Limits: informal definition

Section 2 Limits: informal definition

Objectives

Give an informal definition of the limit of a function \(f\) at a given value.
Investigate the existence and/or value of a function graphically, by examining a a graph.
Use rigorous limit laws to compute limits of functions \(f\) formed from other functions using common arithmetic operations.

Subsection Informal definition of the limit

Many of the questions we will study in regard to functions revolve around how the output of a function \(f\) changes with respect to its input. As such we want to develop useful and precise language and tools for describing this relation. The limit of a function, defined informally here and rigorously later, is one such tool. Before getting to its informal definition, we illustrate with a graphical example the type of statements we wish to formalize with limit language. First, a reminder about what the graph of a function is.

Definition 2.1. Graph of function.

Given a function \(f\colon D\rightarrow \R\text{,}\) where \(D\subseteq \R\text{,}\) its graph is the set

\begin{equation*} \{(x,y)\in \R^2\colon x\in D \text{ and } y=f(x)\}\text{.} \end{equation*}

In other words the graph of \(f\) is the set of all points \(P\) in the \(xy\)-plane of the form \(P=(x,f(x))\text{,}\) where \(x\in D\text{.}\)

Example 2.2. Function behavior.

Suppose \(f\) is the function whose graph is given below.

Graph of complicated function — Figure 2.3. Graph of function \(f\)

Describe how the output \(f(x)\) behaves as the input \(x\) approaches each of the values \(a,b,c,d,e\text{.}\) For some of these values, you might consider how \(x\) approaches from the left and right separately.

Solution.

As \(x\) approaches \(a\text{,}\) either from the left or right, the values \(f(x)\) of the function get arbitrarily large and negative.

As \(x\) approaches \(b\) from the left, the values \(f(x)\) of \(f\) get closer and closer to \(0\text{.}\) However, as \(x\) approaches \(b\) from the right, the values \(f(x)\) get closer and closer to \(A\text{.}\)

As \(x\) approaches \(c\) from the left or right, the values \(f(x)\) of the function approach the real number \(M\text{.}\) We also happen to have \(f(c)=B\text{:}\) i.e., the value of \(f\) at \(x=c\) is equal to \(B\text{.}\)

As \(x\) approaches \(d\) from the left, the values \(f(x)\) of the function approach \(C\text{.}\) However, as \(x\) approaches \(d\) from the right, the values \(f(x)\) oscillate wildly above and below \(C\) and do not seem to approach any single value.

As \(x\) approaches \(e\) from the left or right, the values \(f(x)\) of the function approach \(D\text{.}\) Interestingly, however, the actual value of \(f\) at the input \(x=e\) is not equal to \(D\text{.}\)

Remark 2.4. Large and small, greater than and less thanLarge and small, greater than and less than

This is a good point to elucidate what we mean in mathematics when we talk about a number being large or small. Both these attributes refer to the size or magnitude of the number, and these in turn are defined as the absolute value of the number. Furthermore, geometrically, the absolute value \(\abs{x}\) measures the distance from \(x\) to the origin. Thus, we say a number \(x\in \R\) is large or small depending on whether its absolute value \(\abs{x}\) is large or small, which depends on whether \(x\) is far away from or close to \(0\) on the real line. As a result, we say that both of the following are sequences where the numbers get arbitrarily large:

\begin{align*} 1, 10, 100, 1000, 10000,\dots \amp \amp -1, -10,-100, -1000,-10000,\dots \amp \text{.} \end{align*}

In more detail we say that the first sequence gets arbitrarily large and positive, while the second sequence gets arbitrarily large and negative.

Contrast this with the use of “greater than” and “less than”. These refer specifically to the ordering relation \(<\) defined on \(\R\text{:}\) i.e.,

\begin{align*} x \text{ is less than } y \amp \iff x< y \\ x \text{ is greater than } y \amp \iff x> y \text{.} \end{align*}

The \(<\) relation is related to, but not completely determined by size of numbers. A good way of thinking of the \(<\) relation geometrically is in terms of position on the real number line: we have \(x< y\) if and only if \(x\) lies to the left of \(y\) as points on the number line. Thus, the number \(-100\) is less than the number \(1\) (since \(-100 < 1\)), even though \(1\) is smaller than \(-100\) (since \(\abs{1}< \abs{100}\)).

