Published on: Tue Dec 02 2008
To review my knowledge of limits, I need to think back. A Limit expresses a relationship between two changing but related variables, Epsilon eand Theta strong>d.To review my understanding of limits, the limit is a variable L, which has a relationship to a secondary variable a. a is the change along the x-axis as we approach the limit, which is represented by the variable L. To think back on what I understand, the variable L is bordered on the y-axis by Epsilone. As far as limits are concerned, we are in absolute space. We do not have negative limits. So to start out, I know the limit must be greater than Zero. Both variables, Epsilon e and Theta d must be larger than Zero, or else the limit does not exist. So I can write e > 0. The purpose of this exercise is to describe the relationship between Epsilon and Theta. So what can I write about Theta? Theta d represents the border of variable a along the x-axis. Since both Epsilon and Theta represent both sides of the border, there are absolute values.So what exactly is the relationship between our variables? L, a,e and d, as well as our line, y = f(x) L is the limit. It is reached by x – a. Variable a is found when you look at a limit and there is an equivalency x ->a. Epsilon is a real number, greater than 0. The limit is a point on the line f(x). Epsilon deals with the y vertice, which is described as a function of f(x). So the variable x describes the location along the x-axis, and f(x) describes the location along the y-axis. Lets break this problem off into x and y axis sections. To the x axis, we will give the variables, x,a and d To the y-axis we give the variables f(x), L and Epsilon. x – a should be less than d and f(x) – L should be less than e. The reason our x and y coordinates are less than variables Epsilon and Theta, is because, the Limit is contained inside of the variables Epsilon and Theta, because they form the boundrays. Because the limit is not negative, we need to use absolute values for both of the expressed changes, x-a and f(x)-L so we would say |x-a| < d and |f(x)-L| < e . There is only one more thing I am leaving out, which is that the limit is in positive space, so the whole thing has to be greater than 0. Back at the beginning of this essay, I stated e > 0 I could say the same thing by writing 0 < e . We can include this note about the limit being in positive space, by sticking that Zero Less Than in front of the entire expression to write. 0< |x-a| < d and |f(x)-L| < e Can I use this relationship to find Theta with a known Epsilon. >
The first step is to solve the f(x)-L side, because we know what f(x) is, we know what L is and we know what Epsilon is. Using our known variables we will be able to find the values of x.
So now we have found out an exact value of variable x, it is between 2.00083 and 1.99916. x – a should give us a value for Theta
*Incase you haven't noticed, I'm having some trouble typing out the Greek letter variables. I'll figure that out some other time.