1 / 28

12.2

12.2. TECHNIQUES FOR EVALUATING LIMITS. Graphing Calculator today too. Pick a number between 1 and 10 Multiply it by 3 Add 6 Divide by 2 more than your original number. Dividing Out Technique. Dividing Out Technique.

kmims
Download Presentation

12.2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 12.2 TECHNIQUES FOR EVALUATING LIMITS Graphing Calculator today too Pick a number between 1 and 10 Multiply it by 3 Add 6 Divide by 2 more than your original number

  2. Dividing Out Technique

  3. Dividing Out Technique • We’ve discussed types of functions whose limits can be evaluated by direct substitution. • In this section, we will examine techniques for evaluating limits of functions for which direct substitution fails.

  4. Dividing Out Technique Ex 1 • 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.

  5. Dividing Out Technique Ex 1

  6. 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.

  7. Example 1 – Solution cont’d • You can see this from the graph as well • -Desmos Simplify. Direct substitutionand simplify.

  8. Ex 2 and 3 • 2. • 3. 6 1/6

  9. Example 4 • For the function given by f(x) = x2 –1, find • Solution: • Direct substitution produces an indeterminate form.

  10. Example 4 • For the function given by f(x) = x2 –1, find • Solution: • So, the limit is 6

  11. Rationalizing Technique

  12. 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 is

  13. Example 5 – Rationalizing Technique • Find the limit: • Solution: • By direct substitution, you obtain the indeterminate form . Indeterminate form

  14. Example 5 – Solution cont’d You can reinforce your conclusion that the limit is by constructing a table, as shown below, or by sketching a graph

  15. Example 5 – Solution cont’d • In this case, you can rewrite the fraction by rationalizing the numerator. Multiply. Simplify. Divide out common factor.

  16. Example 5 – Solution cont’d Now you can evaluate the limit by direct substitution. Simplify.

  17. Ex 6 Try on your own 1/6

  18. Using Technology

  19. 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.

  20. Example 7 – 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.

  21. Example 7 – Solution cont’d • 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. • This limit appears to be approximately e  2.71828.

  22. One-Sided Limits

  23. 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.

  24. Example 8 – Evaluating One-Sided Limits • Find the limit as x  0from the left and the limit as x  0 • from the right for • Solution: • From the graph of f, 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–

  25. Example 8 – Solution cont’d • Because f(x) = 2 for all x > 0 the limit from the right is Limit from the right: f (x)  2 as x  0+

  26. 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.

  27. H Dub 12.2 #1-26 cont’d

More Related