Definition 2.5. Function defined on set.

Let \(f\) be a function with domain \(D\text{.}\) Given \(a\in \R\text{,}\) we say that \(f\) is defined at \(a\) if \(a\) is an element of the domain: i.e., \(a\in D\text{.}\) Similarly, given a subset \(A\subseteq \R\text{,}\) we say that \(f\) is defined on \(A\) if \(A\) is included in the domain: i.e., \(A\subseteq D\text{.}\)

Definition 2.6. Limit (informal).

Suppose \(f\) is a function defined everywhere on an open interval containing the point \(a\in \R\text{,}\) except possibly at \(a\) itself. We say that the limit of \(f\) as \(x\) approaches \(a\) exists if there is a value \(L\) such that the function value \(f(x)\) can be made arbitrarily close to \(L\) provided \(x\) is sufficiently close (but not equal) to \(a\text{.}\)

When this is the case, we call \(L\) the limit of \(f\) as \(x\) approaches \(a\) and write

\begin{equation} \lim_{x\to a}f(x)=L\text{.}\tag{2.1} \end{equation}

When the limit does not exist we will say that \(\lim\limits_{x\to a}f(x)\) does not exist.

Example 2.7. Limit description of graph.

Let \(f\) be the function with graph given in Figure 2.3. Use the language and notation of limits to describe the behavior of \(f\) for inputs \(x\) near the values \(a,b,c,d,e\text{.}\)

Solution.

The limit \(\lim\limits_{x\to a}f(x)\) does not exist. There is no value \(L\) that \(f(x)\) approaches for inputs \(x\) sufficiently close to \(a\text{.}\) Indeed, as \(x\) gets arbitrarily close to \(a\text{,}\) \(f(x)\) becomes arbitrarily large and negative.

The limit \(\lim\limits_{x\to b}f(x)\) does not exist. If \(x\) is arbitrarily close to \(b\text{,}\) and less than \(b\) (to the left), then the values \(f(x)\) gets arbitrarily close to \(0\text{;}\) thus if the limit existed it would have to be equal to \(0\text{.}\) And yet for inputs \(x\) arbitrarily close to \(b\) but greater than \(b\) (to the right), the values \(f(x)\) are greater than \(1\text{:}\) i.e., they get no closer than a distance of \(1\) from the value \(0\text{.}\) This implies \(0\) is not the limit, and hence that no limit exists.

We have \(\lim\limits_{x\to c}f(x)=B\text{.}\) Since \(B=f(c)\text{,}\) we have \(\lim\limits_{x\to c}f(x)=f(c)\) in this case.

The limit \(\lim\limits_{x\to d}f(x)\) does not exist. See the explanation in Example 2.2.

We have \(\lim\limits_{x\to e}f(x)=D\text{.}\) Note that in this case \(e\) is an element of the domain and \(\lim\limits_{x\to e}f(x)\ne f(e)\text{.}\)

Example 2.8. Limit as compared to value.

Provide graphs of the following functions on their entire implied domain:

\begin{align*} f(x) \amp =\frac{x^2-1}{x-1} \amp g(x)\amp=\begin{cases} x+1 \amp \text{if } x\ne 1\\ 1 \amp \text{if } x=1 \end{cases} \amp h(x)\amp =x+1\text{.} \end{align*}

For each function discuss the limit behavior at \(x=1\) as compared with the value of the function at \(x=1\text{.}\)

Solution.

Graph of g — Figure 2.10. Graph of \(g\)

Graph of h — Figure 2.11. Graph of \(h\)

Subsection Limit rules

We now state some useful limit formulas and rules. These will give us a means of breaking down the limit computation of a complicated function into limits of simpler functions. Technically speaking we must prove the validity of each of these rules and formulas; however this would be a fool’s errand until we have a rigorous definition of the limit to work with. Such a definition will be provided in the near future, though even then we will not concern ourselves overly with proofs; we are more interested in learning how to make valid use of the rules.

