Use the epsilon-delta definition to prove the limit laws
Describe the epsilon-delta definitions of one-sided limits and infinite limits
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
The triangle inequality states that if [latex]a[/latex] and [latex]b[/latex] are any real numbers, then [latex]|a+b|\le |a|+|b|[/latex].
Proof
We prove the following limit law: If [latex]\underset{x\to a}{\lim}f(x)=L[/latex] and [latex]\underset{x\to a}{\lim}g(x)=M[/latex], then [latex]\underset{x\to a}{\lim}(f(x)+g(x))=L+M[/latex].
Let [latex]\varepsilon >0[/latex].
Choose [latex]\delta_1>0[/latex] so that if [latex]0<|x-a|<\delta_1[/latex], then [latex]|f(x)-L|<\varepsilon/2[/latex].
Choose [latex]\delta_2>0[/latex] so that if [latex]0<|x-a|<\delta_2[/latex], then [latex]|g(x)-M|<\varepsilon/2[/latex].
We now explore what it means for a limit not to exist. The limit [latex]\underset{x\to a}{\lim}f(x)[/latex] does not exist if there is no real number [latex]L[/latex] for which [latex]\underset{x\to a}{\lim}f(x)=L[/latex]. Thus, for all real numbers [latex]L[/latex], [latex]\underset{x\to a}{\lim}f(x)\ne L[/latex]. To understand what this means, we look at each part of the definition of [latex]\underset{x\to a}{\lim}f(x)=L[/latex] together with its opposite. A translation of the definition is given in the table below.
Translation of the Definition of [latex]\underset{x\to a}{\lim}f(x)=L[/latex] and its Opposite
Definition
Opposite
1. For every [latex]\varepsilon >0[/latex],
1. There exists [latex]\varepsilon >0[/latex] so that
2. there exists a [latex]\delta >0[/latex] so that
2. for every [latex]\delta >0[/latex],
3. if [latex]0<|x-a|<\delta[/latex], then [latex]|f(x)-L|<\varepsilon[/latex].
3. There is an [latex]x[/latex] satisfying [latex]0<|x-a|<\delta[/latex] so that [latex]|f(x)-L|\ge \varepsilon[/latex].
Finally, we may state what it means for a limit not to exist. The limit [latex]\underset{x\to a}{\lim}f(x)[/latex] does not exist if for every real number [latex]L[/latex], there exists a real number [latex]\varepsilon >0[/latex] so that for all [latex]\delta >0[/latex], there is an [latex]x[/latex] satisfying [latex]0<|x-a|<\delta[/latex], so that [latex]|f(x)-L|\ge \varepsilon[/latex]. Let’s apply this in the example to show that a limit does not exist.
Example: Showing That a Limit Does Not Exist
Show that [latex]\underset{x\to 0}{\lim}\dfrac{|x|}{x}[/latex] does not exist. The graph of [latex]f(x)=\dfrac{|x|}{x}[/latex] is shown here:
Figure 4.
Show Solution
Suppose that [latex]L[/latex] is a candidate for a limit. Choose [latex]\varepsilon =\frac{1}{2}[/latex].
Let [latex]\delta >0[/latex]. Either [latex]L\ge 0[/latex] or [latex]L<0[/latex]. If [latex]L\ge 0[/latex], then let [latex]x=-\delta/2[/latex]. Thus,
Thus, for any value of [latex]L[/latex], [latex]\underset{x\to 0}{\lim}\frac{|x|}{x}\ne L[/latex].
Watch the following video to see the worked solution to Example: Showing That a Limit Does Not Exist.
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Now that we have proven limits, we can now apply them with actual numbers for [latex]\varepsilon[/latex] and [latex]\delta[/latex]. Think of [latex]\varepsilon[/latex] as the error in the [latex]x[/latex]-direction and [latex]\delta[/latex] to be the error in the [latex]y[/latex]-direction. These have applications in engineering when these errors are considered tolerances. We want to know what the error intervals are, and we are trying to minimize these errors.
Example: Finding deltas algebraically, Part 1
Find an open interval about [latex]x_0[/latex] on which the inequality [latex]|f(x)-L| < 0[/latex] holds. Then give the largest value [latex]\delta > 0[/latex] such that for all [latex]x[/latex] satisfying [latex]0 < |x-x_0| < \delta[/latex] the inequality [latex]|f(x)-L| < \varepsilon[/latex] holds.
First we will need to start with the inequality [latex]|f(x)-L| < \varepsilon[/latex] and plug in our numbers. Then we will solve for x.[latex]|2x-8 - 6| < \varepsilon[/latex][latex]|2x-14| < \varepsilon[/latex][latex]-0.14 < 2x - 14 < 0.14[/latex][latex]13.86 < 2x < 14.14[/latex][latex]6.93 < x < 7.07[/latex]Therefore, the interval is [latex](7.93,8.07)[/latex]For the second answer, we will start with [latex]0 < |x-x_0| < \delta[/latex]. We will plug in our value and solve:[latex]|x-7| < \delta[/latex][latex]-\delta < x-7 < \delta[/latex][latex]7-\delta < x < 7+\delta[/latex]Now we will set each piece equal to the endpoints we found above.[latex]7-\delta=7.93[/latex] and [latex]7+\delta=8.07[/latex]After solving we will get the same answer for each equation: [latex]\delta=0.07[/latex].
