilon\right][/math]

Proof: [ilmath]1\implies 2[/ilmath]

Let [ilmath]\epsilon >0[/ilmath] be given.

We know [ilmath]\exists N'\in\mathbb{N} [/ilmath] such that [ilmath]n>N'\implies d(x_n,x)<\epsilon[/ilmath] by assuming [ilmath]1[/ilmath] is true

Choose [ilmath]N=N'+1[/ilmath]

Now [ilmath]n\ge N\implies[n>N\vee n=N][/ilmath] by definition of [ilmath]\ge[/ilmath], substituting [ilmath]N=N'+1[/ilmath] we get [ilmath]n\ge N'+1\implies[n>N'+1\vee n=N'+1][/ilmath]

• (Case: [ilmath]n>N'+1[/ilmath]) Note that [ilmath]n>N'+1>N'[/ilmath] so by transitivity of [ilmath]>[/ilmath] we see [ilmath]n>N'[/ilmath]

  We know from [ilmath]1[/ilmath] that [ilmath]n>N'\implies d(x_n,x)<\epsilon[/ilmath]

• (Case: [ilmath]n=N'+1[/ilmath]), trivially [ilmath]N'+1>N'[/ilmath] so we have so [ilmath]n>N'[/ilmath]

  We know from [ilmath]1[/ilmath] that [ilmath]n>N'\implies d(x_n,x)<\epsilon[/ilmath]

So in either case we have [ilmath]d(x_n,x)<\epsilon[/ilmath]

• We have shown that if [ilmath]n\ge N[/ilmath] we have [ilmath]d(x_n,x)\epsilon[/ilmath]

• Thus choosing [ilmath]N=N'+1[/ilmath] is "an [ilmath]N[/ilmath] that exists" for the given [ilmath]\epsilon[/ilmath]

This completes the first part of the proof

Proof: [ilmath]2\implies 1[/ilmath]

Let [ilmath]\epsilon>0[/ilmath] be given.

We know [ilmath]\exists N'\in\mathbb{N} [/ilmath] such that [ilmath]n\ge N'\implies d(x_n,x)<\epsilon[/ilmath] by assuming [ilmath]2[/ilmath] is true

Choose [ilmath]N=N'-1[/ilmath]

Now [ilmath]n> N\implies n>N'-1[/ilmath]

• (Case: [ilmath]n=N'[/ilmath]) if this is the case we know that [ilmath]N'>N'-1[/ilmath] so [ilmath]n>N[/ilmath] is satisfied, but also so is [ilmath]n\ge N'[/ilmath] (we have equality)

  We know from [ilmath]2[/ilmath] that this [ilmath]\implies d(x_n,x)<\epsilon[/ilmath]

• (Case: [ilmath]n>N'[/ilmath]) well [ilmath]n\ge N'[/ilmath] means "if [ilmath]n>N'[/ilmath] or [ilmath]n=N'[/ilmath]" so [ilmath]n>N'\implies n\ge N'[/ilmath], thus [ilmath]n\ge N'[/ilmath]

  We know from [ilmath]2[/ilmath] that this [ilmath]\implies d(x_n,x)<\epsilon[/ilmath]

Thus for [ilmath]n>N[/ilmath] we see that [ilmath]d(x_n,x)<\epsilon[/ilmath]

This completes the proof

(End of proof)

Every convergent sequence is Cauchy/Proof

The idea is that defining "tends towards [ilmath]x[/ilmath]" is rather difficult, to sidestep this we just say "we can get as close as we like to" instead. This is the purpose of [ilmath]\epsilon[/ilmath].

We say that "if you give me an [ilmath]\epsilon>0[/ilmath] - as small as you like - I can find you a point of the sequence ([ilmath]N[/ilmath]) where all points after are within [ilmath]\epsilon[/ilmath] of [ilmath]x[/ilmath] (where [ilmath]d(\cdot,\cdot)[/ilmath] is our notion of distance)

• That is after [ilmath]N[/ilmath] in the sequence, so that's [ilmath]x_{n+1},x_{n+1},\ldots[/ilmath] the distance between [ilmath]x_{N+i} [/ilmath] and [ilmath]x[/ilmath] is [ilmath]<\epsilon[/ilmath]

  This is exactly what [ilmath]n>N\implies d(x_n,x)<\epsilon[/ilmath] says, it says that:

  • whenever [ilmath]n>N[/ilmath] we must have [ilmath]d(x_n,x)<\epsilon[/ilmath]

As per the nature of implies we may have [ilmath]d(x_n,x)<\epsilon[/ilmath] without [ilmath]n>N[/ilmath], it is only important that WHENEVER we are beyond [ilmath]N[/ilmath] in the sequence that [ilmath]d(x_n,x)<\epsilon[/ilmath]

Example

Here: [ilmath]x[/ilmath]-axis scale is from [ilmath]0[/ilmath] to [ilmath]12.6[/ilmath], marks are shown every unit. [ilmath]y[/ilmath]-axis scale starts from [ilmath]0[/ilmath] and is marked every [ilmath]0.25[/ilmath] units. The sequence is any sequence of points on the wavy function shown. The limit of this is clearly [ilmath]1[/ilmath] The two horizontal lines show [ilmath]1-\epsilon[/ilmath] and [ilmath]1+\epsilon[/ilmath] The vertical line shows one possible value where every point after it is within [ilmath]\epsilon[/ilmath] of [ilmath]1[/ilmath] due to technical limitations the function [ilmath]f(x)=1+\frac{\sin(\pi x)}{\frac{1}{4}x^2}[/ilmath] is shown The curves are bounds on the function.

Notice that at [ilmath]x=1[/ilmath] that , in fact the curve is within [ilmath]\pm\epsilon[/ilmath] several times before we reach the vertical line, this is the significance of the implies sign, when we write [ilmath]A\implies B[/ilmath] we require that whenever [ilmath]A[/ilmath] is true, [ilmath]B[/ilmath] must be true, but [ilmath]B[/ilmath] may be true regardless of what [ilmath]A[/ilmath] is.

Note that after the vertical line the function is always within the bounds.

Because of this any [ilmath]N'>N[/ilmath] may be used too, as if [ilmath]n>N'[/ilmath] and [ilmath]N'>N[/ilmath] then [ilmath]n>N'>N[/ilmath] so [ilmath]n>N[/ilmath] - this proves that if [ilmath]N[/ilmath] works then any larger [ilmath]N'[/ilmath] will too. There is no requirement to find the smallest [ilmath]N[/ilmath] that'll work, just an [ilmath]N[/ilmath] such that [ilmath]n>N\implies d(x_n,x)<\epsilon[/ilmath]