Module 4  Limit as x Approaches a  
Introduction  Lesson 1  Lesson 2  Lesson 3  SelfTest  
Lesson 4.2: Definition of Limit  
In the previous lesson you found tolerances graphically. In this lesson you will use the TI89 computer algebra system to find these tolerances symbolically. This will prepare you to generalize the tolerances and develop the definition of limit. Tolerances The first tolerance you found in lesson 4.1 was in response to the question
Restate the Question Another way to phrase this question is
Given yTolerance, Find xTolerance
The xtolerance when the ytolerance is 0.1 can be found by solving the
The solution is approximately 2.87 < x < 3.13667. Notice that the solutions are the same as those found in Lesson 4.1 when you used the intersection feature. Finding
Compare 2.87 < x < 3.13667 with the inequality 3 –
< x < 3 +
. It must follow that 3 –
2.87 and 3 +
3.13667 . However, solving these two equations for
yields two different values:
0.13 and
0.13667. As in lesson 4.1, the smaller of the two values is the correct answer:
Smaller Tolerance The second tolerance you found in lesson 4.1 came from answering the question
4.2.1 Rephrase this question with compound inequalities. Click here for the answer. 4.2.2 Use the solve( command to solve the first compound inequality. Click here for the answer. 4.2.3 Compare your answer to 4.2.2 with the inequality 3 – < x < 3 + and find a value of , the xtolerance. Click here for the answer. Finding a Generalized Solution Suppose that you are asked to find values of that correspond to smaller and smaller tolerances for y around y = 3. Rather than going through the same process over and over for various ytolerances, you could solve the problem once with a generalized ytolerance. Call this general ytolerance , the Greek letter "epsilon." The Greek letters delta, , and epsilon, , will be used to represent small positive numbers. The tolerance question then becomes
Rephrasing the Question The question can be rephrased using inequalities as
Solving the Left Inequality: 2 – <
The Greek letter can be entered in the TI89 by pressing . The solution to the equation has two parts: and – 2 0. The second part of the solution, – 2 0, is always true because is a small positive number.
Solving the Right Inequality: < 2 + You can edit the command shown in the Edit Line to solve the right inequality by changing the "–" sign to "+."
The solution is and – 2. Because is a small positive number, the statement – 2 is always true. Combining the solutions to find Combining the two solutions, in order to have 2 – < < 2 + we need Comparing this inequality with 3 – < x < 3 + yields Solving these equations for gives and The smaller of these two values is , which is the general solution. Given any value for you can use this equation to find the corresponding value for . For example, when = 0.01, this equation gives = 0.0133, the same value we found earlier. Limits When the values of the output can be made as close as we like to 2 by taking input values sufficiently close to 3, we say
That is, the value of f(x) = gets closer and closer to 2 as x gets closer and closer to 3. The notation used to indicate this is which is read "the limit of as x approaches 3 is 2." In the previous example we found that guarantees that Because for any positive , a corresponding positive can be found that meets the conditions above, Definition of Limit Formally, if for any > 0, however small, there exists a > 0 such that Conceptually, f(x) L as x a. 4.2.4 Write the inequality with conditions that is associated with the limit. Interpret . Click here for the answer. 

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