Real analysis/Limits

Definition

Recall that a function from a set X to a set Y is a one-to-one mapping $f:X\to Y$. In analysis, we tend to talk about functions from subsets $A\subseteq ℝ$ to $ℝ$.

The definition for the limit of a function is much the same as the definition for a sequence. In fact, as we will see later, it is possible to define functional limits in terms of sequential limits. For the moment, however, let us just give the definition:

Given a subset $A\subset ℝ$ and a function $f:A\to ℝ$, we say $\underset{x\to c}{\lim }f(x)=L$ if $\mathrm{\forall }ϵ:\mathrm{\exists }\delta :0<|x-c|<\delta ⟹|f(x)-L|<ϵ$

The requirement $|x-c|>0$ is somewhat technical. It is an expression of the idea that the behavior of a function near a point shouldn't be affected by its behavior at the point. Thus f(x) need not be defined at c to have a limit there.

This definition gives a lot of people a lot of trouble, so it is best to spend some time puzzling it out, working examples, etc. One way to conceptualize the definition is this: $\underset{x\to c}{\lim }f(x)=L$ means that we can make f(x) as close as we like to L by making x close to c.

We can also define what it means for a function to diverge to infinity, and what it means for a function to have a limit at infinity:

• We say that $\underset{x\to c}{\lim }f(x)=\mathrm{\infty }$ if $\mathrm{\forall }M>0:\mathrm{\exists }\delta :0<|x-c|<\delta ⟹f(x)>M$.

• We say that $\underset{x\to c}{\lim }f(x)=-\mathrm{\infty }$ if $\mathrm{\forall }M>0:\mathrm{\exists }\delta :0<|x-c|<\delta ⟹f(x)<-M$

• We say that $\underset{x\to \mathrm{\infty }}{\lim }f(x)=L$ if $\mathrm{\forall }ϵ>0:\mathrm{\exists }M:x>M⟹|f(x)-L|<ϵ$.

• We say that $\underset{x\to -\mathrm{\infty }}{\lim }f(x)=L$ if $\mathrm{\forall }ϵ>0:\mathrm{\exists }M:x<-M⟹|f(x)-L|<ϵ$.

As an exercise, see if you can define what it means for a function to have limit $\mathrm{\infty }$ as $x\to \mathrm{\infty }$.

Sequential Limits

We might just as well have given the following definition of the limit:

Given a subset $A\subset ℝ$ and a function $f:A\to ℝ$, we say $\underset{x\to c}{\lim }f(x)=L$ if $\mathrm{\forall }(x_n{)}_{n=1}^{\mathrm{\infty }}$ such that $x_n=c,\underset{n\to \mathrm{\infty }}{\lim }(x_n)=c$, and $\underset{n\to \mathrm{\infty }}{\lim }(f(x_n))=L$

Note that the requirement $x_n=c$ corresponds with the requirement $|x-c|>0$.

As an exercise to test your understanding, prove that these two definitions are equivelant. Note that taking the contrapositive gives a good criterion for determining whether or not a function diverges:

If $\mathrm{\exists }(x_n),(y_n):(x_n)\to c,(y_n)\to c$, and $\underset{n\to \mathrm{\infty }}{\lim }(f(x_n))=\underset{n\to \mathrm{\infty }}{\lim }(f(y_n))$, then $\underset{x\to c}{\lim }f(x)$ does not exist.

We will be using this formulation extensively in the examples.

Algebraic Operations/Ordering Theorems

By applying the corresponding theorems for sequential limits, we find that functional limits are unique, that they preserve algebraic operations and ordering, and that a corresponding "Squeeze Theorem" holds. If $\underset{x\to c}{\lim }f(x)=L$, and $\underset{x\to c}{\lim }g(x)=M$, then:

• $\underset{x\to c}{\lim }af(x)=aL$

• $\underset{x\to c}{\lim }f(x)+g(x)=L+M$

• $\underset{x\to c}{\lim }f(x)g(x)=LM$

• $\underset{x\to c}{\lim }\frac{f(x)}{g(x)}=\frac{L}{M}$, assuming

$M$ is non-zero.

