By Jerry R. Shipman
REA’s Algebra and Trigonometry challenge Solver
Each Problem Solver is an insightful and crucial examine and resolution advisor chock-full of transparent, concise problem-solving gemstones. solutions to your whole questions are available in a single handy resource from some of the most relied on names in reference resolution courses. extra worthwhile, more effective, and extra informative, those examine aids are the easiest overview books and textbook partners to be had. they are ideal for undergraduate and graduate studies.
This hugely invaluable reference is the best assessment of algebra and trigonometry at the moment on hand, with 1000s of algebra and trigonometry difficulties that disguise every thing from algebraic legislation and absolute values to quadratic equations and analytic geometry. each one challenge is obviously solved with step by step distinct options.
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Then δ > 0 such that δ < δ1 and > 0 and there exists some 1 |g(x) − M | < 1 whenever 0 < |x − c| < δ < δ1 , M − g(x) |g(x) − M | 1 1 = = − g(x) M g(x)M |g(x)|M 1 1 = · |g(x) − M | M |g(x)| 1 2 < · · 1 M M 21 = 2 M = whenever 0 < |x − c| < δ. This completes the proof of the statement lim x→c 1 1 = g(x) M whenever M > 0. The case for M < 0 can be proven in a similar manner. Now, we can use Part (iv) to prove Part (v) as follows: 1 f (x) = lim f (x) · x→c x→c g(x) g(x) lim = lim f (x) · lim x→c =L· = x→c 1 M L .
Since f (2) = 8, we need to prove that lim x3 = 8 = 23 . x→2 graph Let > 0 be given. Let us concentrate our attention on the open interval 50 CHAPTER 2. LIMITS AND CONTINUITY (1, 3) that contains x = 2 at its mid-point. Then |f (x) − f (2)| = |x3 − 8| = |(x − 2)(x2 + 2x + 4)| = |x − 2| |x2 + 2x + 4| ≤ |x − 2|(|x|2 + 2|x| + 4) (Triangle Inequality |u + v| ≤ |u| + |v|) ≤ |x − 2|(9 + 18 + 4) = 31|x − 2| < Provided |x − 2| < . 31 Since we are concentrating on the interval (1, 3) for which |x − 2| < 1, we need to define δ to be the minimum of 1 and .
LINEAR FUNCTION APPROXIMATIONS 25. f (x) = 2 27. f (x) = 29. f (x) = x−1 x−2 − |x − 1| |x − 2| 2x for x ≤ 0 2 x + 1 for x > 0 x−1 (x − 2)(x − 3) Recall the unit step function u(x) = 26. f (x) = 61 x x2 − 9 if x ≤ 0 if x > 0 28. f (x) = sin x sin x2 30. f (x) = |x − 1| if x ≤ 1 |x − 2| if x > 1 0 if x < 0 1 if x ≥ 0. Sketch the graph of each of the following functions and determine the left hand limit and the right hand limit at each point of discontinuity of f and g. 31. f (x) = 2u(x − 3) − u(x − 4) 32.