By Deborah Hughes- Hallett and et al
Venture : very unlikely
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Additional info for Instructor's Solutions Manual to Calculus: Single and Multivariable
Example text
To find the horizontal asymptote, we look at end behavior. As x → ±∞, the lower-degree terms of f (x) become insignificant, and f (x) becomes approximated by the highest degree term of the numerator and denominator. Thus, as x → ±∞, we see that 5x3 f (x) → 3 = 5. x There is a horizontal asymptote at y = 5. To find the vertical asymptotes, we set the denominator equal to zero. When x3 − 27 = 0, we have x = 3 so there is a vertical asymptote at x = 3. 39. (a) The object starts at t = 0, when s = v0 (0) − g(0)2 /2 = 0.
Then Slope = 2 1 − (−1) = . 5π/2 − 3π/2 π If P had been picked to the right of Q, the slope would have been −2/π. 5 SOLUTIONS 43 35. 47. 900 P 800 700 100 t (months) Jan. 1 Jul. 1 Jan. 47 = 800, amplitude = 900−700 = 100, and period = 12 months, so B = (b) Average value of population = 700+900 2 2 2π/12 = π/6. Since the population is at its minimum when t = 0, we use a negative cosine: P = 800 − 100 cos πt . 6 36. We use a cosine of the form H = A cos(Bt) + C and choose B so that the period is 24 hours, so 2π/B = 24 giving B = π/12.
B) As x approaches 0 from either side, the values of f (x) get closer and closer to 7. ) The limit appears to be about 7. (c) As x approaches 2 from either side, the values of f (x) get closer and closer to 3 on one side of x = 2 and get closer and closer to 2 on the other side of x = 2. Thus the limit does not exist. (d) As x approaches 4 from either side, the values of f (x) get closer and closer to 8. ) The limit appears to be about 8. lim f (x) = 1. 2. (a) x→1− (b) lim f (x) does not exist.