Limit (sequence)

Note: see Limit page for other kinds of limits

Definition

Given a sequence $(x_n)_{n=1}^\infty\subseteq X$, a metric space $(X,d)$ (that is complete) and a point $x\in X$, the sequence $(x_n)$ is said to:

• have limit $x$ or converge to $x$

When:

• $$\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies d(x,x_n)<\epsilon]$$

  (note that $\epsilon\in\mathbb{R}$, obviously - as the co-domain of $d$ is $\mathbb{R}$)

• Read this as:

  for all $\epsilon$ greater than zero, there exists an $N$ in the natural numbers such that for all $n$ that are also natural we have that:

  whenever $n$ is beyond $N$ that $x_n$ is within $\epsilon$ of $x$

Equivalent definitions

Note: where it is not obvious changes have a $\{$ underneath them

[Expand]

$$\lim_{n\rightarrow\infty}(x_n)=x\iff\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}\left[\underbrace{n\ge N}\implies d(x_n,x)<\epsilon\right]$$

---

Here we have two definitions

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

Proof: $1\implies 2$

Let $\epsilon >0$ be given.

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

Choose $N=N'+1$

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

• (Case: $n>N'+1$) Note that $n>N'+1>N'$ so by transitivity of $>$ we see $n>N'$

  We know from $1$ that $n>N'\implies d(x_n,x)<\epsilon$

• (Case: $n=N'+1$), trivially $N'+1>N'$ so we have so $n>N'$

  We know from $1$ that $n>N'\implies d(x_n,x)<\epsilon$

So in either case we have $d(x_n,x)<\epsilon$

• We have shown that if $n\ge N$ we have $d(x_n,x)\epsilon$

• Thus choosing $N=N'+1$ is "an $N$ that exists" for the given $\epsilon$

This completes the first part of the proof

Proof: $2\implies 1$

Let $\epsilon>0$ be given.

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

Choose $N=N'-1$

Now $n> N\implies n>N'-1$

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

  We know from $2$ that this $\implies d(x_n,x)<\epsilon$

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

  We know from $2$ that this $\implies d(x_n,x)<\epsilon$

Thus for $n>N$ we see that $d(x_n,x)<\epsilon$

This completes the proof

(End of proof)

Discussion

Requiring $x\in X$

If $x\notin X$ then $d(x_n,x)$ is undefined, as $d:X\times X\rightarrow\mathbb{R}_{\ge_0}$, that is the distance metric is only defined for things in $X$

Process

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 $\epsilon$.

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

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

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

  • whenever $n>N$ we must have $d(x_n,x)<\epsilon$

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

Example

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 $1-\epsilon$ and $1+\epsilon$ The vertical line shows one possible value where every point after it is within $\epsilon$ of $1$ due to technical limitations the function $f(x)=1+\frac{\sin(\pi x)}{\frac{1}{4}x^2}$ is shown The curves are bounds on the function.

Notice that at $x=1$ that , in fact the curve is within $\pm\epsilon$ several times before we reach the vertical line, this is the significance of the implies sign, when we write $A\implies B$ 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 $n>N\implies d(x_n,x)<\epsilon$

See also

• Cauchy sequence
• Limit