Example: Finding deltas algebraically, Part 2
Find an open interval about [latex]x_0[/latex] on which the inequality [latex]|f(x)-L| < 0[/latex] holds. Then give the largest value [latex]\delta > 0[/latex] such that for all [latex]x[/latex] satisfying [latex]0 < |x-x_0| < \delta[/latex] the inequality [latex]|f(x)-L| < \varepsilon[/latex] holds.
First we will need to start with the inequality [latex]|f(x)-L| < \varepsilon[/latex] and plug in our numbers. Then we will solve for x.[latex]|\sqrt(x+4) - 3| < \varepsilon[/latex][latex]|\sqrt{x+4} - 3| < 1[/latex][latex]-1 < \sqrt{x+4} - 3 < 1[/latex][latex]2 < \sqrt{x+4} < 4[/latex][latex]4 < x + 4 < 16[/latex][latex]0 < x + 4 < 12[/latex]Therefore, the interval is [latex](0,12)[/latex].For the second answer, we will start with [latex]0 < |x-x_0| < \delta[/latex]. We will plug in our value and solve:[latex]|x-5| < \delta[/latex][latex]-\delta < x-5 < \delta[/latex][latex]5-\delta < x < 5+\delta[/latex]Now we will set each piece equal to the endpoints we found above.[latex]5-\delta=0[/latex] and [latex]5+\delta=12[/latex]
After solving we will get [latex]\delta=5 \text{ and }\delta=7[/latex]. Since the question is asking for the smallest interval, we choose the smaller number. Therefore the answer is [latex]\delta=5[/latex].
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 [latex]0
Definition
Limit from the Right: Let [latex]f(x)[/latex] be defined over an open interval of the form [latex](a,b)[/latex] where [latex]a
[latex]\underset{x\to a^+}{\lim}f(x)=L[/latex]
if for every [latex]\varepsilon >0[/latex], there exists a [latex]\delta >0[/latex] such that if [latex]0
Limit from the Left: Let [latex]f(x)[/latex] be defined over an open interval of the form [latex](a,b)[/latex] where [latex]a
[latex]\underset{x\to b^-}{\lim}f(x)=L[/latex]
if for every [latex]\varepsilon >0[/latex], there exists a [latex]\delta >0[/latex] such that if [latex]0
Example: Proving a Statement about a Limit From the Right
Prove that [latex]\underset{x\to 4^+}{\lim}\sqrt{x-4}=0[/latex].
Show Solution
Let [latex]\varepsilon >0.[/latex]
Choose [latex]\delta =\varepsilon^2[/latex]. Since we ultimately want [latex]|\sqrt{x-4}-0|<\varepsilon[/latex], we manipulate this inequality to get [latex]\sqrt{x-4}<\varepsilon[/latex] or, equivalently, [latex]0
Figure 5. This graph shows how we find [latex]\delta[/latex] for the proof of this example.
Watch the following video to see the worked solution to Example: Proving a Statement about a Limit From the Right.
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Find [latex]\delta[/latex] corresponding to [latex]\varepsilon[/latex] for a proof that [latex]\underset{x\to 1^-}{\lim}\sqrt{1-x}=0[/latex].
Hint
Sketch the graph and use Figure 5 as a solving guide.
Show Solution
[latex]\delta =\varepsilon^2[/latex]
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 [latex]\underset{x\to a}{\lim}f(x)=+\infty[/latex], we want the values of the function [latex]f(x)[/latex] to get larger and larger as [latex]x[/latex] approaches [latex]a[/latex]. Instead of the requirement that [latex]|f(x)-L|<\varepsilon[/latex] for arbitrarily small [latex]\varepsilon[/latex] when [latex]0<|x-a|<\delta[/latex] for small enough [latex]\delta[/latex], we want [latex]f(x)>M[/latex] for arbitrarily large positive [latex]M[/latex] when [latex]0<|x-a|<\delta[/latex] for small enough [latex]\delta[/latex]. Figure 6 illustrates this idea by showing the value of [latex]\delta[/latex] for successively larger values of [latex]M[/latex].
Figure 6. These graphs plot values of [latex]\delta[/latex] for [latex]M[/latex] to show that [latex]\underset{x\to a}{\lim}f(x)=+\infty[/latex].
Definition
Let [latex]f(x)[/latex] be defined for all [latex]x\ne a[/latex] in an open interval containing [latex]a[/latex]. Then, we have an infinite limit
if for every [latex]M>0[/latex], there exists [latex]\delta >0[/latex] such that if [latex]0<|x-a|<\delta[/latex], then [latex]f(x)>M[/latex].
Let [latex]f(x)[/latex] be defined for all [latex]x\ne a[/latex] in an open interval containing [latex]a[/latex]. Then, we have a negative infinite limit
