2.3 One-Sided Limits

Learning Objectives

  • Define one-sided limits and provide examples.
  • Explain the relationship between one-sided and two-sided limits.
  • Define a vertical asymptote.

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

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

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

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

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

Definition

We define two types of one-sided limits.

Limit from the left: Let [latex]f(x)[/latex] be a function defined at all values in an open interval of the form z, and let [latex]L[/latex] be a real number. If the values of the function [latex]f(x)[/latex] approach the real number [latex]L[/latex] as the values of [latex]x[/latex] (where [latex]x<a[/latex]) approach the number [latex]a[/latex], then we say that [latex]L[/latex] is the limit of [latex]f(x)[/latex] as [latex]x[/latex] approaches a from the left. Symbolically, we express this idea as

[latex]\underset{x\to a^-}{\lim}f(x)=L[/latex].

Limit from the right: Let [latex]f(x)[/latex] be a function defined at all values in an open interval of the form [latex](a,c)[/latex], and let [latex]L[/latex] be a real number. If the values of the function [latex]f(x)[/latex] approach the real number [latex]L[/latex] as the values of [latex]x[/latex] (where [latex]x>a[/latex]) approach the number [latex]a[/latex], then we say that [latex]L[/latex] is the limit of [latex]f(x)[/latex] as [latex]x[/latex] approaches [latex]a[/latex] from the right. Symbolically, we express this idea as

[latex]\underset{x\to a^+}{\lim}f(x)=L[/latex].

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]

Use a table of functional values to estimate the following limits, if possible.

  1. [latex]\underset{x\to 2^-}{\lim}\frac{|x^2-4|}{x-2}[/latex]
  2. [latex]\underset{x\to 2^+}{\lim}\frac{|x^2-4|}{x-2}[/latex]

Hint

  1. Use [latex]x[/latex]-values 1.9, 1.99, 1.999, 1.9999, 1.9999 to estimate [latex]\underset{x\to 2^-}{\lim}\frac{|x^2-4|}{x-2}[/latex].
  2. Use [latex]x[/latex]-values 2.1, 2.01, 2.001, 2.0001, 2.00001 to estimate [latex]\underset{x\to 2^+}{\lim}\frac{|x^2-4|}{x-2}[/latex].
    (These tables are available from a previous Checkpoint problem.)

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]

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]

Graph [latex]f(x)=\begin{cases} -x-2 & \text{if} \, x<-1 \\ 2 & \text{if} \, x = -1 \\ x^3 & \text{if} \, x > -1 \end{cases}[/latex] and evaluate [latex]\underset{x\to -1^-}{\lim}f(x)[/latex].

Hint

Use the method in (Figure) to evaluate the limit.

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]
The graph of a function f(x) described by the above limits and values. There is a smooth curve for values below x=-2; at (-2, 3), there is an open circle. There is a smooth curve between (-2, 1] with a closed circle at (1,6). There is an open circle at (1,3), and a smooth curve stretching from there down asymptotically to negative infinity along x=3. The function also curves asymptotically along x=3 on the other side, also stretching to negative infinity. The function then changes concavity in the first quadrant around y=4.5 and continues up.

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

A diagram of the unit circle in the x,y plane – it is a circle with radius 1 and center at the origin. A specific point (cos(theta), sin(theta)) is labeled in quadrant 1 on the edge of the circle. This point is one vertex of a right triangle inside the circle, with other vertices at the origin and (cos(theta), 0). As such, the lengths of the sides are cos(theta) for the base and sin(theta) for the height, where theta is the angle created by the hypotenuse and base. The radian measure of angle theta is the length of the arc it subtends on the unit circle. The diagram shows that for 0 < theta < pi/2, 0 < sin(theta) < theta.

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

[latex]\underset{\theta \to 0^+}{\lim} \sin \theta =0[/latex].

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],

[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

[latex]\underset{\theta \to 0}{\lim} \cos \theta =\underset{\theta \to 0}{\lim}\sqrt{1-\sin^2 \theta }=1[/latex].

