Subsection 4.1.1 Definition
Let \(I\) be an open interval, and let \(c\in I\text{.}\) For our first concept (the limit) we consider a function defined on the punctured interval \(I\setminus \{c\}\text{.}\) In Calculus, the limit of \(f(x)\) as \(x\) tends to \(c\) is defined to be the number \(L\) (if it exists) that is approached closer and closer by \(f(x)\) as \(x\) gets closer and closer to \(c\text{.}\)
Our first goal is to make this into a precise definition. Besides the Greek letter
\(\epsilon\) that we first encountered in
Theorem 2.1.6, we will make use of the Greek letter
\(\delta\) (delta), that is also commonly used to denote a positive small number.
Definition 4.1.1.
Let \(f: I\setminus\{c\} \rightarrow {\mathbb R}\text{,}\) where \(c\in I\text{,}\) and let \(L\in {\mathbb R}\text{.}\) We say that the limit of \(f(x)\) as \(x\) tends to \(c\) is \(L\), and write \(\displaystyle \lim_{x\rightarrow c} f(x)=L\text{,}\) if for any given positive number \(\epsilon\text{,}\) we can find some positive number \(\delta\) such that if \(x\) is in \(I\setminus\{c\}\) and \(|x-c|\lt \delta\text{,}\) then \(|f(x)-L|\lt \epsilon\text{.}\) It is often useful to write such statement using the logic symbols:
\begin{equation*}
(\forall \epsilon \gt 0)(\exists \delta \gt 0)(\forall x\in I\setminus \{c\})(0\lt |x-c|\lt \delta \Rightarrow |f(x)-L|\lt \epsilon )
\end{equation*}
We can informally describe in words the above definition as follows: “we can make \(|f(x)-L|\) smaller than any given positive number \(\epsilon\) as long as we keep \(x\) within some sufficiently small distance \(\delta\) of \(c\text{,}\) without actually being equal to \(c\text{.}\)”
A crucial word in the above paragraph is “sufficiently”. It tells us that the number \(\delta\) will need to be taken small enough that it will “work” (meaning that if will result in \(|f(x)-L|\) being less than \(\epsilon\)) for the given \(\epsilon\text{.}\) In other words, we can expect that \(\delta \) will depend on \(\epsilon\text{.}\) If a number \(\delta\) works for some \(\epsilon\text{,}\) and then we take a different \(\epsilon\text{,}\) we can expect that we will need to change our \(\delta\text{.}\) Our first example will illustrate this.
Example 4.1.2.
Let \(f(x)=3x+1\text{,}\) \(c=2\text{.}\) We take the domain to be \({\mathbb R}\setminus \{2\}\text{.}\) So \(f(2)\) is not defined. But of course we expect that if we take \(x\) close enough to \(2\text{,}\) \(f(x)\) will get very close to \(3(2)+1=7\text{.}\) So we take \(L=7\) and wish to prove that
\begin{equation*}
\lim_{x\rightarrow 2}f(x)=7.
\end{equation*}
A good place to start is at the end of
Definition 4.1.1: we want
\(|f(x)-7|\lt \epsilon\text{,}\) or
\(|3x+1-7|\lt \epsilon\text{,}\) that simplifies to
\(3|x-2|\lt \epsilon\text{.}\) So if we make
\(|x-2|\lt \epsilon/3\text{,}\) then
\(3|x-2|\) (that is nothing but
\(|f(x)-7|\)) will be less than
\(\epsilon\text{,}\) and that is what we want. This is telling us that we need to take
\(\delta = \epsilon/3\text{.}\)
We can consider the last paragraph as some “scratch work” for our proof. Now that we have figured out what \(\delta \) will work, we give the official proof:
We prove that \(\displaystyle \lim_{x\rightarrow 2}f(x)=7.\) Let \(\epsilon \gt 0\) be given, and let \(\delta = \epsilon/3\text{.}\) Suppose that \(x\in I\setminus\{2\}\) and \(|x-2|\lt \delta\text{.}\) Then
\begin{equation*}
|f(x)-7| = |3x+1-7|=|3x-6|=3|x-2|\lt 3\delta =3\frac{\epsilon}{3} = \epsilon\text{.}
\end{equation*}
It is important to note that the value of the function at \(x=c\) plays no role at all. In fact, in the example above, we did not even define \(f(2)\text{.}\) And even if it had been defined, we would have made no use of it in the calculation of the limit.
