logging in or signing up LarPCalcLim2_12_02 jwaychoffths Download Post to : URL : Related Presentations : Let's Connect Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Copy embed code: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 73 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: August 18, 2011 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Slide 1: Copyright © Cengage Learning. All rights reserved. 12 Limits and an Introduction to CalculusSlide 2: 12.2 Copyright © Cengage Learning. All rights reserved. TECHNIQUES FOR EVALUATING LIMITSWhat You Should Learn: Use the dividing out technique to evaluate limits of functions. Use the rationalizing technique to evaluate limits of functions. Approximate limits of functions graphically and numerically. Evaluate one-sided limits of functions. Evaluate limits of difference quotients from calculus. What You Should LearnSlide 4: Dividing Out TechniqueDividing Out Technique: Dividing Out Technique You studied several types of functions whose limits can be evaluated by direct substitution. In this section, you will study several techniques for evaluating limits of functions for which direct substitution fails.Dividing Out Technique: Dividing Out Technique Suppose you were asked to find the following limit. Direct substitution produces 0 in both the numerator and denominator. ( –3 ) 2 + ( –3 ) – 6 = 0 –3 + 3 = 0 The resulting fraction, , has no meaning as a real number. It is called an indeterminate form because you cannot, from the form alone, determine the limit. Numerator is 0 when x = –3. Denominator is 0 when x = –3.Dividing Out Technique: Dividing Out Technique By using a table, however, it appears that the limit of the function as x –3 is –5. When you try to evaluate a limit of a rational function by direct substitution and encounter the indeterminate form , you can conclude that the numerator and denominator must have a common factor.Dividing Out Technique: Dividing Out Technique After factoring and dividing out, you should try direct substitution again. Example 1 shows how you can use the dividing out technique to evaluate limits of these types of functions.Example 1 – Dividing Out Technique: Example 1 – Dividing Out Technique Find the limit: Solution: From the discussion, you know that direct substitution fails. So, begin by factoring the numerator and dividing out any common factors . Factor numerator. Divide out common factor.Example 1 – Solution: Example 1 – Solution cont’d Simplify. Direct substitution and simplify.Slide 11: Rationalizing TechniqueRationalizing Technique: Rationalizing Technique Another way to find the limits of some functions is first to rationalize the numerator of the function. This is called the rationalizing technique. Recall that rationalizing the numerator means multiplying the numerator and denominator by the conjugate of the numerator. For instance, the conjugate of isExample 3 – Rationalizing Technique: Example 3 – Rationalizing Technique Find the limit: Solution: By direct substitution, you obtain the indeterminate form . Indeterminate formExample 3 – Solution: Example 3 – Solution In this case, you can rewrite the fraction by rationalizing the numerator. cont’d Multiply. Simplify. Divide out common factor.Example 3 – Solution: Example 3 – Solution cont’d Now you can evaluate the limit by direct substitution. Simplify.Example 3 – Solution: You can reinforce your conclusion that the limit is by constructing a table, as shown below, or by sketching a graph, as shown in Figure 12.12. Example 3 – Solution cont’d Figure 12.12Slide 17: Using TechnologyUsing Technology: Using Technology The dividing out and rationalizing techniques may not work well for finding limits of nonalgebraic functions. You often need to use more sophisticated analytic techniques to find limits of these types of functions.Example 4 – Approximating a Limit: Example 4 – Approximating a Limit Approximate the limit: Solution: Let f ( x ) = (1 + x ) 1/ x . Because you are finding the limit when x = 0, use the table feature of a graphing utility to create a table that shows the values of f for x starting at x = –0.01 and has a step of 0.001, as shown in Figure 12.13. Figure 12.13Example 4 – Solution: Example 4 – Solution Because 0 is halfway between –0.001 and 0.001, use the average of the values of f at these two x -coordinates to estimate the limit, as follows. The actual limit can be found algebraically to be e 2.71828. cont’dSlide 21: One-Sided LimitsOne-Sided Limits: One-Sided Limits You saw that one way in which a limit can fail to exist is when a function approaches a different value from the left side of c than it approaches from the right side of c . This type of behavior can be described more concisely with the concept of a one-sided limit.Example 6 – Evaluating One-Sided Limits: Example 6 – Evaluating One-Sided Limits Find the limit as x 0 from the left and the limit as x 0 from the right for Solution: From the graph of f , shown in Figure 12.17, you can see that f ( x ) = –2 for all x < 0. Therefore, the limit from the left is Limit from the left: f ( x ) –2 as x 0 –Example 6 – Solution: Example 6 – Solution Because f ( x ) = 2 for all x > 0 the limit from the right is cont’d Limit from the right: f ( x ) 2 as x 0 +One-Sided Limits: One-Sided Limits In Example 6, note that the function approaches different limits from the left and from the right. In such cases, the limit of f ( x ) as x c does not exist. For the limit of a function to exist as x c , it must be true that both one-sided limits exist and are equal.Slide 26: A Limit from CalculusA Limit from Calculus: A Limit from Calculus You will study an important type of limit from calculus—the limit of a difference quotient.Example 9 – Evaluating a Limit from Calculus: Example 9 – Evaluating a Limit from Calculus For the function given by f ( x ) = x 2 – 1, find Solution: Direct substitution produces an indeterminate form.Example 9 – Solution: Example 9 – Solution By factoring and dividing out, you obtain the following. So, the limit is 6. cont’dA Limit from Calculus: A Limit from Calculus Note that for any x -value, the limit of a difference quotient is an expression of the form Direct substitution into the difference quotient always produces the indeterminate form . For instance, You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.