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Mathematics LibreTexts

13.2: Definition of a Limit

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    Limits are designed to deal with the following type of problem: We need to know how a certain function \(f(x)\) behaves when x is close to some number c. Perhaps \(f(c)\) is not even defined, so the graph of f looks something like what is shown below; a curve with a hole at a point \((c,L)\).

    Screen Shot 2020-06-21 at 10.22.56 PM.png

    In this picture, for any \(x \ne c\), the corresponding value \(f(x)\) is either greater than L or less than L. But the closer x is to c, the closer \(f(x)\) is to L, as illustrated on the right. We express this as \(\lim_{x \rightarrow c} f(x) = L\). That is, the symbols \(\lim_{x \rightarrow c} f(x)\) stand for the number that \(f(x)\) approaches as x approaches c.

    Your calculus text probably presented an informal, intuitive definition of a limit that likely went something like this.

    Definition 13.1 (Informal definition of a limit)

    Suppose f is a function and c is a number. Then \(\lim_{x \rightarrow c} f(x) = L\) means that \(f(x)\) is arbitrarily close to L provided that x is sufficiently close to c.

    The idea is that no matter how close we want to make \(f(x)\) to L, we can be assured that it will be that close (or closer) if x is close enough to c.

    Definition 13.1 is sufficient for the first few semesters of calculus, but it is not adequate for deeper, more rigorous investigations. The problem is that it is too vague. What, exactly, is meant by close? Saying x is “close” to c is not much better than saying that an integer n is “sort of even.” No proof can be done in the presence of such ambiguity.

    So this section’s first task is to motivate and develop a more rigorous and precise limit definition, the one used in advanced calculus. Achieving this goal forces us the grapple with the imprecise term close. What do we mean by close? Within 0.1 units? Within 0.001 or 0.00001 units, or even closer? We will make the definition precise by introducing a numeric, quantitative measure of closeness.

    Standard practice uses the Greek letters \(\epsilon\) (epsilon) and \(\delta\) (delta) for variables representing how close \(f(x)\) is to L, and x is to c. For instance, x is within a distance of \(\delta\) from c if and only if \(c-\delta < x < c+\delta\), that is, \(-\delta < x-c < \delta\), or \(|x-c| < \delta\). So for any real number \(\delta > 0\) (no matter how small) the statement \(|x-c| < \delta\) means that x is within \(\delta\) units from c.

    Screen Shot 2020-06-21 at 10.28.58 PM.png

    Likewise \(|f(x)-L| < \epsilon\) means that \(f(x)\) is within \(\epsilon\) units from L. Let’s apply these ideas to Definition 13.1, and transform it line by line.

    \[\begin{array}{ccc} {\textbf{Informal Definition}}&{\rightarrow}&{\textbf{Precise Definition}}\\ {\lim_{x \rightarrow c} f(x) = L \text{means that}}&{\rightarrow}&{\lim_{x \rightarrow c} f(x) = L \text{means that}}\\ {\text{f(x) is arbitrarily close to L}}&{\rightarrow}&{\forall \epsilon > 0, |f(x)-L| < \epsilon}\\ {\text{provided that}}&{\rightarrow}&{\text{provided that}}\\ {\text{x is sufficiently close to c}}&{\rightarrow}&{0 < |x-c| < \delta \text{for some} \delta > 0} \nonumber \end{array}\]

    We have arrived at a precise definition of a limit.

    Definition 13.2 (Precise definiton of a limit)

    Suppose \(f : X \rightarrow \mathbb{R}\) is a function, where \(X \subseteq \mathbb{R}\), and \(c \in \mathbb{R}\). Then \(\lim_{x \rightarrow c} f(x) = L\) means that for any real \(\epsilon > 0\) (no matter how small), there is a real number \(\delta > 0\) for which \(|f(x)-L| < \epsilon\) provided that \(0 < |x-c| < \delta\).

    Figure 13.1 illustrates this. For any \(\epsilon > 0\), no matter how small, consider the narrow shaded band of points on the plane whose y-coordinates are between \(y = L-\epsilon\) and \(y = L+\epsilon\). Given this \(\epsilon\), we can find another number \(\delta > 0\) such that the point \((x, f (x))\) is in the shaded band whenever x is within \(\delta\) units from c. In other words, \(|f(x)-L| < \epsilon\) provided that \(0 < |x-c| < \delta\).

    Screen Shot 2020-06-21 at 10.39.08 PM.png

    Figure 13.1. A graphic description of the limit definition.

    Three comments are in order. First, we have slipped into Definition 13.2 the expression \(0 < |x-c| < \delta\) instead of \(|x-c| < \delta\). This is to rule out the possibility \(x = c\), as \(f(c)\) may not be defined, depending of f and c.

