The Definition of a Limit

An Aside about Definitions

Mathematicians seek to establish a body of knowledge that is built on principles of logic. As such, the idea of a definition to a mathematician is more precise than being merely an explanation of the idea or the meaning of the idea. To a mathematician, definitions are given for properties that will be attributed to objects and the definition provides a logical statement that characterizes whether the property applies or not.

For example, consider what it means to say that a set \(S\) of points in the plane is linear. In the sense of natural language, we might say that the set is straight. In the sense of mathematics, the idea of linear is a property of a set that captures the idea that the slope between any two points is always the same. We can formally describe the property if we create a logical statement that matches this idea.

There exists a single value \(m\) (possibly infinite) so that for any points \(P_1 = (x_1,y_1) \in S\) and \(P_2=(x_2,y_2) \in S\), if \(P_1 \ne P_2\) then \(\displaystyle m = \frac{y_2-y_1}{x_2-x_1}\). (Use \(m=\infty\) if \(x_1=x_2\).)

In a loose sense, we could think of this logical statement as a true/false function of a set, maybe named \(\mathrm{IsLinear}\). When we take a set \(S\) as an input to the statement, we check whether \(\mathrm{IsLinear}(S)\) is true or false. When it is true, we are allowed to use the phrase \(S\) is linear. Otherwise, we are obligated to say \(S\) is not linear.

A Mathematical Definition of Limit

When we think about a limit of a function at a point \(c\), we think about a value \(L\) that the output of a function approaches when the input approaches \(c\), perhaps written as \(f(x) \to L\) as \(x \to c\). As an equation, we write \[ \lim_{x \to c} f(x) = L. \] That is, we want to think of a limit as a mathematical object representing a value. Unfortunately, there is not a single rule for how to find the value for limit. So a definition can not be based on how the value is found.

Mathematics requires that we define things using logical statements. In this case, there are three related objects that go into the definition: the function, the input limit \(c\), and the output limit \(L\). So the definition of a limit is actually based on a rule to determine whether we should consider the equation \[ \lim_{x \to c} f(x) = L \] is true or false. That is, for the purposes of the definition, we do not think about the limit in terms of finding its value; we think about what is required for the combination \(f(x)\), \(c\) and \(L\) to make the equation for the limit true.

So for a function or expression \(f(x)\), an input value \(c\), and an output value \(L\), we wish to create a logical statement (essentially a true/false function, say \(\mathrm{IsLimit}(f(x),c,L)\)) that captures whether or not \(L\) is the appropriate limit for \(f(x)\) when \(x \to c\). The standard definition for a limit provides this logical test for the triple \((f(x), c, L)\).

For every value \(\epsilon > 0\), there exists some value \(\delta > 0\) such that if \(x \ne c\) and \(|x-c| < \delta\) then \(|f(x)-L| < \epsilon\).
Ordinarily, the statements \(x \ne c\) and \(|x-c| < \delta\) are combined as \(0 < |x-c| < \delta\).

In this definition (the rule to test \(\mathrm{IsLimit}\)), the use of absolute value captures the idea of distance or accuracy of approximation. We should view the two measures of distance (one for \(x\) and one for \(f(x)\)) as characterizing two sequences. An input sequence \(x_n\) gives us values for \(x = x_1, x_2, x_3, \ldots\), which eventually move closer and closer to \(c\), while the function creates an output sequence \(y_n = f(x_n)\). For \(L\) to be the limit of the approximations \(y_n\), the values \(y_n\) must get closer and closer to \(L\).

The relationship between \(\epsilon\) (epsilon) and \(\delta\) (delta) provides a connection between how close we need the values \(x_n\) to be to \(c\) in order to guarantee that the values of \(y_n\) are close enough to \(L\). The value \(\epsilon\) gives us the requested accuracy between \(y_n\) and \(L\). The value \(\delta\) (which usually depends on \(\epsilon\)) tells us how close to make \(x_n\) from \(c\) to accomplish the \(\epsilon\)-accuracy for \(f(x)\).

