One-Sided Limits

Sometimes indicating that the limit of a function fails to exist at a point does not provide us with enough information about the behavior of the function at that particular point. To see this, we now revisit the function [latex]g(x)=|x-2|/(x-2)[/latex] introduced at the beginning of the section (see (Figure)(b)). As we pick values of [latex]x[/latex] close to 2, [latex]g(x)[/latex] does not approach a single value, so the limit as [latex]x[/latex] approaches 2 does not exist—that is, [latex]\underset{x\to 2}{\lim}g(x)[/latex] DNE. However, this statement alone does not give us a complete picture of the behavior of the function around the [latex]x[/latex]-value 2. To provide a more accurate description, we introduce the idea of a one-sided limit. For all values to the left of 2 (or the negative side of 2), [latex]g(x)=-1[/latex]. Thus, as [latex]x[/latex] approaches 2 from the left, [latex]g(x)[/latex] approaches −1. Mathematically, we say that the limit as [latex]x[/latex] approaches 2 from the left is −1. Symbolically, we express this idea as

Similarly, as [latex]x[/latex] approaches 2 from the right (or from the positive side), [latex]g(x)[/latex] approaches 1. Symbolically, we express this idea as

We can now present an informal definition of one-sided limits.

Evaluating One-Sided Limits

For the function [latex]f(x)=\begin{cases} x+1, & \text{if} \, x < 2 \\ x^2-4, & \text{if} \, x \ge 2 \end{cases}[/latex], evaluate each of the following limits.

1. [latex]\underset{x\to 2^-}{\lim}f(x)[/latex]
2. [latex]\underset{x\to 2^+}{\lim}f(x)[/latex]

Answer:

We can use tables of functional values again (Figure). Observe that for values of [latex]x[/latex] less than 2, we use [latex]f(x)=x+1[/latex] and for values of [latex]x[/latex] greater than 2, we use [latex]f(x)=x^2-4[/latex].

Table of Functional Values for [latex]f(x)=\begin{cases} x+1, & \text{if} \, x < 2 \\ x^2-4, & \text{if} \, x \ge 2 \end{cases}[/latex]

[latex]x[/latex]	[latex]f(x)=x+1[/latex]		[latex]x[/latex]	[latex]f(x)=x^2-4[/latex]
1.9	2.9		2.1	0.41
1.99	2.99		2.01	0.0401
1.999	2.999		2.001	0.004001
1.9999	2.9999		2.0001	0.00040001
1.99999	2.99999		2.00001	0.0000400001

Based on this table, we can conclude that a. [latex]\underset{x\to 2^-}{\lim}f(x)=3[/latex] and b. [latex]\underset{x\to 2^+}{\lim}f(x)=0[/latex]. Therefore, the (two-sided) limit of [latex]f(x)[/latex] does not exist at [latex]x=2[/latex]. (Figure) shows a graph of [latex]f(x)[/latex] and reinforces our conclusion about these limits.

Figure 7. The graph of [latex]f(x)=\begin{cases} x+1, & \text{if} \, x < 2 \\ x^2-4, & \text{if} \, x \ge 2 \end{cases}[/latex] has a break at [latex]x=2[/latex].

Simple modifications in the limit laws allow us to apply them to one-sided limits. For example, to apply the limit laws to a limit of the form [latex]\underset{x\to a^-}{\lim}h(x)[/latex], we require the function [latex]h(x)[/latex] to be defined over an open interval of the form [latex](b,a)[/latex]; for a limit of the form [latex]\underset{x\to a^+}{\lim}h(x)[/latex], we require the function [latex]h(x)[/latex] to be defined over an open interval of the form [latex](a,c)[/latex]. (Figure) illustrates this point.

Evaluating a One-Sided Limit Using the Limit Laws

Evaluate each of the following limits, if possible.

1. [latex]\underset{x\to 3^-}{\lim}\sqrt{x-3}[/latex]
2. [latex]\underset{x\to 3^+}{\lim}\sqrt{x-3}[/latex]

Answer:

(Figure) illustrates the function [latex]f(x)=\sqrt{x-3}[/latex] and aids in our understanding of these limits.

Figure 2. The graph shows the function [latex]f(x)=\sqrt{x-3}[/latex].

1. The function [latex]f(x)=\sqrt{x-3}[/latex] is defined over the interval [latex][3,+\infty)[/latex]. Since this function is not defined to the left of 3, we cannot apply the limit laws to compute [latex]\underset{x\to 3^-}{\lim}\sqrt{x-3}[/latex]. In fact, since [latex]f(x)=\sqrt{x-3}[/latex] is undefined to the left of 3, [latex]\underset{x\to 3^-}{\lim}\sqrt{x-3}[/latex] does not exist.
2. Since [latex]f(x)=\sqrt{x-3}[/latex] is defined to the right of 3, the limit laws do apply to [latex]\underset{x\to 3^+}{\lim}\sqrt{x-3}[/latex]. By applying these limit laws we obtain [latex]\underset{x\to 3^+}{\lim}\sqrt{x-3}=0[/latex].

In (Figure) we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion about a two-sided limit of the same function.

Evaluating a Two-Sided Limit Using the Limit Laws

For [latex]f(x)=\begin{cases} 4x-3 & \text{if} \, x<2 \\ (x-3)^2 & \text{if} \, x \ge 2 \end{cases}[/latex] evaluate each of the following limits:

1. [latex]\underset{x\to 2^-}{\lim}f(x)[/latex]
2. [latex]\underset{x\to 2^+}{\lim}f(x)[/latex]
3. [latex]\underset{x\to 2}{\lim}f(x)[/latex]

Answer:

(Figure) illustrates the function [latex]f(x)[/latex] and aids in our understanding of these limits.

Figure 3. This graph shows the function [latex]f(x)[/latex].