Our first theorem gives us formulas for computing the limits of particular types of functions: constant functions, and the identity function.

Theorem 2.12. Constant and identity functions.

Constant function.

Fix a real number \(c\in \R\) and let \(h\colon \R\rightarrow \R\) be the constant function defined as \(h(x)=c\) for all \(x\in \R\text{.}\) Given any \(a\in \R\) we have

\begin{equation} \lim_{x\to a}h(x)=\lim_{x\to a}c=c\text{.}\tag{2.2} \end{equation}
Identity function.

Let \(\id\colon \R\rightarrow \R\) be the identity function defined as \(\id(x)=x\) for all \(x\in \R\text{.}\) Given any \(a\in \R\) we have

\begin{equation} \lim_{x\to a}\id(x)=\lim_{x\to a}x=a\text{.}\tag{2.3} \end{equation}

In contrast to the last theorem, our next theorem does not provide any formulas per se, but rather gives us rules governing how limits interact with various function operations.

Theorem 2.13. Limit rules.

Let \(f\) and \(g\) be functions, and suppose \(\lim\limits_{x\to a}f(x)\) and \(\lim\limits_{x\to a}g(x)\) exist for the real number \(a\in \R\text{.}\)

Sum rule.
\(\lim\limits_{x\to a} \left(f(x)+g(x)\right)=\lim\limits_{x\to a}f(x)+\lim\limits_{x\to a}g(x)\text{.}\)
Difference rule.
\(\lim\limits_{x\to a}\left(f(x)-g(x)\right)=\lim\limits_{x\to a}f(x)-\lim\limits_{x\to a}g(x)\text{.}\)
Scalar multiple rule.
\(\lim\limits_{x\to a}\left(cf(x)\right)=c\lim\limits_{x\to a}f(x)\) for all \(c\in \R\text{.}\)
Product rule.
\(\lim\limits_{x\to a}\left(f(x)g(x)\right)=\lim\limits_{x\to a}f(x)\, \lim\limits_{x\to a}g(x)\text{.}\)
Quotient rule.
If \(\lim\limits_{x\to a}g(x)\ne 0\text{,}\) then \(\displaystyle \lim\limits_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to a}f(x)}{\lim\limits_{x\to a}g(x)} \text{.}\)
Power rule.
\(\lim\limits_{x\to a}[f(x)]^n=\left[\lim\limits_{x\to a}f(x)\right]^n\) for all positive integers \(n\text{.}\)
Root rule.
\(\lim\limits_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{\lim\limits_{x\to a}f(x)}\) for all positive integers \(n\text{,}\) where we must assume \(\lim\limits_{x\to a}f(x)\) is positive if \(n\) is even.
Replacement rule.
If \(h\) is a function satisfying \(h(x)=f(x)\) for all \(x\ne a\) in an open interval about \(a\text{,}\) then \(\lim\limits_{x\to a}f(x)=\lim\limits_{x\to a}h(x)\text{.}\)

Remark 2.14. Limit rules.

It is useful to think of Theorem 2.13 as giving us a bunch of algebraic rules for computing limits of functions defined using addition, subtraction, multiplication, etc. For the most part these rules tell us that we can “bring the limit into” various operations: for example, the first two rules tell us that we can bring the limit into sums and differences of functions. Or better, using plain English, they tell us that the limit of a sum (of functions) is the sum of the limits, and that the limit of a difference (of functions) is the difference of the limits.

In fact most of the rules in Theorem 2.13 can be nicely summarized in plain English (e.g., “the limit of a product is the product of the limits”, “the limit of an \(n\)-th power is the \(n\)-th power of the limit”), and these summaries are helpful for remembering how the limit interacts with function operations.

Remark 2.15. Quotient rule.

Assume as in Theorem 2.13 that \(\lim_{x\to a}f(x)\) and \(\lim_{x\to a}g(x)\) exist and consider the limit \(\lim_{x\to a}\frac{f(x)}{g(x)}\text{.}\) Mark well that we can only make use of the quotient rule if \(\lim_{x\to a}g(x)\ne 0\text{,}\) in which case we can conclude that the limit of the quotient is the quotient of the limits.

