Proving Limit Laws

We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value.

Definition

Proof

We prove the following limit law: If \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = L\) and \(\underset{x \rightarrow a}{\text{lim}} g \left(\right. x \left.\right) = M ,\) then \(\underset{x \rightarrow a}{\text{lim}} \left(\right. f \left(\right. x \left.\right) + g \left(\right. x \left.\right) \left.\right) = L + M .\)

Let \(\epsilon > 0 .\)

Choose \(\delta_{1} > 0\) so that if \(0 < \left|\right. x - a \left|\right. < \delta_{1} ,\) then \(\left|\right. f \left(\right. x \left.\right) - L \left|\right. < \epsilon / 2 .\)

Choose \(\delta_{2} > 0\) so that if \(0 < \left|\right. x - a \left|\right. < \delta_{2} ,\) then \(\left|\right. g \left(\right. x \left.\right) - M \left|\right. < \epsilon / 2 .\)

Choose \(\delta = \text{min} \left{\right. \delta_{1} , \delta_{2} \left.\right} .\)

Assume \(0 < \left|\right. x - a \left|\right. < \delta .\)

Thus,

\[0 < \left|\right. x - a \left|\right. < \delta_{1} \text{and} 0 < \left|\right. x - a \left|\right. < \delta_{2} .\]

Hence,

\[\begin{aligned} \left|\right. \left(\right. f \left(\right. x \left.\right) + g \left(\right. x \left.\right) \left.\right) - \left(\right. L + M \left.\right) \left|\right. & = \left|\right. \left(\right. f \left(\right. x \left.\right) - L \left.\right) + \left(\right. g \left(\right. x \left.\right) - M \left.\right) \left|\right. \\ & \leq \left|\right. f \left(\right. x \left.\right) - L \left|\right. + \left|\right. g \left(\right. x \left.\right) - M \left|\right. \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{aligned}\]

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We now explore what it means for a limit not to exist. The limit \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right)\) does not exist if there is no real number L for which \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = L .\) Thus, for all real numbers L, \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) \neq L .\) To understand what this means, we look at each part of the definition of \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = L\) together with its opposite. A translation of the definition is given in Table 2.10.

Definition	Opposite
1. For every \(\epsilon > 0 ,\)	1. There exists \(\epsilon > 0\) so that
2. there exists a \(\delta > 0 ,\) so that	2. for every \(\delta > 0 ,\)
3. if \(0 < \left\|\right. x - a \left\|\right. < \delta ,\) then \(\left\|\right. f \left(\right. x \left.\right) - L \left\|\right. < \epsilon .\)	3. There is an x satisfying \(0 < \left\|\right. x - a \left\|\right. < \delta\) so that \(\left\|\right. f \left(\right. x \left.\right) - L \left\|\right. \geq \epsilon .\)

Table 2.10 Translation of the Definition of \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = L\) and its Opposite

Finally, we may state what it means for a limit not to exist. The limit \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right)\) does not exist if for every real number L, there exists a real number \(\epsilon > 0\) so that for all \(\delta > 0 ,\) there is an x satisfying \(0 < \left|\right. x - a \left|\right. < \delta ,\) so that \(\left|\right. f \left(\right. x \left.\right) - L \left|\right. \geq \epsilon .\) Let’s apply this in Example 2.43 to show that a limit does not exist.

Example 2.43

Showing That a Limit Does Not Exist

Show that \(\underset{x \rightarrow 0}{\text{lim}} \frac{\left|\right. x \left|\right.}{x}\) does not exist. The graph of \(f \left(\right. x \left.\right) = \left|\right. x \left|\right. / x\) is shown here:

A graph of a function with two segments. The first exists for x<0, and it is a line with no slope that ends at the y axis in an open circle at (0,-1). The second exists for x>0, and it is a line with no slope that begins at the y axis in an open circle (1,0).

Solution

Suppose that L is a candidate for a limit. Choose \(\epsilon = 1 / 2 .\)

Let \(\delta > 0 .\) Either \(L \geq 0\) or \(L < 0 .\) If \(L \geq 0 ,\) then let \(x = - \delta / 2 .\) Thus,

\[\left|\right. x - 0 \left|\right. = \left|\right. - \frac{\delta}{2} - 0 \left|\right. = \frac{\delta}{2} < \delta\]

and

\[\left|\right. \frac{\left|\right. - \frac{\delta}{2} \left|\right.}{- \frac{\delta}{2}} - L \left|\right. = \left|\right. −1 - L \left|\right. = L + 1 \geq 1 > \frac{1}{2} = \epsilon .\]

On the other hand, if \(L < 0 ,\) then let \(x = \delta / 2 .\) Thus,

\[\left|\right. x - 0 \left|\right. = \left|\right. \frac{\delta}{2} - 0 \left|\right. = \frac{\delta}{2} < \delta\]

and

\[\left|\right. \frac{\left|\right. \frac{\delta}{2} \left|\right.}{\frac{\delta}{2}} - L \left|\right. = \left|\right. 1 - L \left|\right. = \left|\right. L \left|\right. + 1 \geq 1 > \frac{1}{2} = \epsilon .\]

Thus, for any value of L, \(\underset{x \rightarrow 0}{\text{lim}} \frac{\left|\right. x \left|\right.}{x} \neq L .\)