• If $\mathrm{\exists }\delta :f(x)>g(x)\mathrm{\forall }x\in A$, then $L>M$.

• $L=M,f(x)\le h(x)\le g(x)⟹\underset{x\to c}{\lim }h(x)=L$

• If L = 0 and h(x) is bounded, then $\underset{x\to c}{\lim }f(x)h(x)=0$.

Examples

We'll be giving many more examples in the section on continuity. Unfortunately, it is hard to properly define many of the most elementary functions without appealing to derivatives, integrals, and power series, so many of the examples here may seem a bit contrived. We'll start with a relatively nice function and move on to some nastier, unintuitive ones, with the goal being some kind of intuition about what can go right and wrong with limits.

• Let $f(x)=\{\begin{array}{cc}1& \text{if }x=0\\ 0& \text{if }x=0\end{array}$. Then $\underset{x\to 0}{\lim }f(x)=0$.

• Let $f(x)=\{\begin{array}{cc}0& \text{if }x\le 0\\ 1& \text{if }x>0\end{array}$. Then $\underset{x\to 0}{\lim }f(x)$ does not exist.

Consider the sequences $(x_n)=(\frac{1}{n}),(y_n=(\frac{-1}{n})$. Each converges to zero, but $(f(x_n))=1$ and $(f(y_n))=0$, and these have different limits as $n\to \mathrm{\infty }$. Thus the limit does not exist.

The next example is often given as a demonstration of just how nasty functions can get. It is not continuous at any point of its domain.

• Let $f(x)=\{\begin{array}{cc}1& \text{if }x\in \mathbb{Q}\\ 0& \text{if }x\in ℝ\setminus \mathbb{Q}\end{array}$. Then $\underset{x\to c}{\lim }f(x)$ does not exist for any $x\in ℝ$.

Given $x\in R$, let $x_n$ be any rational number in the interval $(\frac{-1}{n},\frac{1}{n})$, and let $y_n$ be any irrational number inthe same interval ($x_n$ and $y_n$ are gauranteed to exist by density of the rationals and irrationals). Given any $ϵ>0,|x_n-0|<\frac{1}{n}$ and $|y_n-0|<\frac{1}{n}$, so $(x_n),(y_n)\to 0$. However, (f(x_n)) = 1 and (f(y_n)) = 0, so their limits are 1 and 0. Since these are not equal, $\underset{y\to x}{\lim }f(y)$ does not exist.

Now we finally get to a limit that actually exists. Don't get too excited, though. The function is still extremely nasty (nastier than the previous two, perhaps), and the fact that it has a limit everywhere is one of its nastier aspects.

• Let $f(x)=\{\begin{array}{cc}\frac{1}{q}& \text{if }x=\frac{p}{q}\in \mathbb{Q}\\ 0& \text{if }x\in ℝ\setminus \mathbb{Q}\end{array}$. Then $\underset{x\to c}{\lim }f(x)=0$ for all $c\in ℝ$.

The idea is to show that the denominators of the rational numbers near f(x) are arbitrarily large. Given $ϵ>0$, let $\frac{1}{N}<ϵ$, and let $S=\frac{p}{q}\in Q:0<|\frac{p}{q}-c|<1,q\le N$. This is a finite set, since the numerators and the denominators are bounded and can therefore only take on a finite number of values. Thus, let $\frac{p}{q}$ be the element of S such that $|\frac{p}{q}-c|$ is minimized, and let $\delta =|\frac{p}{q}-c|=\delta$. Then $\mathrm{\forall }0<|x-c|<\delta :|f(x)-0|=|f(x)|=0<ϵ$(if x is irrational) or $|f(x)-0|=|f(x)|<\frac{1}{N}<ϵ$. Thus $\underset{x\to c}{\lim }f(x)=0$.