More on the limit of functions

 There are a few variations of limits.

Definition (Divergence of a function)

If the function $f(x)$ does not converge to any value as $x\to a$, we say that $f(x)$ diverges as $x\to a$. In particular, if the value of $f(x)$ increases arbitrarily as $x\to a$, we say that $f(x)$ diverges to positive infinity and write

$$\underset{x\to a}{lim}f(x)=\mathrm{\infty }.$$

If the value of $f(x)$ is negative and its absolute value increases arbitrarily, we say that $f(x)$ diverges to negative infinity and write

$$\underset{x\to a}{lim}f(x)=-\mathrm{\infty }.$$

Example. Consider the function

$$f(x)=\frac{1}{(x-1{)}^2}$$

defined on $\mathbb{R}\setminus \{1\}$. We have

$$\underset{x\to 1}{lim}f(x)=\mathrm{\infty }.$$

You should verify this by drawing a graph. □

There are a few variants of the notion of limit.

Definition (Left and right limits)

1. We define the left limit of the function $f(x)$ at $x=a$ $$\underset{x\to a-0}{lim}f(x)=\alpha$$ if the following is satisfied
   • For any $\epsilon >0$, there exists $\delta >0$ such that for any $x\in \text{dom}(f)$, if $0<a-x<\delta$ then $|f(x)-\alpha |<\epsilon$
   This mean $x$ approaches $a$ from the left (so ``$x<a$'' is kept).
2. We define the right limit of the function $f(x)$ at $x=a$ $$\underset{x\to a+0}{lim}f(x)=\alpha$$ if the following is satisfied
   • For any $\epsilon >0$, there exists $\delta >0$ such that for any $x\in \text{dom}(f)$, if $0<x-a<\delta$ then $|f(x)-\alpha |<\epsilon$.
   This mean $x$ approaches $a$ from the right (so ``$a<x$'' is kept).

Remark. We write $x\to +0$ instead of $x\to 0+0$. Similarly $x\to -0$ rather than $x\to 0-0$. □

Example. Consider the function 

$$f(x)=\frac{x(x+1)}{|x|}$$

defined on $\mathbb{R}\setminus \{0\}$. 

If $x>0$, we have

$$f(x)=\frac{x(x+1)}{x}=x+1$$

so that

$$\underset{x\to +0}{lim}f(x)=1.$$

If $x<0$, we have

$$f(x)=\frac{x(x+1)}{-x}=-x-1$$

so that

$$\underset{x\to -0}{lim}f(x)=-1.$$

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Example. For the function $f(x)=\frac{1}{x}$ defined on $\mathbb{R}\setminus \{0\}$, we have

$$\begin{array}{rcl}\underset{x\to +0}{lim}f(x)& =& \mathrm{\infty },\\ \underset{x\to -0}{lim}f(x)& =& -\mathrm{\infty }.\end{array}$$

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Definition (Limits at $\pm \mathrm{\infty }$)

1. We define the limit at $x\to \mathrm{\infty }$ $$\underset{x\to \mathrm{\infty }}{lim}f(x)=\alpha$$ if the following is satisfied
• For any $\epsilon >0$, there exists $\delta >0$ such that for any $x\in \text{dom}(f)$, if $x>\delta$ then $|f(x)-\alpha |<\epsilon$.
2. We define the limit at $x\to -\mathrm{\infty }$ $$\underset{x\to -\mathrm{\infty }}{lim}f(x)=\alpha$$ if the following is satisfied
• For any $\epsilon >0$, there exists $\delta >0$ such that for any $x\in \text{dom}(f)$, if $x<-\delta$ then $|f(x)-\alpha |<\epsilon$.

Theorem

Let $f(x)$ be a function defined on the open interval $I=(a,b)$ $(a<b)$ such that $\underset{x\to a+0}{lim}f(x)=\alpha$ $(\alpha \in \mathbb{R})$.

1. For a given $c\in \mathbb{R}$, if $f(x)\ge c$ for all $x\in I$, then $\alpha \ge c$.
2. For a given $d\in \mathbb{R}$, if $f(x)\le d$ for all $x\in I$, then $\alpha \le d$.

