Download presentation

Presentation is loading. Please wait.

Published byIrving Hammock Modified over 4 years ago

1
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 § 1.4 Limits and the Derivative

2
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 115 Definition of the Limit Finding Limits Limit Theorems Using Limits to Calculate a Derivative Limits as x Increases Without Bound Section Outline

3
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 115 Definition of the Limit

4
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 115 Finding LimitsEXAMPLE SOLUTION Determine whether the limit exists. If it does, compute it. Let us make a table of values of x approaching 4 and the corresponding values of x 3 – 7. x x 3 - 7 4.1 61.9213.9 52.319 4.01 57.4813.99 56.521 4.001 57.0483.999 56.952 4.0001 57.0053.9999 56.995 As x approaches 4, it appears that x 3 – 7 approaches 57. In terms of our notation,

5
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 115 Finding LimitsEXAMPLE SOLUTION For the following function g(x), determine whether or not exists. If so, give the limit. We can see that as x gets closer and closer to 3, the values of g(x) get closer and closer to 2. This is true for values of x to both the right and the left of 3.

6
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 115 Limit Theorems

7
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 115 Finding LimitsEXAMPLE SOLUTION Use the limit theorems to compute the following limit. Limit Theorem VI Limit Theorem II with r = ½ Limit Theorem IV

8
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 115 Finding Limits Limit Theorems I and II CONTINUED Since, we have that:

9
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 115 Limit Theorems

10
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 115 Finding LimitsEXAMPLE SOLUTION Compute the following limit. Since evaluating the denominator of the given function at x = 9 is 8 – 3(9) = -19 ≠ 0, we may use Limit Theorem VIII.

11
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 11 of 115 Using Limits to Calculate a Derivative

12
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 12 of 115 Using Limits to Calculate a DerivativeEXAMPLE SOLUTION Use limits to compute the derivative for the function We must calculate

13
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 13 of 115 Using Limits to Calculate a DerivativeCONTINUED Now that replacing h with 0 will not cause the denominator to be equal to 0, we use Limit Theorem VIII.

14
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 14 of 115 More Work With Derivatives and LimitsEXAMPLE SOLUTION Match the limit with a derivative. Then find the limit by computing the derivative. The idea here is to identify the given limit as a derivative given by for a specific choice of f and x. Toward this end, let us rewrite the limit as follows. Now go back to. Take f (x) = 1/x and evaluate according to the limit definition of the derivative:

15
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 15 of 115 More Work With Derivatives and Limits On the right side we have the desired limit; while on the left side can be computed using the power rule (where r = -1): Hence, CONTINUED

16
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 16 of 115 Limits as x Increases Without BoundEXAMPLE SOLUTION Calculate the following limit. Both 10x + 100 and x 2 – 30 increase without bound as x does. To determine the limit of their quotient, we employ an algebraic trick. Divide both numerator and denominator by x 2 (since the highest power of x in either the numerator or the denominator is 2) to obtain As x increases without bound, 10/x approaches 0, 100/x 2 approaches 0, and 30/x 2 approaches 0. Therefore, as x increases without bound, 10/x + 100/x 2 approaches 0 + 0 = 0 and 1 - 30/x 2 approaches 1 – 0 = 1. Therefore,

17
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 17 of 115 Limits as x Increases Without BoundCONTINUED

Similar presentations

Presentation is loading. Please wait....

OK

U2 L5 Quotient Rule QUOTIENT RULE

U2 L5 Quotient Rule QUOTIENT RULE

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google