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Mathematics 2A: Single-Variable Calculus, Lecture Section C, Fall 2007
There will be no quiz based on this homework. Its material, however, will be included in the Midterms and in the Final Exam. All functions below have the domain consisting of all real numbers excluding those that would lead to division by zero or to taking a square root of a negative number. The abbreviation S&M refers to your textbook (Smith & Minton). Exercises marked “Optional” will not appear as mandatory on any quizzes or exams, but may appear as Extra Credit.
(a) S&M, section 1.5, exercise #1. (b) S&M, section 1.5, exercise #5. (c) S&M, section 1.5, exercise #7. (d) S&M, section 1.5, exercise #9. (e) S&M, section 1.5, exercise #13. (f) limx→π cot(x) (g) limx→π/ 4 tan( x (^2) +1x) (h) S&M, section 1.5, exercise #19. (i) limx→ 0 + |x x| (j) limx→ 0 − |x x| (k) limx→− 3 x
(^2) − 9 x+ (l) limx→ 1 x
(^3) − 1 x− 1 (m) limx→ 4 x
(^2) −x− 12 x− 4 (n) limx→ 0
√x+6−√x x (o) limx→ 0 √x+2x−√x
(p) limx→∞ (^) x (^3) +7^2 xx+5+
(q) limx→∞ 2 x
(^3) + − 4 x^3 − 7 x+ (r) limx→∞ 2 x
(^4) + x^3 − 7 x+ (s) limx→∞ 2 x
(^4) + − 4 x^3 − 7 x+ (t) limx→−∞ √^4 xx− (^2) +2^1
f (x) g(x)
(i.e., if the limit exists and equals L), THEN g is an upper bound of f as x → ∞. If L = 0, one says that g has a higher order of growth than f as x → ∞ and writes f = o(g). (The above definitions and theorem are completely analogous for x → a, where a is a real number; ]M, ∞[ is replaced by an appropriate neighborhood of a.)
(a) Show that g(x) = x^3 is an upper bound for f (x) = 126x^3 + 3047x^2 + x + 3 as x → ∞.
(b) Let n be a positive integer. Show that g(x) = xn^ is an upper bound for every polynomial of degree n or lower. (c) Determine which of the following two functions has a higher order of growth than the other:
f 1 (x) =
x^5 + 42x^4 + 46789, f 2 (x) = 3x^5.^1
[Hint: To keep the huge coefficients from distracting you, let
a = 126 × 1037 , b = 42, c = 46789,
and rewrite f 1 in the form f 1 (x) = ax^5 + bx^4 + c.]
(d) (Optional) Suppose g(x) = f 1 (x) + f 2 (x), and f 1 = o(f 2 ) as x → ∞. Show that g = O(f 2 ) as x → ∞. [Hint: Consider the ratio g/f 2 .] Conclude that, when estimating the growth of a sum of terms, the slower-growing terms can be dropped. You encountered this basic principle when you did 2a and 2b.
(a) the functions g 1 (x) = f 1 (x) + f 2 (x) and g 2 (x) = f 1 (x)f 2 (x) are also right-continuous at a. (b) if, in addition, f 2 (a) 6 = 0, then the function g 3 (x) = f 1 (x)/f 2 (x) is also right-continuous at a.
Same exercise for left-continuity and for continuity.
lim x→a+
f 1 (f 2 (x))
as lim y→f 2 (a)+
f 1 (y)
Use the right-continuity of f 1 to deduce that the latter limit equals f 1 (f 2 (a)).] Same exercise for left-continuity and for continuity.
(a) The particle remains within distance d of a point a at all times. (b) The particle cannot instantly jump a positive distance. (c) The absolute value of the average velocity of the particle over any period of time [t 1 , t 2 ] does not exceed a constant C. (d) The particle eventually stops at position p 0.