Limit (sequence)/Discussion of definition

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

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

  • That is after N in the sequence, so that's xn+1,xn+1, the distance between xN+i and x is <ϵ
    This is exactly what n>Nd(xn,x)<ϵ says, it says that:
    • whenever n>N we must have d(xn,x)<ϵ

As per the nature of implies we may have d(xn,x)<ϵ without n>N, it is only important that WHENEVER we are beyond N in the sequence that d(xn,x)<ϵ

Example
Sequencelimit.gif Here:
  • x-axis scale is from 0 to 12.6, marks are shown every unit.
  • y-axis scale starts from 0 and is marked every 0.25 units.
  • The sequence is any sequence of points on the wavy function shown.
    • The limit of this is clearly 1
  • The two horizontal lines show 1ϵ and 1+ϵ
  • The vertical line shows one possible value where every point after it is within ϵ of 1
  • due to technical limitations the function f(x)=1+sin(πx)14x2 is shown
  • The curves are bounds on the function.

Notice that at x=1 that , in fact the curve is within ±ϵ several times before we reach the vertical line, this is the significance of the implies sign, when we write AB we require that whenever A is true, B must be true, but B may be true regardless of what A is.

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

Because of this any N>N may be used too, as if n>N and N>N then n>N>N so n>N - this proves that if N works then any larger N will too. There is no requirement to find the smallest N that'll work, just an N such that n>Nd(xn,x)<ϵ