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].

The same diagram as the previous one. However, the triangle is expanded. The base is now from the origin to (1,0). The height goes from (1,0) to (1, tan(theta)). The hypotenuse goes from the origin to (1, tan(theta)). As such, the height is now tan(theta). It shows that for 0 < theta < 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 [latex]\sin \theta [/latex] in all parts of the inequality, we obtain

[latex]1 < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta}[/latex].

Equivalently, we have

[latex]1 > \frac{\sin \theta}{\theta} > \cos \theta[/latex].

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,

[latex]\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta}=1[/latex].

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.

Evaluating an Important Trigonometric Limit

Evaluate [latex]\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}[/latex].

Evaluate [latex]\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\sin \theta}[/latex].

Hint

Multiply numerator and denominator by [latex]1+ \cos \theta[/latex].

Special Trigonometric Limits

Summarizing our work from above we have two special limits:      [latex]\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta}=1[/latex]    and   [latex]\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}=0[/latex]

 Evaluating a Limit involving [latex]\frac{\sin \theta}{\theta}[/latex] and [latex]\frac{1-\cos\theta}{\theta}[/latex]

Evaluate [latex]\underset{\theta \to 0}{\lim}\frac{\sin 3\theta}{2\theta}[/latex].

Evaluate [latex]\underset{\theta \to 0}{\lim}\frac{1- \cos 3\theta}{9\theta}[/latex].

Hint

Divide the numerator and denominator by [latex]3[/latex].

Key Equations

  • One-Sided Limits
    [latex]\underset{x\to a^-}{\lim}f(x)=L[/latex]
    [latex]\underset{x\to a^+}{\lim}f(x)=L[/latex]
  • Important Limits
    [latex]\underset{\theta \to 0}{\lim}\frac{\sin \theta}{\theta}=1[/latex]
    [latex]\underset{\theta \to 0}{\lim}\frac{1- \cos \theta}{\theta}=0[/latex]

In the following exercises, set up a table of values to find the indicated limit. Round to eight digits.

1. [T] [latex]\underset{t\to 0^+}{\lim}\frac{\cos t}{t}[/latex]

[latex]t[/latex] [latex]\frac{\cos t}{t}[/latex]
0.1 a.
0.01 b.
0.001 c.
0.0001 d.

2. [T] [latex]\underset{x\to 2^-}{\lim}\frac{1-\frac{2}{x}}{x^2-4}[/latex]

[latex]x[/latex] [latex]\frac{1-\frac{2}{x}}{x^2-4}[/latex] [latex]x[/latex] [latex]\frac{1-\frac{2}{x}}{x^2-4}[/latex]
1.9 a. 2.1 e.
1.99 b. 2.01 f.
1.999 c. 2.001 g.
1.9999 d. 2.0001 h.

In the following exercises, set up a table of values and round to eight significant digits. Based on the table of values, make a guess about what the limit is. Then, use a calculator to graph the function and determine the limit. Was the conjecture correct? If not, why does the method of tables fail?

3. [T] [latex]\underset{\theta \to 0^-}{\lim}\sin (\frac{\pi }{\theta })[/latex]

θ [latex] \sin (\frac{\pi }{\theta })[/latex] θ [latex] \sin (\frac{\pi }{\theta })[/latex]
−0.1 a. 0.1 e.
−0.01 b. 0.01 f.
−0.001 c. 0.001 g.
−0.0001 d. 0.0001 h.

4. [T] [latex]\underset{\alpha \to 0^+}{\lim}\frac{1}{\alpha } \cos (\frac{\pi }{\alpha })[/latex]

[latex]a[/latex] [latex]\frac{1}{\alpha } \cos (\frac{\pi }{\alpha })[/latex]
0.1 a.
0.01 b.
0.001 c.
0.0001 d.

In the following exercises, use direct substitution to obtain an undefined expression. Then, use the method of (Figure) to simplify the function to help determine the limit.