    Second, Definition 13.2 applies only if there is some \(\delta > 0\) for which \((c-\delta,c) \cup (c,c+\delta)\) is a subset of the domain of f. Otherwise the statement

    “\(|f(x)-L|\) provided that \(0 < |x-c| < \delta\)” is meaningless for some x, no matter how small \(\delta\) is. Thus \(\lim_{x \rightarrow c} f(x)\) makes sense only if \(f(x)\) is defined for all \(x \in \mathbb{R}\) that are “close to” c in the sense that \(x \in (c-\delta,c) \cup (c,c+\delta)\) for some \(\delta\). Third, in symbolic form Definition 13.2 says \(\lim_{x \rightarrow c} f(x) = L\) if and only i

    \[\begin{array} {c} {\forall \epsilon > 0, \exists \delta > 0, 0<|x-c|<\delta \Rightarrow |f(x)-L| < \epsilon} \end{array}\]. Thus proving \(\lim_{x \rightarrow c} f(x) = L\) amounts to proving that Statement (13.2.1) is true.
    One strategy for proving Statement (13.2.1) is the direct approach. Begin by assuming \(\epsilon > 0\). Then find a \(\delta\) for which \(0 < |x-c| < \delta \Rightarrow |f(x)-L|< \epsilon\). To find \(\delta\), try to extract a factor of \(|x-c|\) from \(|f(x)-L|\). If you can do this, inspection usually tells you how small \(|x-c|\) needs to be to make \(|f (x)-L| < \epsilon\).

    We will use this strategy in Example 13.1, which proves \(\lim_{x \rightarrow 2}(3x+4) = 10\).

    Here \(f(x) = 3x+4\) and \(L = 10\), so \(|f(x)-L|\) is \(|(3x+4)-10|\). Also \(|x-c|\) is \(|x-2|\).

    Example 13.1

    Prove \(\lim_{x \rightarrow 2}(3x+4) = 10\).

    Proof. Suppose \(\epsilon > 0\). Note that \(|(3x+4)−10| = |3x-6| = |3(x-2)| = 3|x-2|\).

    So if \(\delta = 3\epsilon\), then \(0<|x-2|<\delta\) yields \(|(3x+4)-10|=3|x-2| < 3\delta = 3\frac{\epsilon}{3} = \epsilon\).

    In summary, for any \(\epsilon > 0\), there is a \(\delta = 3\epsilon\) for which \(0 < |x-2| < \delta\) implies \(|(3x+4)-10| < \epsilon\). By Definition 13.2, \(\lim_{x \rightarrow 2}(3x+4) = 10\).

    Example 13.2

    Prove that \(\lim_{x \rightarrow 2} 5x^2 = 20\).

    Proof. Suppose \(\epsilon > 0\). Notice that

    \(|f(x)-L| = |5x^2-20| = |5(x^2-4)| = |5(x-2)(x+2)| = 5 \cdot |x-2| \cdot |x+2|\).

    Now we have a factor of \(|x-2|\) in \(|f(x)-L|\), but it is accompanied with \(|x+2|\). But if \(|x-2|\) is small, then x is close to 2, so \(|x+2|\) should be close to 4. In fact, if \(|x-2| \le 1\), then \(|x+2| = |(x-2)+4| \le |x-2|+|4| \le 1+4 = 5\). (Here we applied the inequality (13.2) from page 245.) In other words, if \(|x-2| \le 1\), then \(|x+2| \le 5\), and the above equation yields

    \(|f(x)-L| = |5x^2-20| = 5 \cdot |x-2| \cdot |x+2| < 5 \cdot |x-2| \cdot 5 = 25|x-2|\).

    Take \(\delta\) to be smaller than both 1 and \(\frac{\epsilon}{25}\). Then \(0 < |x-2| < \delta\) implies

    \(|5x^2-20| < 25 \cdot |x-2| < 25 \delta < 25 \frac{\epsilon}{25} = \epsilon\). By Definition 13.2, \(\lim_{x \rightarrow 2} 5x^2 = 20\).

    The examples above (and the exercises below) involve limits that you probably regard as obvious. Our point is to illustrate Definition 13.2, not to compute difficult limits. Difficult limits come later (mostly in more advanced courses, not in this book) where Definition 13.2 will be used to great effect.


    Exercise \(\PageIndex{1}\)

    Prove that \(\lim_{x \rightarrow 5} (8x-3) = 37\).

    Exercise \(\PageIndex{2}\)

    Prove that \(\lim_{x \rightarrow -1} (4x+6) = 2\).

    Exercise \(\PageIndex{3}\)

    Prove that \(\lim_{x \rightarrow 0} (x+2) = 2\).

    Exercise \(\PageIndex{4}\)

    Prove that \(\lim_{x \rightarrow 8} (2x-7) = 9\).

    Exercise \(\PageIndex{5}\)

    Prove that \(\lim_{x \rightarrow 3} (x^2-2) = 7\).

    Exercise \(\PageIndex{6}\)

    Prove that \(\lim_{x \rightarrow 1} (4x^2+1) = 5\).