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

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

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

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

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,

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.

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

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

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:

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

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.

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

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

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.

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:

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

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,

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

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