When we scrutinize a function or its graph, we might want to talk about its continuity at a point. But first, we must know the value of the function at that point, as well as the limit as we approach it from either direction.
Right-hand limit at a point
Consider the following function, which takes the reciprocal of the input value:
g (x)= 1/x
We can’t define the limit of g (x) as x approaches 0 from the positive direction, because the function blows up as x gets smaller positively, approaching 0. To specify that we approach 0 from the positive direction, we can refine the limit notation by placing a plus sign after the 0, like this:
xLim→ +0 g (x)
This expression reads, “The limit of g (x) as x approaches 0 from the positive direction.” We can also say, “The limit of g (x) as x approaches 0 from the right.” (In most graphs where x is on the horizontal axis, the value of x becomes more positive as we move toward the right.) This sort of limit is called a right-hand limit.
Right-hand continuity at a point
What about some other point, such as where x = 1? As we approach the point where x = 1 from the positive direction, g starts out at positive values smaller than 1 and increases, approaching 1. We can see this with the help of Fig. 2-2, which is a graph of g drawn in the vicinity of the point where x = 1. (Each division on the axes represents 1/2 unit.) This graph tells us that
Limx→1+ g (x)= 1 We can calculate the actual value of g for x = 1, getting
g (1) = 1/1 = 1 Now suppose that:
• We can define the right-hand limit of a function at a certain point
• We can define the actual value of the function at that point
• The limit and the actual value are the same
When all three of these things are true, we say that the function is right-hand continuous at that point. Some texts will tell us that g is continuous on the right at the point. In this example, g is right-hand continuous at x = 1.
Left-hand limit at a point
Let’s expand the domain of g to the entire set of reals except 0, for which g is not defined because 1/0 is not defined. Suppose that we start out with negative real values of x and approach 0 from the left. As we do this, g decreases endlessly, as we can see by looking at Fig. 2-3. (Here, each division on the axes represents 1 unit.) Another way of saying this is that g increases negatively without limit, or that it blows up negatively. Therefore,
xLim→ −0 g (x)
is not defined. We read the above symbolic expression as, “The limit of g (x) as x approaches 0 from the negative direction.” We can also say, “The limit of g (x) as x approaches 0 from the left.” This sort of limit is called a left-hand limit.
Left-hand continuity at a point
Now let’s look again at the point in the graph g where x= 1. Suppose we approach this point from the left, that is, from the negative direction. The value of g starts out at positive values larger than 1. As x increases and approaches 1, g decreases, getting closer and closer to 1. Figure 2-4 illustrates what happens here. (Each division on the axes represents 1/2 unit.) We have
Limx→ −1 g (x)= 1 Each axis
increment is 1/2 unit
g(1) = 1 Limg(x) = 1
x 1+
x g(x)
g(x) = 1/x
Figure 2-2 The limit of the function g (x) = 1/x as x approaches 1 from the right is equal to the value of the function when x = 1. In this graph, each axis division represents 1/2 unit.
Continuity at a Point 25
x
Limg(x)
x 0–
is not defined Each axis increment is 1 unit
g(x)
g(x) = 1/x
Figure 2-3 As x approaches 0 from the negative direction, the value of 1/x increases negatively without limit. In this graph, each axis division represents 1 unit.
Each axis increment is 1/2 unit
g(1) = 1
g(x) = 1/x
x g(x)
Limg(x) = 1
x 1–
Figure 2-4 The limit of the function g (x) = 1/x as x approaches 1 from the left is equal to the value of the function when x = 1. In this graph, each axis division represents 1/2 unit.
We already know that g (1) = 1. Just as we did when approaching from the right, we can say thatg is left-hand continuous at the point where x = 1. Some texts will say that g is continuous on the left at that point.
Are you confused?
It’s easy to get mixed up by the meanings of “negative direction” and “positive direction,” and how these relate to the notions of “left-hand” and “right-hand.” These terms are based on the assumption that we’re talking about the horizontal axis in a graph, and that this axis represents the independent variable. In most graphs of this type, the value of the independent variable gets more negative as we move to the left, and more positive as we move to the right. This is true no matter where on the axis we start.
As we travel along the horizontal axis, we might be in positive territory the whole time; we might be in negative territory the whole time; we might cross over from the negative side to the positive side or vice-versa. Whenever we come toward a point from the left, we approach from the negative direction, even if that point corresponds to something like x = 567. Whenever we come toward a point from the right, we approach from the positive direction, even if the point is at x =−53,535. The location of the point doesn’t matter. The important thing is the direction from which we approach.
“Total” continuity at a point
Now that we’ve defined right-hand and left-hand continuity at a point, we can define conti-nuity at a point in a “total” sense. When a function is both left-hand continuous and right-hand continuous at a point, we say that the function is continuous at that point. Conversely, whenever we say that a function is continuous at a point, we mean that it’s continuous as we approach and then reach the point from both the left-hand side and the right-hand side.
Here’s a challenge!
Suppose we modify the function g (x) = 1/x by changing the value for x= 1. Let’s call this new function g∗. The domain remains the same: all real numbers except 0. The only difference between g∗ and g is that g∗ (1) is not equal to 1, but instead is equal to 4, as shown in Fig. 2-5. (In this graph, each axis division is 1 unit.) We have seen that the original function g is both right-hand continuous and left-hand continuous at (1,1), so we know that g is “totally” continuous at (1,1). But what about g∗? Is this modified function continuous at the point where x= 1? Is it left-hand continuous there? Is it right-hand continuous there?
Solution
The answer to each of these three questions is “No.” The function g∗ is not continuous at the point where x= 1. It’s not right-hand continuous or left-hand continuous there. The limit of g∗ as we approach x= 1 from either direction is equal to 1. That is,
Limx→ +1 g∗ (x)= 1
Continuity at a Point 27
and
Limx→ −1 g∗ (x)= 1
As we move toward the point where x= 1 from either direction along the curve, it seems as if we should end up at the point (1,1) when the value of x reaches 1. But when we look at the actual value of g∗ at x= 1, we find that it’s not equal to 1. There is a discontinuity in g∗ at the point where x= 1. We can also say that g∗ is not continuous at the point where x= 1, or that g∗ is discontinuous at the point where x= 1.
Here’s another challenge!
Look again at the step function we saw in Fig. 1-11 near the end of Chap. 1. Is this function right-hand continuous at point where x= 3? Is it left-hand continuous at that point?
Solution
A portion of that function is reproduced in Fig. 2-6, showing only the values in the vicinity of our point of interest. Let’s call the function s (x). We can see from this drawing that
xLim→ +3 s (x)= 3
x Each axis
increment is 1 unit
g(x)
(1,4)
g* (x) = 1/xwhenx≠ 1 = 4 when x= 1
Figure 2-5 A modified function g∗, which is the same as g except that the point (1,1) has been moved to (1,4).
and
Limx→ −3 s (x)= 2
The value of the function at the point where x= 3 is s (3) = 3. The right-hand limit and the actual value are the same, so s is right-hand continuous at the point where x= 3. But s is not left-hand continuous at the point where x= 3, because the left-hand limit and the actual value are different. The same situation occurs at every integer value of x in the step function illustrated in Fig. 1-11. Because s is not continuous from both the right and the left at any point where x is an integer, s is not continuous at any such point.
This function has an infinite number of discontinuities!