Proof. We prove only part 1 and prove it by contradiction. Part 2 is similar.

Suppose $\alpha <c$ and let $\epsilon =c-\alpha >0$. Since $\underset{x\to a+0}{lim}f(x)=\alpha$, we can find $\delta >0$ such that $0<x-a<\delta$ implies $|f(x)-\alpha |<\epsilon$. But then $f(x)<\alpha +\epsilon =c$, which is a contradiction. ■

Theorem

Let $f(x)$ be a function defined on an interval that includes $x=a$.

1. If $\underset{x\to a}{lim}f(x)=\alpha$ exists, then the corresponding left and right limits also exist and $\underset{x\to a+0}{lim}f(x)=\underset{x\to a-0}{lim}f(x)=\alpha$.
2. If both $\underset{x\to a+0}{lim}f(x)$ and $\underset{x\to a-0}{lim}f(x)$ exist and $\underset{x\to a+0}{lim}f(x)=\underset{x\to a-0}{lim}f(x)=\alpha$, then $\underset{x\to a}{lim}f(x)=\alpha$.

Proof. Exercise. ■

Example. Let us prove the following:

$$e=\underset{x\to 0}{lim}(1+x{)}^{\frac{1}{x}}=\underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=\underset{x\to -\mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x.$$

Before proving this, recall that we have

$$\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{1}{n})}^n=e$$

where $n=1,2,3,\cdots$ (i.e., natural numbers).

See also: $e=2.178\cdots$: Napier's constant.

In order to show $e=\underset{x\to 0}{lim}(1+x{)}^{\frac{1}{x}}$, we need to show $e=\underset{x\to +0}{lim}(1+x{)}^{\frac{1}{x}}$ and $e=\underset{x\to -0}{lim}(1+x{)}^{\frac{1}{x}}$. By changing variables $x=\frac{1}{y}$, $\underset{x\to +0}{lim}(1+x{)}^{\frac{1}{x}}=\underset{y\to \mathrm{\infty }}{lim}{(1+\frac{1}{y})}^y$ and $\underset{x\to -0}{lim}(1+x{)}^{\frac{1}{x}}=\underset{y\to -\mathrm{\infty }}{lim}{(1+\frac{1}{y})}^y$. Furthermore,  by changing variables $y=-t$, assuming $t>1$, we have

$$\begin{array}{rcl}\underset{y\to -\mathrm{\infty }}{lim}{(1+\frac{1}{y})}^y& =& \underset{t\to \mathrm{\infty }}{lim}{(1-\frac{1}{t})}^{-t}\\ & =& \underset{t\to \mathrm{\infty }}{lim}{\left(\frac{t}{t-1}\right)}^t\\ & =& \underset{t\to \mathrm{\infty }}{lim}{\left(\frac{t}{t-1}\right)}^{t-1}\left(\frac{t}{t-1}\right).\end{array}$$

Since $\underset{t\to \mathrm{\infty }}{lim}(1+\frac{1}{t-1})=1$, we can see that we only need to show $e=\underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x$.

For any $x\ge 1$, we can find $n\in \mathbb{N}$ such that

$$n\le x\le n+1.$$

Then we have

$$1<1+\frac{1}{n+1}\le 1+\frac{1}{x}\le 1+\frac{1}{n}.$$

And hence,

$${(1+\frac{1}{n+1})}^n\le {(1+\frac{1}{x})}^x\le {(1+\frac{1}{n})}^{n+1}.$$

Now,

$$\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{1}{n+1})}^n=\underset{n\to \mathrm{\infty }}{lim}\frac{{(1+\frac{1}{n+1})}^{n+1}}{(1+\frac{1}{n+1})}=e,$$

and

$$\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{1}{n})}^{n+1}=\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{1}{n})}^n(1+\frac{1}{n})=e.$$

Therefore, by the Squeeze Theorem,

$$\underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x})}^x=e.$$

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In a previous post, we have seen Cauchy's theorem for convergence of sequences (i.e., a Cauchy sequence converges and vice versa). 

See also: Monotone sequences and Cauchy sequences.

A similar result hold for the convergence of functions.

