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,
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,
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.
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.
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
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
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