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Section 14.1 The Limit Rules (Theorems)

Subsection 14.1.1 The Limit of a Composition and Continuity at a Point

In the old days when people invented a new function they had something useful in mind. Now, they invent them deliberately just to invalidate our ancestors’ reasoning, and that is all they are ever going to get out of them.
Henri Poincarè
 1 
https://mathshistory.st-andrews.ac.uk/Biographies/Poincare
(1854–1912)
In this section we will state several theorems about limits which we will need in the sections following. The limit concept is very subtle and our understanding of it is still quite intuitive. We are not yet quite prepared to prove these theorems so we will leave these theorems unproven for now. Our immediate goal is simply to understand what they say and learn how to use them. In the next section we will begin using these theorems to show how the limit in Definition 13.2.3 allows us to recapture all of the major results we used in Part I of this text.
In Chapter 17 we will finally discard our intuitive definition of a limit (Definition 12.1.1) and formally define both a limit at infinity (Definition 17.2.19 and Definition 17.2.13) and a limit at a point (Definition 17.3.1). Then we will return to the theorems in this section and (finally) prove rigorously that they are, in fact, true. Until then any result which relies on the theorems in this section should be considered contingent.
In Chapter 12 we stated the following three theorems about limits “at infinity.”
  1. the limit of a sum is the sum of the limits (Theorem 12.1.6),
  2. the limit of a product is the product of the limits (Theorem 12.1.7) and,
  3. the limit of a quotient is the quotient of the limits (Theorem 12.1.8).
All three of these theorems remain true if \(x\) is approaching some finite number, \(a\text{,}\) instead of infinity.
Notice that in addition to changing \(\infty\) to some real number, \(a\text{,}\) we have added two qualifications to the statement of each of these theorems from Chapter 12:
  • “Suppose that the functions \(f(x)\) and \(g(x)\) are defined on some open interval about \(a\)” and,
  • “except, possibly, at \(a\) itself”
To see why these are necessary recall that we’re going to use limits to define the derivative as in Definition 13.2.3 so we’ll need to evaluate the limit \(f^\prime(x)=\tlimit{h}{0}{\frac{f(x+h)-f(x)}{h}}\text{.}\) Clearly the expression \(\frac{f(x+h)-f(x)}{h}\) is not defined at \(h=0\text{.}\) But we’re only interested in its value in the limit as \(h\rightarrow0\) which means that \(h\) must be able to get close to \(0\text{.}\) That is, there must be an open interval around \(0\) where the expression \(\frac{f(x+h)-f(x)}{h}\) is defined.
But we don’t care if \(\frac{f(x+h)-f(x)}{h}\) is defined at \(h=0\) or not. That is irrelevant to our purpose. So we state explicitly that we do not consider whether \(h=0\text{.}\)
Ok, but why did we insert the word “possibly”? Wouldn’t it be enough to simply say “except at \(a\)”?
We need to say “possibly” because these theorems, like all theorems, are stated with as much generality as possible. For example, consider the function \(f(x)=2x\text{.}\) Had we not included “possibly” in the conditions of our theorems the limit: \(\tlimit{x}{3}{2x}\text{,}\) which is clearly equal to \(6\text{,}\) would have to be considered undefined because \(f(x)=2x\) is defined at \(x=3\text{.}\) This distinction may seem like a very fussy, and unimportant detail right now, but it will be important when we discuss the meaning of continuity in Subsection 14.1.1.

Example 14.1.4.

Suppose
\begin{equation*} f(x)= \begin{cases} 3x \amp \text{if } x\neq2\\ 10 \amp \text{if } x=2 \end{cases}. \end{equation*}
Then
\begin{equation*} \limit{x}{2}{f(x)}= 6. \end{equation*}
In particular the limit is not \(10\text{.}\) Here is how we would evaluate this limit using the tools we currently have at our disposal. We’re interested in the limit as \(x\rightarrow2\) so in particular we do not need to consider the case when \(x=2\text{.}\) But as long as \(x\neq2\) we have \(f(x)=3x\) so
\begin{equation*} \limit{x}{2}{f(x)}= \limit{x}{2}{3x}. \end{equation*}
As \(x\) gets close to \(2\) it is clear that \(3x\) gets close to \(6\text{.}\) Therefore
\begin{equation*} \limit{x}{2}{f(x)}= 6. \end{equation*}
Notice that our reasoning is a little vague in the last step because we had to resort to the phase “gets close to,” and we know from our work in Section 12.4 that this is not a precise phrase. This is the best we can do now because we have not yet rigorously defined a limit. We will do that in Chapter 17.

Problem 14.1.5.

