Mathematics 136 – Advanced Placement Calculus Solutions to Ungraded Problems on Problem Set 2 September 22, 2009

2.3/5. We compute:

lim_{x→8}(1 + x^{1/3})(2 − 6x^{2} + x^{3}) = (lim_{x→8} 1 + x^{1/3}) · (lim_{x→8} 2 − 6x^{2} + x^{3}) by Limit Law 4

= (1 + lim_{x→8} x^{1/3}) · (2 − 6 lim_{x→8} x^{2} + lim_{x→8} x^{3}) by Limit Laws 1, 3

= (1 + 2)(2 − 6 · 64 + 512) by Limit Law 6

= 3 · 130 = 390.

2.3/42. We have by Limit Law 11:

lim_{v→c^{−}} \sqrt{1 − v^{2}/c^{2}} = \sqrt{lim_{v→c^{−}}(1 − v^{2}/c^{2})} = 0.

This says that the length of the object appears (to any observer) to go to zero as its speed relative to that observer approaches the speed of light. There are two ways to answer the last part: Physically, no object can move faster than c, so we must take the limit as v → c^{−}. Mathematically, the domain of the L function only consists of v with −c ≤ v ≤ c. So again, the limit must be taken as v → c^{−} (or else we get sqare roots of negative numbers).

2.4/6. There are infinitely many correct graphs here. Any graph showing lim_{x→−1^{−}} g(x) = g(−1) and lim_{x→4^{+}} g(x) = g(4) is OK. For instance we might have jump discontinuities at x = −1, 4 and the value of g at −1 matches the limit from the left, while the value of g at 4 matches the limit from the right.

2.5/5. Again, infinitely many correct graphs. There should be a vertical asymptote at x = 0, where f(x) → −∞ from both sides (notice the form of the limit as x → 0). There should also be horizontal asymptotes at y = 5 to the left and y = −5 to the right.

2.5/26. We have

lim_{x→∞} \frac{x + 2}{\sqrt{9x^{2} + 1}} = lim_{x→∞} \frac{x + 2}{\sqrt{9x^{2} + 1}} · \frac{1/x}{1/x}

= lim_{x→∞} \frac{1 + 2/x}{\sqrt{9 + 1/x^{2}}}

= \frac{1}{3},

since 2/x → 0 and 1/x^{2} → 0 as x → ∞.

2.6/17. Of these numbers only g′(0) is negative (the slope of the tangent at 0), g′(−2) looks larger (tangent is steeper) than g′(2) or g′(4). Hence the order is

g′(0) < 0 < g′(4) < g′(2) < g′(−2).

2.7/9. Since the graph of f is made up of straight line segments, the graph y = f′(x) will be “piecewise constant.” Starting from the left, y = f′(x) should be a negative constant c_{1} up to x = −a (say), a positive constant c_{2} from x = −a to x = 0, the negative constant −c_{2} from x = 0 to x = a, and −c_{1} from x = a on. The derivative graph has jump discontinuities at x = −a, 0, a, and f′(x) is undefined there (so open circles on both sides on y = f′(x)).