5. [latex]\underset{x\to 1^+}{\lim}\frac{2x^2+7x-4}{x^2+x-2}[/latex]

6. [latex]\underset{x\to -2^-}{\lim}\frac{2x^2+7x-4}{x^2+x-2}[/latex]

7. [latex]\underset{x\to -2^+}{\lim}\frac{2x^2+7x-4}{x^2+x-2}[/latex]

8. [latex]\underset{x\to 1^-}{\lim}\frac{2x^2+7x-4}{x^2+x-2}[/latex]

In the following exercises, use a calculator to draw the graph of each piecewise-defined function and study the graph to evaluate the given limits.

9. [T] [latex]g(x)=\begin{cases} 3-x & x \le 2 \\ \frac{x}{2} & x > 2 \end{cases}[/latex]

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

10. [T] [latex]f(x)=\begin{cases} x^2 & x \le 3 \\ x+4 & x > 3 \end{cases}[/latex]

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

11. [T] [latex]g(x)=\begin{cases} x^3 - 1 & x \le 0 \\ 1 & x > 0 \end{cases}[/latex]

  1. [latex]\underset{x\to 0^-}{\lim}g(x)[/latex]
  2. [latex]\underset{x\to 0^+}{\lim}g(x)[/latex]

12. [T] [latex]h(x)=\begin{cases} x^2-2x+1 & x < 2 \\ 3 - x & x \ge 2 \end{cases}[/latex]

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

In the following exercises, use the following graphs and the limit laws to evaluate each limit.

Two graphs of piecewise functions. The upper is f(x), which has two linear segments. The first is a line with negative slope existing for x < -3. It goes toward the point (-3,0) at x= -3. The next has increasing slope and goes to the point (-3,-2) at x=-3. It exists for x > -3. Other key points are (0, 1), (-5,2), (1,2), (-7, 4), and (-9,6). The lower piecewise function has a linear segment and a curved segment. The linear segment exists for x < -3 and has decreasing slope. It goes to (-3,-2) at x=-3. The curved segment appears to be the right half of a downward opening parabola. It goes to the vertex point (-3,2) at x=-3. It crosses the y axis a little below y=-2. Other key points are (0, -7/3), (-5,0), (1,-5), (-7, 2), and (-9, 4).

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

14. [latex]\underset{x\to -3^-}{\lim}(f(x)-3g(x))[/latex]

In the following exercises, consider the graph of the function [latex]y=f(x)[/latex] shown here. Which of the statements about [latex]y=f(x)[/latex] are true and which are false? Explain why a statement is false.

A graph of a piecewise function with three segments and a point. The first segment is a curve opening upward with vertex at (-8, -6). This vertex is an open circle, and there is a closed circle instead at (-8, -3). The segment ends at (-2,3), where there is a closed circle. The second segment stretches up asymptotically to infinity along x=-2, changes direction to increasing at about (0,1.25), increases until about (2.25, 3), and decreases until (6,2), where there is an open circle. The last segment starts at (6,5), increases slightly, and then decreases into quadrant four, crossing the x axis at (10,0). All of the changes in direction are smooth curves.

15. [latex]\underset{x\to 10^-}{\lim}f(x)=0[/latex]

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

17. [latex]\underset{x\to -8^+}{\lim}f(x)=f(-8)[/latex]

18. [latex]\underset{x\to 6}{\lim}f(x)=5[/latex]

In the following exercises, use the following graph of the function [latex]y=f(x)[/latex] to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with two segments. The first segment exists for x <=1, and the second segment exists for x > 1. The first segment is linear with a slope of 1 and goes through the origin. Its endpoint is a closed circle at (1,1). The second segment is also linear with a slope of -1. It begins with the open circle at (1,2).

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

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

21. [latex]\underset{x\to 1}{\lim}f(x)[/latex]

In the following exercises, use the graph of the function [latex]y=f(x)[/latex] shown here to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with two segments. The first is a linear function for x < 0. There is an open circle at (0,1), and its slope is -1. The second segment is the right half of a parabola opening upward. Its vertex is a closed circle at (0, -4), and it goes through the point (2,0).