Subsets and Visualizing the Definition

The implication \( 0 < |x-c| < \delta \Rightarrow |f(x)-L| < \epsilon\) can be interpreted as a statement about subsets. Suppose \(\delta\) and \(\epsilon\) are fixed numbers. The inequality \(0 < |x-c| < \delta\) has a solution set that is sometimes called a punctured neighborhood of \(c\) given by \(N(c,\delta)=(c-\delta, c) \cup (c, c+\delta)\). We could also solve the inequality \(|f(x)-L| < \epsilon\) and obtain its solution set. The implication tells us that for every \(x\) in the punctured neighborhood, that \(x\) must also be in the solution set. So the punctured neighborhood is a subset of the solution set to \(|f(x)-L| < \epsilon\).

c-δ c c+δ (Illustration of a punctured neighborhood, \(N(c,\delta)\))

The inequality \(|f(x)-L| < \epsilon\) is related to the inequality \(|y-L| < \epsilon\), which in the \((x,y)\)-plane corresponds to a horizontal band between \(y=L-\epsilon\) and \(y=L+\epsilon\). The solution set to \(|f(x)-L| < \epsilon\) corresponds to the set of \(x\)-values for which \(y=f(x)\) appears inside this horizontal band. In other words, \(f(x)\) is between the thresholds of \(L-\epsilon\) and \(L+\epsilon\).

Example: Consider \(f(x) = x^2 - 2x\). Since \(f(4)=8\), we expect \[ \lim_{x \to 4} f(x) = 8. \] This example will illustrate the relationships between the punctured neighborhoods of \(c=4\) and the threshold bands around \(L=8\).

A graph of \(y=f(x)\) along with threshold bands around \(y=8\) with \(\epsilon = 2\) is illustrated. The solution set to \(|f(x)-8| < 2\), which is a union of two intervals, can be found by solving for the intersection points with the function and the threshold lines, \[ \begin{align*} x^2-2x &= 6 \qquad &\Leftrightarrow \qquad x &= 1 \pm \sqrt{7}, \\ x^2-2x &= 10 \qquad &\Leftrightarrow \qquad x &= 1 \pm \sqrt{11}. \\ \end{align*} \] So the solution set is \((1-\sqrt{11}, 1-\sqrt{7}) \cup (1+\sqrt{7}, 1+\sqrt{11})\), which to 4 decimal places is approximately \((-2.3166,-1.6458) \cup (3.6458, 4.3166)\).

This solution set is illustrated below using a marked number line.
-4 -3 -2 -1 0 1 2 3 4
Notice that one of the intervals contains our reference point \(c=4\). We will now focus only on that subinterval and look at the punctured neighborhoods around \(c=4\). Notice that \(N(4,\delta=0.2)\) is a subset while \(N(4,\delta=0.4)\) is not.
N(4,0.2) N(4,0.4) 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8
Consequently, we have just seen that the inequality \(|f(x)-8| < 2\) is true whenever \(x \in N(4,\delta=0.2)\), or in other words \(0 < |x-4| < 0.2\).

The value of \(\delta\) that we found is not the largest possible value that works. Because the solution set includes the entire interval \((1+\sqrt{7},1+\sqrt{11})\), we might seek the largest value of \(\delta\) so that \(N(4,\delta) \subseteq (1+\sqrt{7},1+\sqrt{11})\). An approximate answer is easily seen using decimal approximations and seeing how far each end-point is from the neighborhood center at \(c=4\). \[ \begin{align*} |4 - (1+\sqrt{7})| &\approx |4 - 3.6458| = 0.3542 \\ |4 - (1+\sqrt{11})| &\approx |4 - 4.3166| = 0.3166 \end{align*} \] Let us look at the corresponding punctured neighborhoods.
N(4,0.3542) N(4,0.3166) 3.6 3.7 3.8 3.9 4.0 4.1 4.2 4.3 4.4
We see that the smaller distance from 4 to the solution's edge gives the size of the largest punctured neighborhood that remains completely inside the solution set. So the largest value of \(\delta\) that makes \(N(4,\delta) \subseteq (1+\sqrt{7},1+\sqrt{11})\) is \(\delta_\max = \sqrt{11}-3 \approx 0.3166\).

The most common way of visualizing this relationship between the \(\epsilon\)-inequality and the \(\delta\)-neighborhood is to show the \(y = L \pm \epsilon\) thresholds with the graph \(y=f(x)\). Then add vertical lines at the edges of the punctured neighborhood, at \(x=c \pm \delta\) and \(x=c\). The subset relation will be true, as needed, if the graph \(y=f(x)\) is contained between the \(\epsilon\)-thresholds for each \(x\) in the punctured neighborhood. For the previous example, the illustration would be as given to the right. If \(\delta\) is chosen too large, then the graph \(y=f(x)\) will appear above or below the thresholds.

