Download Adopting IT food program sponsor discovers it's no picnic by John M. Anderson and William H. Gwinn. PDF

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By John M. Anderson and William H. Gwinn.

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Using (1) with f (x) = x−1 and P (3, 2), x−2 x−1 x − 1 − 2(x − 2) −2 f (x) − f (a) x − 2 x−2 m = lim = lim = lim x→a x→3 x→3 x−a x−3 x−3 = lim x→3 3−x −1 −1 = lim = = −1 (x − 2)(x − 3) x→3 x − 2 1 Tangent line: y − 2 = −1(x − 3) ⇔ y − 2 = −x + 3 ⇔ y = −x + 5 √ √ √ √ x− 1 ( x − 1)( x + 1) x−1 1 1 √ √ = lim = lim = lim √ = . x→1 x→1 (x − 1)( x + 1) x→1 (x − 1)( x + 1) x→1 x−1 2 x+1 7. Using (1), m = lim Tangent line: y − 1 = 12 (x − 1) ⇔ y = 12 x + 1 2 9. (a) Using (2) with y = f (x) = 3 + 4x2 − 2x3 , f (a + h) − f (a) 3 + 4(a + h)2 − 2(a + h)3 − (3 + 4a2 − 2a3 ) = lim h→0 h→0 h h m = lim 3 + 4(a2 + 2ah + h2 ) − 2(a3 + 3a2 h + 3ah2 + h3 ) − 3 − 4a2 + 2a3 h→0 h = lim 3 + 4a2 + 8ah + 4h2 − 2a3 − 6a2 h − 6ah2 − 2h3 − 3 − 4a2 + 2a3 h→0 h = lim 8ah + 4h2 − 6a2 h − 6ah2 − 2h3 h(8a + 4h − 6a2 − 6ah − 2h2 ) = lim h→0 h→0 h h = lim = lim (8a + 4h − 6a2 − 6ah − 2h2 ) = 8a − 6a2 h→0 (b) At (1, 5): m = 8(1) − 6(1)2 = 2, so an equation of the tangent line (c) is y − 5 = 2(x − 1) ⇔ y = 2x + 3.

Then 0 < |x − 3| < δ of a limit, lim x→3 x 3 − <ε ⇔ 5 5 ⇒ |x − 3| < 5ε ⇒ 1 5 |x − 3| < ε ⇔ |x − 3| < 5ε. |x − 3| <ε ⇒ 5 x 3 − < ε. By the definition 5 5 x 3 = . 5 5 21. Given ε > 0, we need δ > 0 such that if 0 < |x − 2| < δ, then (x + 3)(x − 2) −5 <ε x−2 Then 0 < |x − 2| < δ ⇔ x2 + x − 6 −5 < ε ⇔ x−2 |x + 3 − 5| < ε [x 6= 2] ⇔ |x − 2| < ε. So choose δ = ε. ⇒ |x − 2| < ε ⇒ |x + 3 − 5| < ε ⇒ (x + 3)(x − 2) − 5 < ε [x 6= 2] ⇒ x−2 x2 + x − 6 x2 + x − 6 − 5 < ε. By the definition of a limit, lim = 5.

71828, which is approximately e. 6 we will see that the value of the limit is exactly e. 37. 001465 It appears that lim f (x) = 0. 001. x→0 46 ¤ CHAPTER 2 LIMITS AND DERIVATIVES 39. No matter how many times we zoom in toward the origin, the graphs of f (x) = sin(π/x) appear to consist of almost-vertical lines. This indicates more and more frequent oscillations as x → 0. 90 41. 24. To find the exact equations of these asymptotes, we note that the graph of the tangent function has vertical asymptotes at x = must have 2 sin x = π 2 + πn, or equivalently, sin x = π 4 π 2 + πn.

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