22. [latex]\underset{x\to 0^-}{\lim}f(x)[/latex]

23. [latex]\underset{x\to 0^+}{\lim}f(x)[/latex]

24. [latex]\underset{x\to 2}{\lim}f(x)[/latex]

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

In the following exercises, use the graph of the function [latex]y=f(x)[/latex] shown here to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with three segments, all linear. The first exists for x < -2, has a slope of 1, and ends at the open circle at (-2, 0). The second exists over the interval [-2, 2], has a slope of -1, goes through the origin, and has closed circles at its endpoints (-2, 2) and (2,-2). The third exists for x>2, has a slope of 1, and begins at the open circle (2,2).

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

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

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

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

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

31. [latex]\underset{x\to 2}{\lim}f(x)[/latex]

In the following exercises, use the graph of the function [latex]y=g(x)[/latex] shown here to find the values, if possible. Estimate when necessary.

A graph of a piecewise function with two segments. The first exists for x>=0 and is the left half of an upward opening parabola with vertex at the closed circle (0,3). The second exists for x>0 and is the right half of a downward opening parabola with vertex at the open circle (0,0).

32. [latex]\underset{x\to 0^-}{\lim}g(x)[/latex]

33. [latex]\underset{x\to 0^+}{\lim}g(x)[/latex]

34. [latex]\underset{x\to 0}{\lim}g(x)[/latex]

In the following exercises, use the graph of the function [latex]y=h(x)[/latex] shown here to find the values, if possible. Estimate when necessary.

A graph of a function with two curves approaching 0 from quadrant 1 and quadrant 3. The curve in quadrant one appears to be the top half of a parabola opening to the right of the y axis along the x axis with vertex at the origin. The curve in quadrant three appears to be the left half of a parabola opening downward with vertex at the origin.

35. [latex]\underset{x\to 0^-}{\lim}h(x)[/latex]

36. [latex]\underset{x\to 0^+}{\lim}h(x)[/latex]

37. [latex]\underset{x\to 0}{\lim}h(x)[/latex]

In the following exercises, use the graph of the function [latex]y=f(x)[/latex] shown here to find the values, if possible. Estimate when necessary.

A graph with a curve and a point. The point is a closed circle at (0,-2). The curve is part of an upward opening parabola with vertex at (1,-1). It exists for x > 0, and there is a closed circle at the origin.

38. [latex]\underset{x\to 0^-}{\lim}f(x)[/latex]

39. [latex]\underset{x\to 0^+}{\lim}f(x)[/latex]

40. [latex]\underset{x\to 0}{\lim}f(x)[/latex]

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

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

In the next exercises, evaluate the limit.

43. [latex]\underset{\theta \to 0}{\lim}\frac{\sin 2\theta}{5\theta}[/latex]

44. [latex]\underset{\theta \to 0}{\lim}\frac{\sin 4\theta}{3\theta}[/latex]

45. [latex]\underset{\theta \to 0}{\lim}\frac{1-\cos 2\theta}{9\theta}[/latex]

46. [latex]\underset{\theta \to 0}{\lim}\frac{1-\cos 2\theta}{3\theta}[/latex]

47. [latex]\underset{\theta \to 0}{\lim}\frac{\cos 2\theta+\sin 5\theta-1}{2\theta}[/latex]

48. [latex]\underset{\theta \to 0}{\lim}\frac{\cos 7\theta+\sin 2\theta-1}{7\theta}[/latex]

49. [latex]\underset{x \to 0}{\lim}\frac{x^2-x-\sin x}{3x}[/latex]

50. [latex]\underset{x \to 0}{\lim}\frac{x^2+x+\sin x}{4x}[/latex]

51. [latex]\underset{x \to 0}{\lim}\frac{\sin 5x}{\sin 4x}[/latex]

52. [latex]\underset{x \to 0}{\lim}\frac{\sin x}{\sin 2x}[/latex]