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# 13.3: Limits That Do Not Exist

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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. 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 $$\PageIndex{6}$$

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