The Relationship between Delta and Epsilon

So far, we have considered a particular value for \(\epsilon\) and focused on what values of \(\delta\) make the implication

If \(0 < |x-c| < \delta\) then \(|f(x)-L| < \epsilon\)
a true statement. That is, the punctured neighborhood of size \(\delta\) is a subset of the solution set, \[ N(c,\delta) = \{ x : 0 < |x-c| < \delta \} \subseteq \{ x : |f(x)-L| < \epsilon \}. \] In the limit definition of a limit, we are interested in the set of all possible choices of \(\epsilon\) and \(\delta\) that make this true.

So consider the pairs \((\epsilon, \delta)\) that satisfy the limit implication for a given function \(f(x)\), point \(c\) and value \(L\), \[ \{ (\epsilon, \delta) : 0 < |x-c| < \delta \Rightarrow |f(x)-L| < \epsilon \}. \] The limit definition tells us that \(f(x) \to L\) as \(x \to c\) if for every \(\epsilon > 0\), there will be a value \(\delta > 0\) so that \((\epsilon, \delta)\) is in this set where the implication is true.

Example: Using the same function \(f(x)=x^2-2x\), \(c=4\) and \(L=8\), visualize and interpret the set \[ \{ (\epsilon, \delta) : 0 < |x-4| < \delta \Rightarrow |(x^2-2x) - 8| < \epsilon \}. \]

The set of \((\epsilon, \delta)\) is shown to the right. For example, the case where \(\epsilon = 2\) is the example we studied above. The three points illustrated with \(\epsilon = 2\) correspond to the three cases we analyzed. The point \(2, 0.4\) is not in the set because we saw that \(N(4,0.4) \not \subseteq \{ |f(x)-8| < 2 \}\). The point \(2, 0.2\) is in the set because \(N(4,0.2) \subseteq \{ |f(x)-8| < 2 \}\). The highest point with \(\epsilon=2\) is \((2, \sqrt{11}-3)\), which is also in the set.

Notice that for the function \(f(x)=x^2-2x\), \(c=4\) and \(L=8\), the set of \((\epsilon, \delta)\) making the limit implication true includes points for every \(\epsilon > 0\). This is the meaning of the definition of the limit statement \[ \lim_{x \to 4} [ x^2-2x ] = 8 \] which is the logical statement

For every \(\epsilon > 0\), there exists a value \(\delta > 0\) so that \(0 < |x-4| < \delta\) implies \(|(x^2-2x) - 8| < \epsilon\).

In order to understand how the limit definition is mathematically meaningful, it is useful to explore some examples where a value \(L\) is not the limit value.

Example: Using the function \(f(x)=x^2-2x\), \(c=2\) and \(L=1\), visualize and interpret the set \[ \{ (\epsilon, \delta) : 0 < |x-2| < \delta \Rightarrow |(x^2-2x) - 1| < \epsilon \}. \]

The actual limit should have been \(f(2) = 2^2-2(2)=0\). So when we draw thresholds, if \(\epsilon\) is small enough, the graph \(y=f(x)\) is not going to be between the lines at \(y=1 \pm \epsilon\) for any punctured neighborhood of \(c=2\). It may be possible that when \(\epsilon\) is large enough, there will be some overlap.

The three graphs below illustrate three different situations. When \(\epsilon < 1\), the solution set to \(|f(x)-1| < \epsilon\) is completely away from \(x=2\). So there are no punctured neighborhoods of \(x=2\) that are subsets of the solution. When \(1 < \epsilon < 2\), the solution set \(\{|f(x)-1| < \epsilon\}\) is the union of two intervals and the right-most interval does contain a neighborhood of \(x=2\). Finally, when \(\epsilon > 2\), the set \(\{|f(x)-1|<\epsilon\}\) is a single connected interval which contains neighborhoods of \(x=2\).


The relation of all pairs \((\epsilon, \delta)\) such that the limit implication \[ 0 < |x-2| < \delta \Rightarrow |(x^2-2x) - 1| < \epsilon \] is true is shown to the right. Because there are no points with \(\epsilon < 1\) in this set, \(L=1\) is clearly not the limit of \(f(x)\) as \(x \to 2\).

Question: Based on the graphs above, you should be able to find formulas that describe the boundary of the relation for \((\epsilon, \delta)\) by using the quadratic formula to find intersection points, and then using these solutions to compute \(\delta\). What are the formulas?

The previous example illustrated what happens when you have the wrong value for the limit. For a piecewise function with a jump discontinuity, we have this problem on one side but not the other, as illustrated in the next example.

