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 $g(x)=|x-2|/(x-2)$ introduced at the beginning of the section (see (Figure)(b)). As we pick values of $x$ close to 2, $g(x)$ does not approach a single value, so the limit as $x$ approaches 2 does not exist—that is, $\underset{x\to 2}{\lim}g(x)$ DNE. However, this statement alone does not give us a complete picture of the behavior of the function around the $x$-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), $g(x)=-1$. Thus, as $x$ approaches 2 from the left, $g(x)$ approaches −1. Mathematically, we say that the limit as $x$ approaches 2 from the left is −1. Symbolically, we express this idea as

Similarly, as $x$ approaches 2 from the right (or from the positive side), $g(x)$ 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 $f(x)=\begin{cases} x+1, & \text{if} \, x < 2 \\ x^2-4, & \text{if} \, x \ge 2 \end{cases}$, evaluate each of the following limits.

1. $\underset{x\to 2^-}{\lim}f(x)$
2. $\underset{x\to 2^+}{\lim}f(x)$

Answer:

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

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

$x$	$f(x)=x+1$		$x$	$f(x)=x^2-4$
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. $\underset{x\to 2^-}{\lim}f(x)=3$ and b. $\underset{x\to 2^+}{\lim}f(x)=0$. Therefore, the (two-sided) limit of $f(x)$ does not exist at $x=2$. (Figure) shows a graph of $f(x)$ and reinforces our conclusion about these limits.

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

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 $\underset{x\to a^-}{\lim}h(x)$, we require the function $h(x)$ to be defined over an open interval of the form $(b,a)$; for a limit of the form $\underset{x\to a^+}{\lim}h(x)$, we require the function $h(x)$ to be defined over an open interval of the form $(a,c)$. (Figure) illustrates this point.

Evaluating a One-Sided Limit Using the Limit Laws

Evaluate each of the following limits, if possible.

1. $\underset{x\to 3^-}{\lim}\sqrt{x-3}$
2. $\underset{x\to 3^+}{\lim}\sqrt{x-3}$

Answer:

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

Figure 2. The graph shows the function $f(x)=\sqrt{x-3}$.

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

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

1. $\underset{x\to 2^-}{\lim}f(x)$
2. $\underset{x\to 2^+}{\lim}f(x)$
3. $\underset{x\to 2}{\lim}f(x)$

Answer:

(Figure) illustrates the function $f(x)$ and aids in our understanding of these limits.

Figure 3. This graph shows the function $f(x)$.

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

evaluating one-sided limits from a graph

Use the graph of $f(x)$ in (Figure) to determine each of the following values:

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

Figure 10. The graph shows $f(x)$.

Answer:

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

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

We will now recall the Squeeze Theorem to prove two special limits.  The first of these limits is $\underset{\theta \to 0}{\lim} \sin \theta$. Consider the unit circle shown in (Figure). In the figure, we see that $\sin \theta$ is the $y$-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 $0<\theta <\frac{\pi }{2}, \, 0 < \sin \theta < \theta$.

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

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

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

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

We now take a look at a limit that plays an important role in later chapters—namely, $\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta}$. 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 $\theta$ in this new triangle is $\tan \theta$. Thus, we see that for $0 < \theta < \frac{\pi}{2}, \, \sin \theta < \theta < \tan \theta$.

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

By dividing by $\sin \theta$ in all parts of the inequality, we obtain

Equivalently, we have

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

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