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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.

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12.2

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  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

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