# Instructor's Solution Manuals to Calculus Early by C. Henry Edwards,David E. Penney

By C. Henry Edwards,David E. Penney

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Additional resources for Instructor's Solution Manuals to Calculus Early Transcendentals

Example text

The graph of the tangent line and the graph of f (x) = 25 − x2 are shown next. y 5 -5 5 x The numerical evidence suggests that the tangent line has slope − 34 , so that its equation is √ 3x + 4y = 25. The graph of the tangent line and the graph of f (x) = 25 − x2 are shown next. 48: y 5 -5 3 5 x 57 Copyright © 2012 Pearson Education, Inc. 2: x→3 x→−2 x→1 x3 − 3x + 3 · lim x→−2 x→−2 x2 + 2x + 5 = 1 · 5 = 5. t→−2 lim (t + 2) t+2 0 t→−2 = = = 0. 10: lim y→4 lim lim x3 − 25 3z 2 + 2z + 1 x→1 3/2 z3 + 5 lim = lim (27 − y→4 10 lim = 5 z→−1 3z 2 + 2z + 1 lim z3 + 5 √ = 83/2 = 16 2.

17: 3 lim t→−4 6 lim (t + 1) = 9. t→−4 x+2 = (x − 2)2 3 lim 6 3 (t + 1) = x→−2 3 lim x→−2 1/3 (x + 2) = 0. 19: x+1 x+1 1 1 = lim = lim =− . x2 − x − 2 x→−1 (x + 1)(x − 2) x→−1 x − 2 3 y→5 lim x→−1 = 64 27 = 4 . 20: lim t2 − 9 (t − 3)(t + 3) = lim = lim (t + 3) = 6. 21: lim x2 + x − 2 (x + 2)(x − 1) x+2 3 = lim = lim =− . 23: lim y→−1/2 lim t→−3 4y 2 − 1 (2y − 1)(2y + 1) 2y − 1 2 = lim = lim = − = −1. 4y 2 + 8y + 3 y→−1/2 (2y + 3)(2y + 1) y→−1/2 2y + 3 2 t2 + 6t + 9 (t + 3)(t + 3) t+3 = lim = lim = 0.

30524, . . 30259. This is fair evidence that L = ln 10 ≈ 65 Copyright © 2012 Pearson Education, Inc.