As seen in the example, the number \(\delta\) will usually depend on \(\epsilon\text{.}\) In that example, it was \(\delta= \epsilon/3\text{.}\) So taking a smaller \(\epsilon\) will result in a smaller \(\delta\text{.}\)
In fact, we can describe the proof that \(\displaystyle \lim_{x\rightarrow c}f(x)=L\) as the task of finding a function \(\delta:(0,\infty)\rightarrow (0,\infty)\) with the property that
\begin{equation*}
0\lt |x-c|\lt \delta(\epsilon)\Rightarrow |f(x)-L|\lt \epsilon
\end{equation*}
for all \(\epsilon \in (0,\infty)\text{.}\) In the last example, \(\delta(\epsilon)=\epsilon/3\text{.}\)
Compare with the proof that \(\displaystyle \lim_{n\rightarrow \infty} a_n=L\text{,}\) where \((a_n)\) is a sequence. In both cases, we start with a positive number \(\epsilon\text{.}\) For a sequence, we need to find a positive integer \(n_0\) (that will depend on \(\epsilon\)) with the property that \(|a_n-L|\lt \epsilon \) when \(n\geq n_0\text{.}\) For a function, we need to find a positive number \(\delta\) (that will depend on \(\epsilon\)) with the property that \(|f(x)-L|\lt \epsilon\) when \(|x-c|\lt \delta\text{.}\)
In the next example, finding the right \(\delta\) requires more work. We will see how the triangle inequality is essential in many of our proofs.
Example 4.1.3.
Let \(f(x)=x^2\text{,}\) \(c=3\text{,}\) with domain \({\mathbb R}\setminus\{3\}\text{.}\) Once again, we do not define \(f\) at the point where we do the limit. Because we are interested in the behavior of \(f\) when \(x\) is very close to \(3\text{,}\) not at \(3\text{.}\) We expect of course that as \(x\) gets very close to \(3\text{,}\) \(x^2\) will get very close to \(9\text{.}\) So we want to prove that
\begin{equation*}
\lim_{x\rightarrow 3}x^2=9.
\end{equation*}
We start with some \(\epsilon \gt 0\text{,}\) as usual. We first do some scratch work: we want to have \(|x^2-9|\lt \epsilon\) by controlling the size of \(|x-3|\text{.}\) Remember how in the previous example the factor \(x-2\) appeared as soon as we simplified \(f(x)-7\text{.}\) This time we use a basic factoring identity: \(x^2-9 =(x-3)(x+3)\text{.}\) One of the two factors is what we want: \(x-3\text{.}\) But if we try to solve for it in the inequality \(|(x-3)(x+3)|\lt \epsilon\text{,}\) we find \(|x-3|\lt \epsilon/|x+3|\text{.}\) We cannot use \(\epsilon/|x+3|\) as our \(\delta\text{,}\) because it is not a positive real number depending only on \(\epsilon\text{.}\)
Instead, we need to use some different reasoning: to make \(|(x-3)(x+3)|\lt \epsilon\text{,}\) we can make the size of the first factor \(|x-3|\) as small as we want by choosing a small enough \(\delta\text{.}\) But how small should it be? It depends on how big the other factor \(|x+3|\) is going to get. If for example it does not get any bigger than \(M\text{,}\) then we can take \(\delta\) to be \(\epsilon/M\text{,}\) so that \(|x-3||x+3|\) will be at most \(\displaystyle \frac{\epsilon}{M} M=\epsilon\text{.}\)
So we see that it is essential that we find an upper bound for \(|x+3|\text{.}\) We only need an upper bound when \(x\) is near \(3\text{.}\) So we first take \(\delta_1=1\) and note that if \(|x-3|\lt \delta_1=1\text{,}\) then \(|x+3|=|x-3+6|\leq |x-3| +6\leq 1+6=7\text{.}\) So \(7\) is the upper bound we will work with. It works as soon as \(|x-3|\leq \delta_1=1\text{.}\) Now we need to make sure that the product \(|x-3||x+3|\) is less than \(\epsilon\text{.}\) As noted before, we will need \(|x-3|\) to be less than \(\delta_2=\epsilon/7\text{.}\) So in order for the upper bound \(7\) for \(|x+3|\) to be valid and the product \(|x+3||x-3|\) to be less than \(\epsilon\text{,}\) we will need \(|x-3|\) to be both less than \(1\) and less than \(\epsilon/7\text{.}\) We can do this by taking \(\delta\) to be the minimum of \(\delta_1=1\) and \(\delta_2=\epsilon/7\text{.}\) We are now ready to do the official proof.
Let \(\epsilon \gt 0\) be given. Let \(\delta =\min(1,\epsilon/7)\text{.}\) Then if \(0\lt |x-3|\lt \delta\text{,}\) we have \(|x-3|\lt 1\text{,}\) and so \(|x+3|=|x-3+6|\leq |x-3|+6\leq 1+6=7\text{,}\) and since we also have \(|x-3|\lt \epsilon/7\text{,}\) we find \(|x^2-9|=|x-3||x+3|\lt \displaystyle \frac{\epsilon}{7} 7 = \epsilon\text{.}\)
