One-Sided and Infinite Limits

Just as we first gained an intuitive understanding of limits and then moved on to a more rigorous definition of a limit, we now revisit one-sided limits. To do this, we modify the epsilon-delta definition of a limit to give formal epsilon-delta definitions for limits from the right and left at a point. These definitions only require slight modifications from the definition of the limit. In the definition of the limit from the right, the inequality \(0 < x - a < \delta\) replaces \(0 < \left|\right. x - a \left|\right. < \delta ,\) which ensures that we only consider values of x that are greater than (to the right of) a. Similarly, in the definition of the limit from the left, the inequality \(- \delta < x - a < 0\) replaces \(0 < \left|\right. x - a \left|\right. < \delta ,\) which ensures that we only consider values of x that are less than (to the left of) a.

Definition

Limit from the Right: Let \(f \left(\right. x \left.\right)\) be defined over an open interval of the form \(\left(\right. a , b \left.\right)\) where \(a < b .\) Then,

\[\underset{x \rightarrow a^{+}}{\text{lim}} f \left(\right. x \left.\right) = L\]

if for every \(\epsilon > 0 ,\) there exists a \(\delta > 0\) such that if \(0 < x - a < \delta ,\) then \(\left|\right. f \left(\right. x \left.\right) - L \left|\right. < \epsilon .\)

Limit from the Left: Let \(f \left(\right. x \left.\right)\) be defined over an open interval of the form \(\left(\right. b , a \left.\right)\) where \(b < a .\) Then,

\[\underset{x \rightarrow a^{-}}{\text{lim}} f \left(\right. x \left.\right) = L\]

if for every \(\epsilon > 0 ,\) there exists a \(\delta > 0\) such that if \(- \delta < x - a < 0 ,\) then \(\left|\right. f \left(\right. x \left.\right) - L \left|\right. < \epsilon .\)

Example 2.44

Proving a Statement about a Limit From the Right

Prove that \(\underset{x \rightarrow 4^{+}}{\text{lim}} \sqrt{x - 4} = 0 .\)

Solution

Let \(\epsilon > 0 .\)

Choose \(\delta = \epsilon^{2} .\) Since we ultimately want \(\left|\right. \sqrt{x - 4} - 0 \left|\right. < \epsilon ,\) we manipulate this inequality to get \(\sqrt{x - 4} < \epsilon\) or, equivalently, \(0 < x - 4 < \epsilon^{2} ,\) making \(\delta = \epsilon^{2}\) a clear choice. We may also determine \(\delta\) geometrically, as shown in Figure 2.42.

A graph showing how to find delta for the above proof. The function f(x) = sqrt(x-4) is drawn for x > 4. Since the proposed limit is 0, lines y = 0 + epsilon and y = 0 – epsilon are drawn  in blue. Since only the top blue line corresponding to y = 0 + epsilon intersects the function, one red line is drawn from the point of intersection to the x axis. This x value is found by solving sqrt(x-4) = epsilon, or x = epsilon squared + 4. Delta is then the distance between this point and 4, which is epsilon squared.
Figure 2.42 This graph shows how we find δ for the proof in Example 2.44.

Assume \(0 < x - 4 < \delta .\) Thus, \(0 < x - 4 < \epsilon^{2} .\) Hence, \(0 < \sqrt{x - 4} < \epsilon .\) Finally, \(\left|\right. \sqrt{x - 4} - 0 \left|\right. < \epsilon .\)

Therefore, \(\underset{x \rightarrow 4^{+}}{\text{lim}} \sqrt{x - 4} = 0 .\)

Checkpoint 2.30

Find \(\delta\) corresponding to ε for a proof that \(\underset{x \rightarrow 1^{-}}{\text{lim}} \sqrt{1 - x} = 0 .\)

We conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by pursuing a formal definition of infinite limits. To have \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = + \infty ,\) we want the values of the function \(f \left(\right. x \left.\right)\) to get larger and larger as x approaches a. Instead of the requirement that \(\left|\right. f \left(\right. x \left.\right) - L \left|\right. < \epsilon\) for arbitrarily small ε when \(0 < \left|\right. x - a \left|\right. < \delta\) for small enough \(\delta ,\) we want \(f \left(\right. x \left.\right) > M\) for arbitrarily large positive M when \(0 < \left|\right. x - a \left|\right. < \delta\) for small enough \(\delta .\) Figure 2.43 illustrates this idea by showing the value of \(\delta\) for successively larger values of M.

Two graphs side by side. Each graph contains two curves above the x axis separated by an asymptote at x=a. The curves on the left go to infinity as x goes to a and to 0 as x goes to negative infinity. The curves on the right go to infinity as x goes to a and to 0 as x goes to infinity. The first graph has a value M greater than zero marked on the y axis and a horizontal line drawn from there (y=M) to intersect with both curves. Lines are drawn down from the points of intersection to the x axis. Delta is the smaller of the distances between point a and these new spots on the x axis. The same lines are drawn on the second graph, but this M is larger, and the distances from the x axis intersections to point a are smaller.
Figure 2.43 These graphs plot values of \(\delta\) for M to show that \(\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = + \infty .\)

Definition

Let \(f \left(\right. x \left.\right)\) be defined for all \(x \neq a\) in an open interval containing a. Then, we have an infinite limit

\[\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = + \infty\]

if for every \(M > 0 ,\) there exists \(\delta > 0\) such that if \(0 < \left|\right. x - a \left|\right. < \delta ,\) then \(f \left(\right. x \left.\right) > M .\)

Let \(f \left(\right. x \left.\right)\) be defined for all \(x \neq a\) in an open interval containing a. Then, we have a negative infinite limit

\[\underset{x \rightarrow a}{\text{lim}} f \left(\right. x \left.\right) = − \infty\]

if for every \(M > 0 ,\) there exists \(\delta > 0\) such that if \(0 < \left|\right. x - a \left|\right. < \delta ,\) then \(f \left(\right. x \left.\right) < − M .\)

This lesson is part of:

Limits

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