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Mathematics LibreTexts

13.3: Limits That Do Not Exist

  • Page ID
    24879
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    Given a function f and a number c, there are two ways that \(\lim_{x \rightarrow c} f(x) = L\) can be false. First, there may be a different number \(M \ne L\) for which \(\lim_{x \rightarrow c} f(x) = M\).

    Second, it may be that Statement (13.4) is false for all \(L \in \mathbb{R}\). In such a case we say that \(\lim_{x \rightarrow c} f(x)\) does not exist. Contradiction is one way to prove that \(\lim_{x \rightarrow c} f(x)\) does not exist. Assume \(\lim_{x \rightarrow c} f(x) = L\) and produce a contradiction.

    Example 13.3

    Suppose \(f(x) = \frac{x}{2}+\frac{|x-2|}{x-2}+2\). Prove \(\lim_{x \rightarrow 2} f(x)\) does not exist.

    Proof. Notice that f (2) is not defined, as it involves division by zero. Also, \(f(x)\) behaves differently depending on whether x is to the right or left of 2.

    If \(x > 2\), then \(x-2\) is positive, so \(|x-2| = x-2\) and \(\frac{|x-2|}{x-2} = 1\), so \(f(x) = \frac{x}{2}+3\).

    If \(x < 2\), then \(x-2\) is negative, so \(|x-2| = -(x-2)\) and \(\frac{|x-2|}{x-2} = -1\), so \(f(x) = \frac{x}{2}+1\).

    Therefore f, graphed below, is a piecewise function \(f(x) = \left\{ \begin{array}{ll} {\frac{1}{2}x+3} & {\text{if} x > 2}\\ {\frac{1}{2}x+1} & {\text{if} x < 2} \nonumber \end{array} \nonumber \right.\)

    Suppose for the sake of contradiction that \(\lim_{x \rightarrow 2} f(x) = L\), where L is a real number. Let \(\epsilon = \frac{1}{2}\).

    By Definition 13.2, there is a real number \(\delta > 0\) for which \(0<|x-2|<\delta\) implies \(f(x)-L< \frac{1}{2}\).

    Put \(a = 2−\frac{\delta}{2}\), so \(0 < |a-2| < \delta\). Hence \(|f(a)-L| < \frac{1}{2}\).

    Put \(b = 2+\frac{\delta}{2}\), so \(0 < |b-2| < \delta\). Hence \(|f(b)-L| < \frac{1}{2}\).

    Further, \(f(a) < 2\) and \(f(b) > 4\), so \(2<|f(b)-f(a)|\). With this and the help of the inequality (13.1), we get a contradiction \(2 < 1\), as follows:

    \(2 < |f(b)-f(a)| = |(f(b)-L)-(f(a)-L)| \le |f(b)-L|+|f(a)-L| < \frac{1}{2}+\frac{1}{2} = 1\).

    Our next limit is a classic example of a non-existent limit. It often appears in first-semester calculus texts, where it is treated informally.

    Example 13.4

    Prove that \(\lim_{x \rightarrow 0} sin(\frac{1}{x})\) does not exist.

    As x approaches 0, the number \(\frac{1}{x}\) grows bigger, approaching infinity, so \(sin(\frac{1}{x})\) just bounces up and down, faster and faster the closer x gets to 0.

    Screen Shot 2020-06-22 at 2.55.19 PM.png

    Intuitively, we would guess that the limit does not exist, because \(sin(\frac{1}{x})\) does not approach any single number as x approaches 0. Here is a proof.

    proof. Suppose for the sake of contradiction that \(\lim_{x \rightarrow 0} sin(\frac{1}{x}) = L\) for \(L \in \mathbb{R}\).

    Definition 13.2 guarantees a number \(\delta\) for which \(0 < |x-0| < \delta\) implies

    \(|sin(\frac{1}{x})-L| < \frac{1}{4}\)

    Select \(k \in \mathbb{N}\) large enough so that \(\frac{1}{k\pi} < \delta\). As \(0 < \frac{1}{k\pi}-0 < \delta\) we have \(|sin(\frac{1}{k\pi})-L| < \frac{1}{4}\), and this yields \(|sin(k\pi)-L|=|0-L| = |L|< \frac{1}{4}\).

    Next, take \(l \in \mathbb{N}\) large enough so that \(\frac{1}{\frac{\pi}{2}+2l \pi} < \delta\). Then \(0 < |\frac{1}{\frac{\pi}{2}+2l \pi}-0| < \delta\), so we have \(|sin(\frac{1}{\frac{1}{\frac{\pi}{2}+2l \pi}}-L| < \frac{1}{4}\), which simplifies to \(|sin(\frac{\pi}{2}+2l \pi)-L| = |1-L| < \frac{1}{4}\).

    Above we showed \(|L| \le 4\) and \(|1-L| \le \frac{1}{4}\). Now apply the inequality (13.2) to get the contradiction \(1<\frac{1}{2}\), as \(1 = |L+(1-L)| \le |L+1|-|L| < \frac{1}{4}+\frac{1}{4} = \frac{1}{2}\).

    Example 13.5

    Investigate \(\lim_{x \rightarrow 0} x sin(\frac{1}{x})\).

    This is like the previous example, except for the extra x. Because \(|sin(\frac{1}{x})| \le 1\), we expect \(x sin(\frac{1}{x})\) to go to 0 as x goes to 0. Indeed, we prove \(\lim_{x \rightarrow 0} x sin(\frac{1}{x}) = 0\).

    Proof. Given \(\epsilon > 0\), let \(\delta = \epsilon\). Suppose \(0 < |x-0| < \delta\). Simplifying, \(|x| < \delta\), which is the same as \(|x| < \epsilon\). We get \(|x sin(\frac{1}{x})-0| = |xsin(\frac{1}{x})|= |x| \cdot |sin(\frac{1}{x})| \le \epsilon |sin(\frac{1}{x})| \le \epsilon \cdot 1 = \epsilon\). From this, Definition 13.2 gives \(\lim_{x \rightarrow 0} x sin(\frac{1}{x}) = 0\).

    One final point. We remarked on page 248 that for \(\lim_{x \rightarrow c} f(x) = L\) to make sense, there must be a \(\delta\) for which \(f(x)\) is defined for all \(x \in (c-\delta, c) \cup (c, c+\delta)\). Thus, for example, following Definition 13.2 to the letter, we have to say that \(\lim_{x \rightarrow c} \sqrt{x}\) does not exist because \(\sqrt{x}\) is not defined for all \(x \in (-\delta, 0) \cup (0, \delta)\). Your calculus text probably introduced a right-hand limit \(\lim_{x \rightarrow 0^{+}} \sqrt{x} = 0\). Though this notion is not programmed into our Definition 13.2, you may revisit such embellishments in later courses.

    Exercise

    Prove that the following limits do not exist.

    Exercise \(\PageIndex{1}\)

    \(\lim_{x \rightarrow 0} log_{10}|x|\)

    Exercise \(\PageIndex{2}\)

    \(\lim_{x \rightarrow 0} \frac{|x|}{x}\)

    Exercise \(\PageIndex{3}\)

    \(\lim_{x \rightarrow 0} \frac{1}{x^2}\)

    Exercise \(\PageIndex{4}\)

    \(\lim_{x \rightarrow \frac{\pi}{2}} cos(\frac{1}{x})\)

    Exercise \(\PageIndex{5}\)

    \(\lim_{x \rightarrow 0} xcot(\frac{1}{x})\)

    Exercise \(\PageIndex{6}\)

    \(\lim_{x \rightarrow 1} \frac{1}{x^2-2x+1}\)