Example: The piecewise function \(f\) defined below has no limit at \(x=1\) because the left- and right-limits disagree. \[ f(x) = \begin{cases} x^2, & x < 1, \\ 2, & x = 1, \\ 4-x, & x > 1. \end{cases} \]

Suppose we thought the limit at \(x=1\) was based on the formula \(x^2\) so that \(L=1^2 = 1\). Let us explore the inequality \(|f(x) - 1| < \epsilon\) for small \(\epsilon\). The graph of \(y=f(x)\) along with thresholds around \(L=1\) using \(\epsilon = 0.5\) is illustrated to the right. Because of the jump in the function, the solution to \(|f(x)-L|<\epsilon\) does not include a punctured neighborhood of \(x=1\). (If \(\epsilon\) is made large enough, there will be an included neighborhood; but a limit requires a neighborhood for all values of \(\epsilon\) no matter how small.) So \(L=1\) is not the limit. However it does include a left-neighborhood, \((1-\delta, 1)\), with \(\delta = 1-\frac{1}{\sqrt{2}} \approx 0.293\).

Next, suppose we thought the limit at \(x=1\) was based on the formula \(4-x\) so that \(L=4-1 = 3\). If we explore the inequality \(|f(x) - 3| < \epsilon\) for small \(\epsilon\), again using \(\epsilon = 0.5\), then we need to consider the graph of \(y=f(x)\) and the thresholds around \(L=3\) using \(\epsilon = 0.5\) as shown to the right. Again because of the jump in the function, the solution to \(|f(x)-L|<\epsilon\) does not include any punctured neighborhood of \(x=1\). So \(L=3\) is not the limit. Yet this time, the inequality solution includes a right-neighborhood, \((1, 1+\delta)\), with \(\delta = 0.5\).

Finally we might think the limit at \(x=1\) is just the value \(L=f(1)=2\). So we explore the inequality \(|f(x) - 2| < \epsilon\). The graph of \(y=f(x)\) along with thresholds around \(L=2\) using \(\epsilon = 0.5\) is shown. This time, we are not using a value matching either the left or the right side. So the solution to \(|f(x)-L|<\epsilon\) does not include a neighborhood to the left or to the right of \(x=1\), let alone a punctured neighborhood around \(x=1\). Consequently, \(L=2\) is also not the limit.

This function has no limit at \(x=1\) since the inequality \(|f(x)-L| < \epsilon\) for any value of \(L\) will not include a punctured neighborhood around \(x=1\). However, for \(L=1\), the inequality will have a left interval \((1-\delta, 1)\) that is in the solution set. For this reason, we say that \(L=1\) is a left-limit of \(f(x)\) at \(x=1\), which is written \[ \lim_{x \to 1^-} f(x) = 1. \] Similarly, for \(L=3\), the inequality will have a right interval \((1,1+\delta)\) that is in the solution set. Thus, we say that \(L=3\) is a right-limit of \(f(x)\) at \(x=1\), written \[ \lim_{x \to 1^+} f(x) = 3. \]

We now consider the relation of values \((\epsilon, \delta)\) that make the limit implication true, except that we consider the left and right neighborhoods separately. In particular, the left-limit implication corresponds to the logical statement \[ x < 1 \hbox{ and } |x-1|<\delta \quad \Rightarrow \quad |f(x)-L| < \epsilon. \] The right-limit implication corresponds to the logical statement \[ x > 1 \hbox{ and } |x-1|<\delta \quad \Rightarrow \quad |f(x)-L| < \epsilon. \]

For the value \(L=1\), which is the limit on the left, we see that the relation between \(\epsilon\) and \(\delta\) includes points for all values of \(\epsilon > 0\), but only for the left-limit implication. That is, for every value \(\epsilon > 0\), there are values \(\delta > 0\) so that \((1-\delta, 1)\) is a subset to the solution \(|f(x)-1| < \epsilon\). The right-limit implication is not satisfied for small values of \(\epsilon\).

For the value \(L=3\), which is the limit on the right, we see that the relation between \(\epsilon\) and \(\delta\) includes points for all values of \(\epsilon > 0\), but only for the right-limit implication. That is, for every value \(\epsilon > 0\), there are values \(\delta > 0\) so that \((1, 1+\delta)\) is a subset to the solution \(|f(x)-1| < \epsilon\). This time, the left-limit implication is not satisfied for small values of \(\epsilon\).

For the value \(L=2\), which is the value of the function \(f(1)=2\), this value is neither a left-limit or a right-limit. The relation shows that for small values of \(\epsilon\), namely \(\epsilon < 1\), there are no values \(\delta > 0\) so that either the left or right intervals satisfy the inequality \(|f(x) - 2| < \epsilon\).

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