Note further that in the case where \(\lim_{x\to a}g(x)=0\text{,}\) we cannot automatically conclude that the limit \(\lim_{x\to a}\frac{f(x)}{g(x)}\) does not exist; it simply the case that we cannot make use of the quotient rule to evaluate this limit. When this happens, we must look to other means for investigating the limit. See Example 2.19.

Example 2.16. Using limit rules.

Compute the limit below. Your answer should be a chain of equalities with steps justified by limit rules.

\begin{equation*} \lim_{x\to -2}\frac{x^3+x+1}{\sqrt[3]{4x^2+11}} \end{equation*}

Solution.

\begin{align*} \lim_{x\to -2}\frac{x^3+x+1}{\sqrt[3]{4x^2+11}} \amp = \frac{\lim\limits_{x\to -2}(x^3+x+1)}{\lim\limits_{x\to -2}\sqrt[3]{4x^2+11}} \amp \text{(quot. rule)}\\ \amp =\frac{(\lim\limits_{x\to -2}x)^3+\lim\limits_{x\to -2}x+\lim\limits_{x\to -2}1}{\sqrt[3]{\lim\limits_{x\to -2}(4x^2+11)}} \amp \text{(sum, power rule/root rule)}\\ \amp =\frac{(-2)^3-2+1}{\sqrt[3]{\lim_{x\to -2}(4x^2)+\lim\limits_{x\to -2}11}} \amp \text{(ident., const. rule/sum rule)}\\ \amp =\frac{-9}{\sqrt[3]{4(\lim\limits_{x\to -2}x)^2+11}} \amp \text{(scal. mult., power, const. rule)}\\ \amp =-\frac{9}{\sqrt[3]{4(-2)^2+11}} \amp \text{(ident. rule)}\\ \amp =-\frac{9}{\sqrt[3]{27}}=-3\text{.} \end{align*}

Corollary 2.17. Evaluation (polynomials and rational functions).

Polynomial evaluation.

Let \(f(x)=\anpoly\) be a polynomial. Given any \(a\in \R\text{,}\) we have

\begin{equation*} \lim\limits_{x\to a}f(x)=f(a)\text{.} \end{equation*}
Rational evaluation.

Let \(f\) and \(g\) be polynomials. For any \(a\in \R\) satisfying \(g(a)\ne 0\text{,}\) we have

\begin{equation*} \lim\limits_{x\to a}\frac{f(x)}{g(x)}=\frac{f(a)}{g(a)}\text{.} \end{equation*}

Example 2.18. Polynomial evaluation.

Give a simplified computation of the limit in Example 2.16 using Corollary 2.17.

Solution.

The polynomial evaluation formula allows us to skip a few steps in our computation from Example 2.16:

\begin{align*} \lim_{x\to -2}\frac{x^3+x+1}{\sqrt[3]{4x^2+11}} \amp = \frac{\lim\limits_{x\to -2}(x^3+x+1)}{\lim\limits_{x\to -2}\sqrt[3]{4x^2+11}} \amp \text{(quot. rule)}\\ \amp =\frac{(-2)^3+(-2)+1}{\sqrt[3]{\lim\limits_{x\to -2}(4x^2+11)}} \amp \text{(poly. eval./root rule)}\\ \amp =\frac{-9}{\sqrt[3]{4(-2)^2+11}} \amp \text{(poly eval.)}\\ \amp =-\frac{9}{\sqrt[3]{27}}=-3\text{.} \end{align*}

Example 2.19. Quotient rule does not apply.

Compute the limit below. Your answer should be a chain of equalities with each step justified.

\begin{equation*} \lim_{x\to 1}\frac{x^2-1}{x-1} \end{equation*}

Solution.

Note that since the limit of the denominator function is 0, we are not able to use the quotient (or rational function) rule. We begin instead with some algebra:

\begin{align*} \lim_{x\to 1}\frac{x^2-1}{x-1}\amp =\lim_{x\to 1}\frac{(x-1)(x+1)}{x-1} \\ \amp = \lim_{x\to 1}x+1 \amp \text{(repl. rule)}\\ \amp =2 \amp \text{(poly eval.)}\text{.} \end{align*}

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