Theorem (Cauchy criterion for the right limit of a function)

Let $f(x)$ be a function on $I=(a,b]$. The right limit $\underset{x\to a+0}{lim}f(x)$ exists if and only if the following condition is satisfied:

• For any $\epsilon >0$, there exists a $\delta >0$ such that for all $x,y\in I$, if $a<x<a+\delta$ and $a<y<a+\delta$ then $|f(x)-f(y)|<\epsilon$.     ($\star$)

In a logical form:

$$\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0,\mathrm{\forall }x,y\in I(x,y\in (a,a+\delta )⟹|f(x)-f(y)|<\epsilon ).$$

Remark. Similarly, the left limit $\underset{x\to b-0}{lim}f(x)$ exists if and only if

• For any $\epsilon >0$, there exists a $\delta >0$ such that for all $x,y\in I$, if $b-\delta <x<b$ and $b-\delta <y<b$ then $|f(x)-f(y)|<\epsilon$.

The limit $\underset{x\to \mathrm{\infty }}{lim}f(x)$ exists if and only if

• For any $\epsilon >0$, there exists a $\delta >0$ such that for all $x,y\in I$, if $x>\delta$ and $y>\delta$ then $|f(x)-f(y)|<\epsilon$.

The limit $\underset{x\to -\mathrm{\infty }}{lim}f(x)$ exists if and only if

• For any $\epsilon >0$, there exists a $\delta >0$ such that for all $x,y\in I$, if $x<-\delta$ and $y<-\delta$ then $|f(x)-f(y)|<\epsilon$.

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Proof. ($\Rightarrow$) Suppose $\underset{x\to a+0}{lim}f(x)$ exists. Then there exists some real number $\alpha$ such that $\underset{x\to a+0}{lim}f(x)=\alpha$. For any $\epsilon >0$, there exists $\delta >0$ such that, for all $x\in I$, if $a<x<a+\delta$ then $|f(x)-\alpha |<\frac{\epsilon }{2}$. Let $x$ and $y$ be arbitrary real numbers such that $a<x<a+\delta$ and $a<y<a+\delta$. Then

$$\begin{array}{rcl}|f(x)-f(y)|& =& |f(x)-\alpha +\alpha -f(y)|\\ & \le & |f(x)-\alpha |+|f(y)-\alpha |\\ & <& \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .\end{array}$$

Therefore the condition ($\star$) is satisfied.

($\Leftarrow$) Suppose ($\star$) is satisfied. Let us define a sequence $\{a_n\}$ by $a_n=a+\frac{b-a}{n}$, $n\in \mathbb{N}$. For any $\epsilon >0$,  choose a $\delta >0$ satisfying ($\star$). By Archimedes' principle, we can find an $N\in \mathbb{N}$ such that $N\delta >b-a$. Then for any $n\ge N$, $a_n-a=\frac{b-a}{n}<\frac{b-a}{N}<\delta$ so that $a<a_n<a+\delta$. Therefore by ($\star$), for all $n,m>N$, $|f(a_n)-f(a_m)|<\epsilon$ so that $\{f(a_n)\}$ is a Cauchy sequence, and hence this sequence converges: $\underset{n\to \mathrm{\infty }}{lim}f(a_n)=\alpha$ for some $\alpha \in \mathbb{R}$.

Now, let us show that $\underset{x\to a+0}{lim}f(x)=\alpha$. By ($\star$), for any $\epsilon >0$, we can find $\delta >0$ such that for all $x,y\in (a,a+\delta )$, $|f(x)-f(y)|<\frac{\epsilon }{2}$. Also, since $\underset{n\to \mathrm{\infty }}{lim}f(a_n)=\alpha$, we can find an $N\in \mathbb{N}$ such that for all $n\ge N$, $|f(a_n)-\alpha |<\frac{\epsilon }{2}$. If $a<x<a+\delta$, choose $n\ge N$ such that $a<a_n<a+\delta$. Then

$$\begin{array}{rcl}|f(x)-\alpha |& =& |f(x)-f(a_n)+f(a_n)-\alpha |\\ & \le & |f(x)-f(a_n)|+|f(a_n)-\alpha |\\ & <& \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .\end{array}$$

Therefore,

$$\underset{x\to a+0}{lim}f(x)=\alpha .$$

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