By reasoning in a manner similar to Example 14.1.4 show that \(\tlimit{x}{3}{f(x)}=9\) for each function.
(a)
\(f(x)=x^2\)
(b)
\(f(x)= \begin{cases} x^2 \amp \text{if } x\neq3\\ \text{undefined } \amp \text{if } x=3 \end{cases}\)
It will be tedious to write (and to read) the phrase “Suppose that \(f(x)\) is defined on some open interval about \(a\) except, possibly, at \(a\) itself” every time we need it so it is customary to say something more like “Suppose \(f(x)\) is defined near \(a\)” instead. Because we are trying to be as precise, and rigorous as possible we will formalize this by redefining the word near.

Definition 14.1.6. Near.

We say that \(f(x)\) has some property near \(x=a\) if \(f(x)\) has that property on an open interval about \(x=a\text{,}\) except possibly at \(a\) itself.
Notice that this is not what “near” means in ordinary speech. This is one of the things that makes it difficult to read mathematics. We routinely co-opt words from natural languages (like English) and redefine them to fit our needs. In this case our purpose requires that we change the definition of “near” slightly as you’ve seen. Because “near” is a common word and you have a lifetime of experience using it, it can be very difficult to cast off your preconceptions. The familiar definition you learned in childhood will intrude and cause confusion. It is hard to overcome this. Refer back to the definition frequently until you have internalized the mathematical definition.
We have the following theorem.
Notice that Theorem 14.1.7 would be considerably less useful had we not required that \(f(x)=L\) for \(x\) near \(a\text{,}\) rather than \(f(x)=L\) on its entire domain. For example, as stated Theorem 14.1.7 allows us to conclude that if
\begin{equation*} H(x) = \begin{cases} 1\amp \text{ if } x\ge0\\ -1\amp \text{ if } x\lt0 \end{cases} \end{equation*}
then
\begin{equation*} \limit{x}{5}{H(x)}=1 \end{equation*}
and
\begin{equation*} \limit{x}{-5}{H(x)}=-1 \end{equation*}
because there are open intervals about \(x=5\) and \(x=-5\) where \(H(x)=1\text{,}\) and \(H(x)=-1\text{,}\) respectively.
Since there is no such interval about \(x=0\text{,}\) \(\tlimit{x}{0}{H(x)}\) is undefined.

Drill 14.1.8.

  1. Find an open interval about \(5\) where \(H(x)=1\text{.}\)
  2. Find an open interval about \(-5\) where \(H(x)=-1\text{.}\)

Drill 14.1.9.

Suppose
\begin{equation*} f(x)= \begin{cases} \ \ 1;\amp \text{ if } x\gt 1 \text{ or if } x=-2\\ -2;\amp \text{ if } x\le1 \text{ and } x\ne-2. \end{cases} \end{equation*}
Determine whether the following statements are true or false.
  1. \(f(x)= 1\) near \(x= 4\text{.}\)
  2. \(f(x)= 1\) near \(x= 1\text{.}\)
  3. \(f(x)= -2\) near \(x=1 \text{.}\)
  4. \(f(x)= 1\) near \(x=-2 \text{.}\)
  5. \(f(x)= -2\) near \(x=-2 \text{.}\)
  6. \(f(x)= 0\) near \(x= 0\text{.}\)

Problem 14.1.10.

Explain, that the following statements are true by citing Theorem 14.1.1 through Theorem 14.1.7 as needed.
(a)
\(\limit{x}{0}{\left(\frac{5x}{x}+\frac{\pi x}{x}\right)}=5+\pi\)
(b)
\(\limit{x}{2}{\left(\frac{x-2}{x-2}+\frac{3x-6}{x-2}\right)}=4\)
(c)
\(\limit{x}{5}{\left(2x+3x^2\right)}=85\)
(d)
\(\limit{x}{-1}{\left(\frac{x^2-1}{x+1}+\frac{x^2+3x+2}{x+1}\right)}=-1\)

Problem 14.1.11.

Notice that neither \(\tlimit{x}{0}{\frac{1}{x}}\) nor \(\tlimit{x}{0}{\left(\frac{-1}{x}\right)}\) exists. However their sum,
\begin{equation*} \tlimit{x}{0}{\left(\frac{1}{x}+\frac{-1}{x}\right)}=\tlimit{x}{0}{0}=0 \end{equation*}
does exist. Explain why this does not contradict Theorem 14.1.1.

Problem 14.1.12.

