Limits of Functions

Definition

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Let $A\subset X$ and

$$f:A\to Y$$

Let $p\in X$ be a limit point of $A$ ($p$ need not be in $A$). Let $q\in Y$. We say

$$\underset{x\to p}{\lim }f(x)=q\text{ or }f(x)\stackrel{x\to p}{\to }q$$

if $\forall \epsilon >0,\exists \delta >0$ such that $0<d_X(x,p)<\delta$ for all $x\in A$ with $d_Y(f(x),q)<\epsilon$

As we get closer to $p$ by $x$, we get closer to $q$ by $f(x)$.

Example 1

Let $f:A=(0,1)\subset \mathbb{R}\to \mathbb{R}=Y$ where $x\mapsto x$. Then $0$ is a limit point of $A$ but not in $A$. However, we can claim ${\lim }_{x\to 0}f(x)=0$.

Proof: Let $\epsilon >0$. Choose $\delta =\epsilon$. If $x\in A=(0,1)$ and $d(x,0)=|x|<\delta$ then

$$d(f(x),0)=|f(x)|=|x|<\delta =\epsilon$$

and so we are done.

Example 2

$$f:(0,1)\to \mathbb{R}f(x)=1/x$$

then ${\lim }_{x\to 0}f(x)$ does not exist.

Proof: Suppose by contradiction the limit is $M\in \mathbb{R}$. Let $\epsilon =1/2$. Then $\exists \delta >0$ such that

$$\forall x\in A,0<d(x,0)<\delta ⟹d(f(x),M)<\epsilon =1/2$$

By Archimedean Property, $\exists n\in \mathbb{N}$ such that $1/\delta <n⟹1/n<\delta$. By picking $n$ larger, we can get $M+1<n$. Then $x=1/n$, note $1/n\ne 0$ and $1/n\in N_{\delta }(0)$, so

$$d(f(1/n),M)=|n-m|=n-M<1/2$$

which is a contradiction.

Lemma (Sequential Characterization of Limits)

Let $A\subset X,Y$ be metric spaces. Let $p\in A$ be a limit point, $f:A\to Y$. Then

$$\underset{x\to p}{\lim }f(x)=q⟺(\forall (a_n)\subset A)(a_n\ne p\wedge a\to p),f(a_n)\to q$$

Proof: $(⟹)$ Suppose ${\lim }_{x\to p}f(x)=q$. Let $\epsilon >0$. Then $\exists \delta >0$ such that $\forall x\in A,0<d(x,p)<\delta$. We have $d(f(x),q)<\epsilon$.

We’ve chosen some $\epsilon$, and rewrote the definition of function limits. The goal is that if $(a_n)\subset A$ and $a_n\ne p$, $a_n\to p$, then $f(a_n)\to q$. (Unfold definitions.)

Since $a_n\to p,\exists N\in \mathbb{N}$ such that $\forall n\ge N$, $d(a_n,p)<\delta$ and $0<d(a_n,p)$ since $a_n\ne p$. Hence, by definition of ${\lim }_{x\to p}f(x)=q$, we have that $d(f(a_n),q)<\epsilon$. Then $\forall n\ge N,f(a_n)\to q$.

$(⟸)$ Suppose $\forall (a_n)\subset A,a_n\ne p,a_n\to p$ and that $f(a_n)\to q$.

The goal is to show ${\lim }_{x\to p}f(x)=q$. We show by contradiction since the goal is static.

Suppose ${\lim }_{x\to p}f(x)\ne q$. Then

$$(\exists {\epsilon }_0>0)(\forall \delta >0)(\exists x\in A,0<d(x,p)<\delta \wedge d(f(x),q)\ge {\epsilon }_0)$$

Apply for $n\in \mathbb{N}$, this negation to $\delta =1/n$. Then $\forall n\in \mathbb{N},\exists a_n\in A,0<d(a_n,p)<1/n$ but

$$d(f(a_n),q)\ge {\epsilon }_0$$

Now we have $(a_n)\subset A$, since $d(a_n,p)>0$ then $a_n\ne p$ and $a_n\to p$. So by assumption, $d(f(a_n),q)<{\epsilon }_0$. This is a contradiction.

Corollary (Uniqueness of Limits)

If $f$ has a limit at $p$, then the limit is unique.

Proof: We apply the previous lemma and the uniqueness of of limits of Sequences.

Corollary (Algebra of Limits)

Let $A\subset X,f:A\to \mathbb{C},g:A\to \mathbb{C}$. Suppose

1. ${\lim }_{x\to p}f(x)={\ell }_1$
2. ${\lim }_{x\to p}g(x)={\ell }_2$ Then
3. ${\lim }_{x\to p}(f\pm g)(x)={\ell }_1\pm {\ell }_2$
4. ${\lim }_{x\to p}(f\cdot g)(x)={\ell }_1\cdot {\ell }_2$
5. If ${\ell }_2\ne 0$, ${\lim }_{x\to p}(f/g)(x)={\ell }_1/{\ell }_2$

Proof: This follows from Lemma (Sequential Characterization of Limits) and Lemma (Algebra of Sequences).

Lemma (Continuity = Limits)

If $p\in A$ and $p$ is a limit point of $A$, then

$$f\text{ is continuous at }p⟺\underset{x\to p}{\lim }f(x)=f(p)$$

Proof: $(⟹)$ For every $\epsilon >0,\exists \delta >0$ such that $\forall x\in A$, if $d(x,p)<\delta$, then $d(f(x),f(p))<\epsilon$. But then this is the definition of limits as the limit $L=f(p)$.

$(⟸)$ Let $\epsilon >0$. Choose $\delta$ as in the definition of limit. Suppose now that $x\in A,d(x,p)<\delta$. We have 2 cases:

1. $x=p$. Then $d(x,p)<\delta$. Moreover, $d(f(x),f([]))=d(f(p),f(p))=0<\delta$
2. $x\ne p$. Then $d(x,p)>0$. By definition of limits, $d(f(x),f(p))<\epsilon$.

Corollary (Sequential Characterization of Continuity)

Let $f:A\to Y$, $p\in A$. Then $f$ is continuous at $p$ $⟺\forall (a_n)\subset A,a_n\to p$ we have $f(a_n)\to f(p)$.

Corollary (Algebra of Continuous Functions)

Let $f,g:X\to \mathbb{C}$ where $X,Y$ are metric spaces. Then if $f,g$ are continuous at $p\in X$, then

1. $f+g$ is continuous at $p$.
2. $f\cdot g$ is continuous at $p$.
3. $f/g$ is continuous at $p$ if $g(p)\ne 0$