Arkansas Tech University MATH 2914: Calculus I Dr. Marcel B. Finan 1.3 The Limit of a Function The aim of this section is to develop the concept of the limit of a function. We will use intuitive approaches ,i.e., a geometric approach and a numerical approach, via the concepts of left-hand and right-hand limits. The rigorous δ definition is not covered. One-Sided Limits Consider the piecewise defined function x2; x < 1 f(x) = 3 − x; x ≥ 1 whose graph is shown in Figure 1.3.1. Figure 1.3.1 Imagine that x and f(x) are two moving objects that move simultaneously with x moving along the horizontal axis and f(x) moving along the curve. For example, if x moves toward x = 1 from the right, we see that f(x) moves toward the value 2. We express this statement with the notation lim f(x) = 2 x!1+ 1 which reads \the limit of f(x) as x approaches 1 from the right" is equal to 2. The notation x ! 1+ means that x gets really close to 1 from the right but it will never reach the value 1. The above is an example of what we call a right-hand limit. In a similar way, we can define the concept of a left-hand limit. For exam- ple, suppose that x approaches 1 from the left side. In this case, f(x) is approaching the value 1 and we write lim f(x) = 1: x!1− We call such a limit, the left-hand limit. The Limit of a Function sin θ Now, let's find the left-hand limit and the right-hand limit of f(θ) = θ whose graph is given in Figure 1.3.2 near 0. Figure 1.3.2 From the graph we see lim f(θ) = 1 and lim f(θ) = 1: θ!0− θ!0− Since the left-hand limit is the same as the right-hand limit, we write lim f(θ) = 0 θ!0 which reads \the limit of f(θ) as θ approaches 0 (from either direction) is 1". Again, the notation θ ! 0 means that θ can be very close to 0 from either sides but will never assume the value 0. This is an example of the limit of a function. 2 Example 1.3.1 Show, graphically, that lim 1 does not exist. x!0 x Solution. 1 The graph of f(x) = x is shown in Figure 1.3.3. Figure 1.3.3 From the figure, we have limx!0+ f(x) = 1 and limx!0− f(x) = −∞: Thus, both left-hand and right-hand limits do not exist. Hence, lim 1 does not x!0 x exist Besides the graphical approach of finding limits, one can also find limits numerically as suggested by the following example. Example 1.3.2 2 2 Evaluate numerically lim 16(2+t) −16(2) : t!0 t Solution. We first create the following chart. t −0:1 −0:01 −0:001 −0:0001 0 0:0001 0:001 0:01 0.1 16(2+t)2−16(2)2 t 62.4 63.84 63.984 63.9984 undefined 64.0016 64.016 64.16 65.6 From the table we conclude that 16(2 + t)2 − 16(2)2 16(2 + t)2 − 16(2)2 lim = lim = 64: t!0− t t!0+ t That is 16(2 + t)2 − 16(2)2 lim = 64 t!0 t 3 Example 1.3.3 Show, numerically, that lim sin 1 does not exist. x!0 x Solution. Consider the following tables 1 1 x sin x x sin x 2 2 − π −1 π 1 2 2 − 3π 1 3π −1 2 2 − 5π −1 5π 1 2 2 − 7π 1 7π −1 . It follows that both left-hand and right-hand limits do not exist. Hence, lim sin 1 does not exist x!0 x 4.