1. Since [latex]f(x)=4x-3[/latex] for all [latex]x[/latex] in [latex](−\infty,2)[/latex], replace [latex]f(x)[/latex] in the limit with [latex]4x-3[/latex] and apply the limit laws:
   [latex]\underset{x\to 2^-}{\lim}f(x)=\underset{x\to 2^-}{\lim}(4x-3)=5[/latex].
2. Since [latex]f(x)=(x-3)^2[/latex] for all [latex]x[/latex] in [latex](2,+\infty)[/latex], replace [latex]f(x)[/latex] in the limit with [latex](x-3)^2[/latex] and apply the limit laws:
   [latex]\underset{x\to 2^+}{\lim}f(x)=\underset{x\to 2^-}{\lim}(x-3)^2=1[/latex].
3. Since [latex]\underset{x\to 2^-}{\lim}f(x)=5[/latex] and [latex]\underset{x\to 2^+}{\lim}f(x)=1[/latex], we conclude that [latex]\underset{x\to 2}{\lim}f(x)[/latex] does not exist.

evaluating one-sided limits from a graph

Use the graph of [latex]f(x)[/latex] in (Figure) to determine each of the following values:

1. [latex]\underset{x\to -4^-}{\lim}f(x); \, \underset{x\to -4^+}{\lim}f(x); \, \underset{x\to -4}{\lim}f(x); \, f(-4)[/latex]
2. [latex]\underset{x\to -2^-}{\lim}f(x); \, \underset{x\to -2^+}{\lim}f(x); \, \underset{x\to -2}{\lim}f(x); \, f(-2)[/latex]
3. [latex]\underset{x\to 1^-}{\lim}f(x); \, \underset{x\to 1^+}{\lim}f(x); \, \underset{x\to 1}{\lim}f(x); \, f(1)[/latex]

Figure 10. The graph shows [latex]f(x)[/latex].

Answer:

Using (Figure) and the graph for reference, we arrive at the following values:

1. [latex]\underset{x\to -4^-}{\lim}f(x)=0; \, \underset{x\to -4^+}{\lim}f(x)=0; \, \underset{x\to -4}{\lim}f(x)=0; \, f(-4)=0[/latex]
2. [latex]\underset{x\to -2^-}{\lim}f(x)=3; \, \underset{x\to -2^+}{\lim}f(x)=3; \, \underset{x\to -2}{\lim}f(x)=3; \, f(-2)[/latex] is undefined
3. [latex]\underset{x\to 1^-}{\lim}f(x)=6; \, \underset{x\to 1^+}{\lim}f(x)=3; \, \underset{x\to 1}{\lim}f(x)=DNE[/latex]; [latex]f(1)=6[/latex]

We will now recall the Squeeze Theorem to prove two special limits.  The first of these limits is [latex]\underset{\theta \to 0}{\lim} \sin \theta[/latex]. Consider the unit circle shown in (Figure). In the figure, we see that [latex] \sin \theta [/latex] is the [latex]y[/latex]-coordinate on the unit circle and it corresponds to the line segment shown in blue. The radian measure of angle θ is the length of the arc it subtends on the unit circle. Therefore, we see that for [latex]0<\theta <\frac{\pi }{2}, \, 0 < \sin \theta < \theta[/latex].

Figure 6. The sine function is shown as a line on the unit circle.

Because [latex]\underset{\theta \to 0^+}{\lim}0=0[/latex] and [latex]\underset{\theta \to 0^+}{\lim}\theta =0[/latex], by using the Squeeze Theorem we conclude that

To see that [latex]\underset{\theta \to 0^-}{\lim} \sin \theta =0[/latex] as well, observe that for [latex]-\frac{\pi }{2} < \theta <0, \, 0 < −\theta < \frac{\pi}{2}[/latex] and hence, [latex]0 < \sin(-\theta) < −\theta[/latex]. Consequently, [latex]0 < -\sin \theta < −\theta[/latex] It follows that [latex]0 > \sin \theta > \theta[/latex]. An application of the Squeeze Theorem produces the desired limit. Thus, since [latex]\underset{\theta \to 0^+}{\lim} \sin \theta =0[/latex] and [latex]\underset{\theta \to 0^-}{\lim} \sin \theta =0[/latex],

Next, using the identity [latex] \cos \theta =\sqrt{1-\sin^2 \theta}[/latex] for [latex]-\frac{\pi}{2}<\theta <\frac{\pi}{2}[/latex], we see that

We now take a look at a limit that plays an important role in later chapters—namely, [latex]\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta}[/latex]. To evaluate this limit, we use the unit circle in (Figure). Notice that this figure adds one additional triangle to (Figure). We see that the length of the side opposite angle [latex]\theta[/latex] in this new triangle is [latex]\tan \theta[/latex]. Thus, we see that for [latex]0 < \theta < \frac{\pi}{2}, \, \sin \theta < \theta < \tan \theta[/latex].

Figure 7. The sine and tangent functions are shown as lines on the unit circle.

By dividing by [latex]\sin \theta [/latex] in all parts of the inequality, we obtain

Equivalently, we have

Since [latex]\underset{\theta \to 0^+}{\lim}1=1=\underset{\theta \to 0^+}{\lim}\cos \theta[/latex], we conclude that [latex]\underset{\theta \to 0^+}{\lim}\frac{\sin \theta}{\theta}=1[/latex]. By applying a manipulation similar to that used in demonstrating that [latex]\underset{\theta \to 0^-}{\lim}\sin \theta =0[/latex], we can show that [latex]\underset{\theta \to 0^-}{\lim}\frac{\sin \theta}{\theta}=1[/latex]. Thus,

In (Figure) we use this limit to establish [latex]\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}=0[/latex]. This limit also proves useful in later chapters.