Suppose \(g(x)\neq0\) near \(x=a\) and \(\tlimit{x}{a}{\frac{f(x)}{g(x)}}\) exists. Use Theorem 14.1.2 to show that if \(\limit{x}{a}{g(x)}=0\) then \(\limit{x}{a}{f(x)}=0\text{.}\)
Hint.
Consider \(f(x)=\frac{f(x)}{g(x)}\cdot g(x)\) for \(x\) near \(a\text{.}\)
The following Corollary says that if \(f(x)\) is approaching \(L_f\) and we multiply \(f(x)\) by a number, \(k\text{,}\) then the product \(kf(x)\) approaches \(kL_f\text{.}\) It follows from Theorem 14.1.7 and Theorem 14.1.2.
The concept of continuity is essential to Calculus, but you may have noticed that we have carefully avoided it as much as possible until now. This is because defining continuity is similar to defining the line tangent to a curve (Definition 5.2.4). We need to think carefully about what we want the term continuous to mean, and then craft our definition to capture that meaning. This would have been very difficult to do without a fairly sophisticated understanding of the limit concept.
So stop and think about this for a moment. What do we mean when we say a curve is continuous? A first, intuitive definition usually goes something like this: “A function is continuous if you can draw its graph without lifting your pencil from the paper,” but this is unsatisfactory for a number of reasons. In particular, it is impossible to apply in most cases. Think about it. How often have you seen the entire graph of any function? Usually we just draw the part neat the origin and put arrowheads on both ends of the graph. We need something more precise.
At the end of Example 12.4.31 we remarked that it is only when \(f(x)\) is continuous at \(x=g(a)\) that \(\textcolor{blue}{\tlimit{x}{a}{\textcolor{red}{f(g(x))}}}\) is equal to \(\textcolor{red}{f}\textcolor{blue}{\left(\tlimit{x}{a}{g(x)}\right)}\text{,}\) but we did not discuss the matter any further. It is time for that discussion.
First, notice that when you think closely about the statement “\(f(x)\) is continuous at \(g(a)\)” it appears to be nonsense, because \(g(a)\) is the value of \(f\) at the single value \(g(a)\text{.}\) Does it make sense to you that a curve can be continuous at a single value of its domain? In ordinary usage the concept of continuity requires an interval to be continuous on, doesn’t it?
Since we need the concept of “continuity at a point,” we define it.

Definition 14.1.15. Continuity at a Point.

A function \(f\text{,}\) whose domain is an interval in \(\RR\text{,}\) is continuous at \(x=a\) in the interval, if and only if \(\limit{x}{a}{f(x)}=f(a)\text{,}\) (alternatively, if \(\limit{h}{0}{f(a+h)}=f(a)\)).
If \(f\) is continuous at every point in its domain we’ll just call it a continuous function.
The sketch below shows that Definition 14.1.15 recovers the intuitive notion that a function is continuous if we can draw its graph without lifting pen from page. Both of the functions, \(f(x)\) and \(g(x)\) are identical everywhere except at \(x=1\text{.}\) Clearly, we must lift our pen from the page to draw the graph of \(g(x)\text{,}\) which is discontinuous at \(x=1\text{.}\) But this is not true of the graph of \(f(x)\text{,}\) which is continuous.
So intuitively, the continuity of \(f(x)\) at \(a\) means that as \(x\) approaches \(a\text{,}\) \(f(x)\) should approach \(f(a)\text{.}\) For \(g(x)\) in the sketch above this does not happen when \(a=1\text{.}\) From Definition 14.1.15 it is clear that if \(f(x)\) is continuous at \(a\) then
\begin{equation*} \limit{x}{a}{f(x)} = f(a) = f\left(\limit{x}{a}{x}\right)\text{.} \end{equation*}
This may seem to be a little silly observation but it emphaizes the fact that the continuity of \(f(x)\) at \(a\) amounts to saying that we can interchange the name of the function and the limit symbol. This is the essence of Theorem 14.1.16 below.
Essentially this says that we can interchange the function \(f\) and the “\(\lim\)” symbols if \(f\) is continuous at \(g(a)\text{.}\)
The following lemma is true and the proof will be valid once the limit theorems have been proven in Chapter 17.
When we were studying horizontal asymptotes in Section 12.2 we encountered Theorem 12.2.2 (the Squeeze Theorem “at” Infinity). But the Squeeze Theorem is also valid if \(x\rightarrow a\text{,}\) where \(a\) is a real number.
Theorem 14.1.19 is illustrated below, but a formal proof will not be given until Chapter 17. Notice that on the interval \((c,d)\) \(f(x)\) is caught between (“squeezed by”) \(\alpha{(x)}\) and \(\beta{(x)}\text{.}\)

Problem 14.1.20.

Consider the two functions defined in the sketch below:
(a)
Use Theorem 14.1.19 to show that \(U(x)\) is continuous at \(x=0\text{.}\)
Hint.
What functions is \(U(x)\) caught between?
(b)
Use Definition 14.1.15 to show that \(T(x)\) is not continuous at \(x=0\text{.}\)
Hint.
Try the substitution \(z=\frac1x\) for \(x\neq0\text{.}\) What would \(\limit{x}{0}{T(x)}\) look like in terms of \(z\text{?}\)