Answers Exercises Limits, Continuity and derivatives (Stewart Ch. 2)
3 b. 4 c. 2 d. – (does not exist) e. 3
2 b. 3 c. – (does not exist) d. 4 e. -1 b. -2 c. - d. 2 e. 0 f. - g. 1
-∞ b. ∞ c. -∞ d. ∞ e. x = -3, x = 2, x = 5
exists for all a  ± 1
0.677419, 0,666778, 0.655172, 0.666566: our guess is that the limit equals 2/3
0.594885, 0.501671, 0.426123, 0.498337: our guess is that the limit equals ½
a. e.g. f(0.99) = -33.7 and f(1.01) = 33.0 b. if x is slightly smaller than 1, then x3 - 1 will be a
negative number close to zero: its reciprocal is large negative. So
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
a.
a.
a.
a.
2.9
2.10
2.11
2.12
2.13
2.14
2.15
2.16
2.17
Use the a. sum law b. quotient and sum law resp. c. product and power law d. root law.
a. (x – 2) can only be cancelled if x  2 b. the limit as x→2 means x  2!
a. 5 b. does not exist c. 4/5 d. 8 e. 12 f. 6 g. -1/16 h. 108
a. 7 b. 2
Use the squeeze theorem and -1 ≤ cos (y) ≤ 1
a. No, limit from the left (-2) and from the right (2) are not equal b. No (∞) c. Yes limit = 0
a. 1 b. 1 c. 3 d. -2 e. -1
f. does not exist (limits from the left and the right are not equal)
a. asymptotic discontinuity b. asymptotic disc. c. jump disc. d.
a. Polynomials and root functions are continuous on their domain and so is the sum of the two
b. Trigonometric functions and polynomials are continuous on and so is the quotient on its domain.
c. logarithm and polynomials are continuous on their domain and so is the composition of them.
a. discontinuous at x = 0 (but continuous from the left)
b. discontinuous at x = 0 and x = 3, but continuous from the left at x = 0 and from the right at x = 3
c. discontinuous at x = 0 and x = 1, but continuous from the left at 1 and from the right at 0
a. f(1) = -1 and f(2) = 15
b. f(x) = cos(x) – x: f(0) = 1 and f(1) = cos(1) -1 < 0
c. f(x) =ln(x) – ex : f(1) = – e-1 < 0 and f(x) =ln(2) – e2 > 0
d. f(x) = ex – 2 + x: f(0) = -1 and f(3) = e3 + 1 > 0
f(x) = x2/ 2x : f(1) = 0.5, f(10) = 0.097656, f(20) = 0.00038147, f(100) = 7.8886 10-27 (limit ≈ 0)
Start by dividing numerator and denominator by x2 : apply quotient and addition laws of limits:
2.18
2.19
2.20
2.21
2.22 a. 0 b. 3 c. -3 d. ½ e. 3 f. does not exist g. 1/6 h. ∞ i. -∞ use
2.23
2.24
2.25
2.26
2.27
2.28
2.29
2.30
j. -½ k. 0: squeeze
a. y = 1, x = -4 b. y = -1, x = -1, x = 0, x = 1
c. x = 5 (x = 1 is a removable discontinuity)
5 (apply the squeeze theorem)
a. m = f `(1) = ½ : y = ½ x + ½
b. m = f `(3) = -1 : y = -x + 5
The velocity (the derivative of the position function) at t = 2 is – 9.6
a. -2 + 8a b. 5/ (t +3)2
The derivative at x = 5 is -26/25 (the velocity is 26/25 m/s)
(a) → II , (b) → IV, (c) → I, (d) → III
a. f `(x) = ½, Df = Df ` =
b. f `(t) =15 -18 t, Df = Df ` =
c. f `(x) =1+
, Df = [0, ∞) and Df ` = (0, ∞)
2.31 a. False b. False c. True d. True e. False f. False
g. True h. False i. True j. False k. False l. False