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Intro+to+the+derivative+notes.pdf

# F has an infinite discontinuity at x a if lim x a f x

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f has an infinite discontinuity at x = a if lim x a f ( x ) = ±∞ . Example 3.17. The discontinuities illustrated in Figures (a) and (c) are removable, while the discontinuity in Figure (b) is infinite. In (a) and (c) making f (2) = 3, the new function is continuous at x = 2.

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3. Introduction to the derivative: Limits and Continuity 17 Example 3.18. Is the following function continuous at - 1 ? If not, specify the type of discontinuity and redefine f (if possible) to make it continuous at - 1 . f ( x ) = ( x 2 - 1) / ( x + 1) if x 6 = - 1 1 if x = - 1 Solution: In the figure below, we could see the graph of f . From it, we can deduce that lim x →- 1 f ( x ) = - 2. -1 1 2 0 -2 -1 1 x y In fact: lim x →- 1 f ( x ) = lim x →- 1 x 2 - 1 x + 1 = lim x →- 1 ( x + 1)( x - 1) x + 1 = lim x →- 1 ( x - 1) = - 1 - 1 = - 2 Since lim x →- 1 f ( x ) exists and lim x →- 1 f ( x ) = - 2 6 = 1 = f ( - 1), we have that f has a removable discontinuity at x = - 1. In order to make f continuous at x = - 1, we redefine f ( - 1) = - 2. See also Examples 2 and 3 of the textbook (pages 711 - 713)! Exercise 3.19. Find the value of b that would make the f continuous at 1. f ( x ) = bx + 1 if x < 1 x 2 + 1 if x 1
18 J. S´ anchez-Ortega 3.4 Limits at Infinity of a Rational Function Let f ( x ) be a rational function, that is, a quotient of polynomial functions. In this section, we will learn how to evaluate the limits at infinity of f . Suppose that f ( x ) has the following form f ( x ) = c n x n + c n - 1 x n - 1 + . . . + c 1 x + c 0 d m x m + d m - 1 x m - 1 + . . . + d 1 x + d 0 with the c i and d i constants ( c n 6 = 0 and d m 6 = 0). Then we can calculate the limit of f ( x ) as x → ±∞ by ignoring all the powers of x except the highest in both the numerator and denominator. Thus lim x →±∞ f ( x ) = lim x →±∞ c n x n d m x m = c n /d n if n = m ±∞ if n > m 0 if n < m Examples 3.20. 1. lim x + 2 x 2 - 4 x x 2 - 1 = lim x + 2 x 2 x 2 = lim x + 2 1 = 2 2. lim x + - x 3 - 4 x 2 x 2 - 1 = lim x + - x 3 2 x 2 = lim x + - x 2 = -∞ 3. lim x + 2 x 2 - 4 x 5 x 3 - 3 x + 5 = lim x + 2 x 2 5 x 3 = lim x + 2 5 x = 0

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3. Introduction to the derivative: Limits and Continuity 19 Some Comments and Remarks about Determinate and Indeterminate Forms Notice that when computing limits it happens quite frequently that after evaluating at x = a , we end up with an expression of the form 0 / 0 , which indicates to us that further work is needed since we have no information about the limit . A similar situation arises letting x approach ±∞ , when we obtain the expression / . Those are called indeterminate forms. We could also ended up with the determinate forms k/ 0 , which will always yield ±∞ depending on the sign of the overall expression as x a (here k 6 = 0). For a more precise information, see the table below. * The signs gets switched in these forms if k is negative. Example 3.21. In case of an indetermination, for example, 0 / 0, the limit lim x a f ( x ) can exist or not. For instance, lim x 0 | x | /x does not exist, while lim x 0 sin x/x = 1.
20 J. S´ anchez-Ortega 3.5 Average Rate of Change Let y be a quantity that depends on another quantity x , that is, y is a function of x and we write y = f ( x ). Assume that

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