11.2: Techniques for Evaluating Limits Pre-Calculus Honors 11.2: Techniques for Evaluating Limits HW: p.767 (10-22 even) Copyright © Cengage Learning. All rights reserved.

Dividing Out Technique Suppose you were asked to find the following limit and direct substitution cannot be used.

Example 1 – Dividing Out Technique Find the limit. Solution: Begin by factoring the numerator and dividing out any common factors. Factor numerator.

Example 1 – Solution (x – 2) = –3 – 2 = –5 cont’d Divide out common factor. Simplify. Direct substitution Simplify.

Dividing Out Technique The validity of this technique stems from the fact that when two functions agree at all but a single number c, they must have identical limit behavior at x = c. In Example 1, the functions given by f (x) and g (x) = x – 2 agree at all values of x other than x = –3. So, you can use g (x) to find the limit of f (x).

Dividing Out Technique The dividing out technique should be applied only when direct substitution produces 0 in both the numerator and the denominator. An expression such as has no meaning as a real number. It is called an indeterminate form because you cannot, from the form alone, determine the limit.

11.2: Techniques for Evaluating Limits Pre-Calculus Honors 11.2: Techniques for Evaluating Limits HW: p.767 (6, 8, 24-36: mult of 4, 38) Copyright © Cengage Learning. All rights reserved.

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.

Example 3 – Rationalizing Technique Find the limit. Solution: By direct substitution, you obtain the indeterminate form . Indeterminate form

Example 3 – Solution cont’d In this case, you can rewrite the fraction by rationalizing the numerator. Multiply. Simplify. Divide out common factor. Simplify.

Example 3 – Solution cont’d Now you can evaluate the limit by direct substitution.

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 Numerically Approximate the limit: . Solution: Let f (x) = (1+x)1/x. Because 0 is halfway between –0.001 and 0.001 (see Figure 11.13), use the average of the values of f at these two x-coordinates to estimate the limit. (1 + x)1/x  = 2.71825 Create a table that shows values of f (x) for several x-values near 0. Figure 11.13

One-Sided Limits The limit of f (x) as x  c does not exist when the function f (x) approaches a different number 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. f (x) = L1 or f (x)  L1 as x  c– f (x) = L2 or f (x)  L2 as x  c+ Limit from the left Limit from the right

Example 6 – Evaluating One-Sided Limits Find and So, the limit from the left is = –2. and the limit from the right is = 2.

One-Sided Limits

Review for Quiz 11.1-11.2 11.1 Approximate the limit of a function analyzing the graph or table with a calculator or given graph/table. Approximate/find the limit of a function graphically without a calculator (easier type of graphing problems). Graphically determine when a limit does not exist. Find a limit using direct substitution. (when can this be done??) 11.2 Find a limit using direct substitution after dividing out, simplifying complex fraction, or rationalizing numerator. Approximating the limit of a function graphically by averaging. Find limits from the left and right. Textbook questions: p.757-758 (5-67 any odd), p.767-768 (5-59 any odd, 71)

A Limit from Calculus A Limit from Calculus In the next section, you will study an important type of limit from calculus—the limit of a difference quotient.

Example 9 – Evaluating a Limit from Calculus For the function given by f (x) = x2 – 1, find Solution: Direct substitution produces an indeterminate form.

Example 9 – Solution cont’d By factoring and dividing out, you obtain the following.

Example 9 – Solution cont’d = (6 + h) = 6 + 0 = 6 So, the limit is 6.