Calculus! #14: Early Transcendentals 2.1B

Chapter 2: Limits and Derivatives

You did it. You made it to Calculus. For many people, this is the real c-word. Take some pride in the fact that you’ve been willing to come this far. Although you’re only at the gates of MATH GLORY, you’ve still come much farther than most other people.

Okay. Let’s get into it:

Section 2.1: The Tangent and Velocity Problems

Draw a curve.

Did you do it?

Now, draw two points on it (A and B). Then, draw a right triangle using those two points as the hypotenuse. Note two features:

1) That the hypotenuse has a certain slope.

2) That there is a certain length of the bottom face of the triangle, given by B – A = L.

Now, cut L in half by moving B toward A  and draw a new triangle. Notice you get a new slope. Now cut L in half and draw another new triangle. And again, and again, and again.

Eventually, you’ll see that the slope of that triangle gets closer and closer to the slope of the curve at a particular point. To say it in a mathy way: As B – A approaches 0, the slope of the line made by the hypotenuse approaches the slope of the curve at A.

The line given by that slope is said to be “tangent” to the curve at that point.

The “tangent  problem” is to figure out what that slope is. Well, you know from the exercise we just did that by taking the slope between two points on the curve you can get better and better approximations. Fortunately, you don’t have to do everything by hand. You can just pack two points and find the slope. And you can keep narrowing the space between the two points.

What you’ll find is that as you bring the points closer and closer you get the slope closer and closer to a particular number. That number you’re approaching is called the “limit.”

Let’s do an example:

Find the slope at x = 1 of y = x^3.

At x = 1, y = 1. So that’s our first point: (1,1)

At x = 2, y = 8. So, that’s our second point. (2,8)

So, we’ll say the slope from 1 to 2 is 7/1, or 7.

Now, let’s tighten things up.

From 1 to 1.5 we get 1 and 3.375.

So that slope is 4.75.

Here are more:

1 to 1.25: 3.8125

1 to 1.15: 3.4725

1 to 1.05: 3.1525

1 to 1.01: 3.0301

1 to 1.00001: 3.00003

So, it really looks like we’re approaching 3. But, how can we prove it? Here’s where limits come in.

You may have noticed that for the above I’m following a simple algorithm.

Slope = \dfrac{x^3 - 1}{x -1}

Makes sense, right? Slope is the rise over the run. The first point is always (1,1), and the second point is given by (x,y). According to the function, y=x^3.

We could also have said it this way: The starting point is (A,B). The second point is always some form of (x, f(x)). So, at each point, the slope is going to be \dfrac{f(x) - B}{x - A}

We also note that as x gets closer and closer to 1, we get closer and closer to a certain value. In this case, that value is 3.

The way we phrase this mathematically would be:

\lim_{x\to 1}\dfrac{x^3 - 1}{x - 1} = 3

Read aloud, that’s “The limit of as x approaches 1 of \dfrac{x^3 - 1}{x - 1} is equal to 3.

Now, of course, later you’ll want to be able to solve limits of equations without looking at a giant list of numbers. This one was easy since the equation was simple, the solution was an integer, and we weren’t looking at any large numbers. Obviously it gets a bit more complex later.

In the next blog, we’ll go over an important example of the utility of limits: the velocity problem.


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