Intro+to+the+derivative+notes.pdf

3 introduction to the derivative limits and

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tinuous functions on their respective domains.
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3. Introduction to the derivative: Limits and Continuity 11 Examples 3.10. Evaluate the following limits and justify each step. a. lim x 5 (2 x 2 - 3 x + 4) b. lim x →- 2 x 3 + 2 x 2 - 1 5 - 3 x Solution: a. lim x 5 (2 x 2 - 3 x + 4) L2, L1 = lim x 5 (2 x 2 ) - lim x 5 (3 x ) + lim x 5 4 L3, L7 = = 2 lim x 5 x 2 - 3 lim x 5 x + 4 L9, L8 = = 2(5 2 ) - 3(5) + 4 = 39 b. Let us first compute the limit of the numerator and denominator: lim x →- 2 ( x 3 + 2 x 2 - 1) L2, L1 = lim x →- 2 ( x 3 ) + lim x →- 2 (2 x 2 ) - lim x →- 2 1 L9, L3, L7 = ( - 2) 3 + 2 lim x →- 2 ( x 2 ) - 1 L9 = - 8 + 2( - 2) 2 - 1 = - 1 lim x →- 2 (5 - 3 x ) L2 = lim x →- 2 5 - lim x →- 2 (3 x ) L7, L3 = 5 - 3 lim x →- 2 x L8 = = 5 - 3( - 2) = 11 Since the limits of the numerator and denominator exist and the limit of the denominator is not 0, an application of Law 5, allows us to conclude that lim x →- 2 x 3 + 2 x 2 - 1 5 - 3 x = - 1 11 Remark 3.11. Direct substitution does not always work! Let us consider again the function f ( x ) = x - 1 x 2 - 1 In Example 3.3 we saw that lim x 1 f ( x ) = 0 . 5. Note that we can not find the limit by substituting x = 1 it does not make sense to divide by zero. We
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12 J. S´ anchez-Ortega can not apply either the Quotient Law, because the limit of the denominator is 0. Instead, we proceed as follows: f ( x ) = x - 1 x 2 - 1 = x - 1 ( x - 1)( x + 1) = 1 x + 1 We can now apply the limits laws and conclude that lim x 1 x - 1 x 2 - 1 = lim x 1 1 x + 1 = 1 1 + 1 = 1 2 , as expected. Notice that we have been able to compute the limit by replacing the given function f ( x ) = ( x - 1) / ( x 2 - 1) by a simpler function, g ( x ) = x +1, with the same limit. This is valid because f ( x ) = g ( x ) except for x = 1, and as said in computing a limit as x approaches 1, we do not consider what happens when x = 1. F This situation is an example of a more general fact: Suppose that we are interested in computing lim x a f ( x ) but for many distinct reasons, we would like to replace our function f ( x ) by another function g ( x ), simpler than f in terms of computing the limit when x ap- proaches a . If g ( x ) equals f ( x ) for all x in an open interval containing a , expect possible at a (we do not care whether g ( a ) equals f ( a ); like in the preceding example our function f might not be even defined when x = a ) then lim x a f ( x ) = lim x a g ( x ) , provided the limits exist. Example 3.12. Find lim x 1 f ( x ) where f ( x ) = x + 1 if x 6 = 1 3 if x = 1 Solution: As said before, the value of a limit as x approaches 1 does not depend on the value of the function at 1. Here the given function f ( x ) = g ( x ) for all x 6 = 1, where g ( x ) = x + 1.
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3. Introduction to the derivative: Limits and Continuity 13 1 2 3 0 1 2 3 x y y = f ( x ) Applying the property above we can conclude that lim x 1 f ( x ) = lim x 1 g ( x ) = lim x 1 ( x + 1) = 1 + 1 = 2 . Example 3.13. Sketch the graph of an example of a function that satisfies all the given conditions. a. lim x 1 - f ( x ) = 2 , lim x 1 + f ( x ) = - 2 , f (1) = 2 b. lim x 0 - f ( x ) = 2 , lim x 0 + f ( x ) = 0 , f (0) = 2 , lim x 4 - f ( x ) = 3 , lim x 4 + f ( x ) = 0 , f (4) = 1 Solution: a. The following function satisfies the required conditions: -2 -1 1 2 0 -2 -1 1 2 x y b. Do it yourself! You are welcome to email me your answer :-)
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14 J. S´ anchez-Ortega 3.3 Continuity
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