The Calculus Diaries - Jennifer Ouellette [106]
Finding the Limit. We’ll start with some handy tricks for finding the limit of a given function (assuming the limit exists; sometimes there is no limit). Trust me, this will come in handy when we get to derivatives. Earlier we looked at the function f(x) = 2x + 5, representing a straight line with a slope of 2 and a y-intercept of 5. The limit of f(x) as x approaches 3 equals 11. This just means that as we plug in values for x that are closer and closer to 3, the height of the graphed function gets closer and closer to y = 11 (aka the limit).
How do we know this? Well, it becomes fairly obvious if you plug in a series of values that get closer and closer to 3. For example, x = 2.9 gives a limit of 10.8, while x = 2.95 gives a limit of 10.9, and x = 2.99999 gives a limit of 10.99998. The closer the value of x is to 3, the closer the answer is to 11. Ergo, 11 is the limit of this particular function when x = 3.
But this is a tedious and time-consuming process that merely approximates the limit; we’d prefer to determine the limit precisely. The simplest strategy is called the substitution method: You just plug in the value of whatever number is specified under the “lim” notation. For example, let’s find the limit of a parabolic function, f(x) = x2, as x approaches Plug 2 into the equation, and we get 4. So
Similarly, to find the answer to , make x = 4, so that 42 − 4 + 2 = 14. So
You can verify this by using the graph of the function f(x) = (x2 − x + 2): another parabola. Simply plug in a few values both above and below 4, and you should see the results come closer and closer to 4 as those values trend closer and closer to 4.
Alas, it is not always that simple. Sometimes when you substitute the number specified under the “lim” notation, you get a nonsensical result, such as a 0 in the denominator, which is a mathematical taboo. In that case, you could use the factoring method to simplify things a little. Let’s say we want the answer to. If we try to plug the value −3 into the equation, we end up with This is not helpful.
So we switch tactics and factor the numerator; x2 and 9 both happen to be perfect squares. (There’s a reason I chose this particular example.) The result is
Aha! We learned in high school algebra that if you have the same expression in the numerator and denominator, they cancel each other out: In this case, we have (x + 3) in both the numerator and denominator. Cross them out, and that gives us a far simpler problem:
Now we can revert back to the substitution method and plug in 3. This time we get: −3 − 3 = −6. So As Sean explains, “The limit of the function is well defined at x = −3, even though the function itself is not.”
Finding the Slope of a Straight Line. In chapter 2 we went on a road trip from Los Angeles to Las Vegas, using highly idealized parameters to illustrate the fundamental concepts of what is essentially precalculus. For illustrative purposes, we’ll use another idealization here: that of a car accelerating to the speed limit, then traveling for a while at a constant speed, before braking suddenly to avoid an obstacle in the road. If we graphed our changing velocity as a function of time, the resulting curve would look like this:
Yes, this shape technically is still a “curve,” despite its straight edges. Because we are dealing with straight lines, it is pretty easy to determine the slopes of S1, S2, and S3. We simply pick any two points at random on the line of interest, (a,b) and (c,d ), and plug those values into this handy formula: let’s pick the beginning and ending points (0,0) and (3,5). Plug those values into our formula and we get . It’s simple arithmetic to determine that S1 = 1. We can follow the same process for S3, using points (6,5) and (9,0). We get The fact that the slope is negative means we were slowing down.
The slope of S2 is 0 because it is perfectly horizontal, or flat. Because we are traveling at a constant speed, there will be no difference between the values of those two points.