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MATH 100, Section 103 Week 2: Marked Homework Solutions 2009 Sep 25 1. [8] The function Ax2 + Bx + C is continuous on −∞ < x < 0 for any A, B and C (it is a polynomial, so it is continuous for all x), and x3/2 cos(1/x) is continuous on 0 < x < ∞ (the three functions x3/2 , 1/x and cos(x) are all continous on 0 < x < ∞; sums, products and compositions of continuous functions are continuous; the function cos(1/x) is the composition of the continuous functions cos(x) and 1/x so it is continuous on 0 < x < ∞ (Theorem 9, p. 130); the product of the continuous functions x3/2 and cos(1/x) is continuous). Therefore it remains only to check continuity at x = 0. This means we have to check that i) limx→0− f (x), ii) limx→0+ f (x) and iii) f (0) all exist and iv) are all equal to the same number. i) limx→0− f (x) = limx→0− (Ax2 + Bx + C) = C, ii) The cosine of any angle is between 1 and −1, therefore −1 ≤ cos(1/x) ≤ 1 for all x near 0 √ and multiplying by x3/2 = x x (which is positive, for x > 0), we get −x3/2 ≤ x3/2 cos(1/x) ≤ x3/2 for all x near 0. We easily evaluate the right-hand limits of the expressions on the left and right above: lim −x3/2 = 0, lim x3/2 = 0. x→0+ x→0+ Then by the Squeeze Theorem for limits, we have lim x3/2 cos(1/x) = 0. x→0+ iii) f (0) = C by the deﬁnition of the function f at 0. Therefore f (x) is continuous at x = 0 if and only if C = 0. Since f (x) is already known to be continuous on −∞ < x < 0 and on 0 < x < ∞, it follows that f is continuous at all points on the real line if and only if C = 0. A and B can be any real numbers, their values do not aﬀect the continuity of f . 2. [4] See the graph below. It appears that there is an intersection of the graphs of y = tan x and y = −x between x = 3π/2 (a vertical asymptote for y = tan x) and x = 2π (where tan 2π = 0). Note that 3π/2 ≈ 4.712388981 and 2π ≈ 6.283185308, so a = 4.72 is larger than 3π/2 and b = 6.28 is smaller than 2π. Consider the closed interval [a, b] = [4.72, 6.28]. To prove that an intersection in [a, b] exists, we rewrite tan x = −x as x + tan x = 0, which is equivalent. Let f (x) = x + tan x, and look for a solution of f (x) = 0 in [a, b]. Evaluate f (a) = f (4.72) ≈ −126.7 which is negative, and f (b) = f (6.28) ≈ 6.28 which 1 is positive. Observe that f is continuous on the closed interval [4.72, 6.28], because f is continuous on the open interval (3π/2, 5π/2), and [4.72, 6.28] is contained in (3π/2, 5π/2). Since N = 0 is between the endpoint values f (4.72) and f (6.28), by the Intermediate Value Theorem there is at least one number c between a = 4.72 and b = 6.28 such that f (c) = 0, i.e., tan x = −x has at least one solution x = c between 4.72 and 6.28. (From the graph, it looks like there is only one such number, but the Intermediate Value Theorem cannot be used to prove that.) √ q √ 1 5 2+ x + 1 √ 2x2 +5x+1 2x2 +5x+1 x2 2 3. [3 each part] (a) limx→∞ 4x+7 = limx→∞ x 1 (4x+7) = limx→∞ 7 4+ x = 4 . x 1 (b) limx→∞ (x3 − x5 ) = limx→∞ x5 x2 − 1 = limx→∞ x5 (−1) = −∞. (c) limt→∞ cos(t) does not exist. As t → ∞, cos(t) oscillates inﬁnitely often between +1 and −1 so it does not get close to a single ﬁnite number L. (d) limt→∞ cos(1/t) = cos (limt→∞ (1/t)) = cos(0) = 1 because the cosine function is contin- uous at 0 (Theorem 8, p. 129). √ √ √1 √1 x+√x2 −4x x+ x2 −4x (e) limx→+∞ x− x2 −4x = limx→∞ x− x 2 −4x · x+ x2 −4x = limx→∞ x2 −(x2 −4x) = √ q √ q √ 1 1+ 1 x2 −4x 1+ 1 (x2 −4x) x+ x2 −4x 1+ x x2 −4x x2 x2 limx→∞ = limx→∞ = limx→∞ = limx→∞ = √ 4 4x 4 4 4 1+ 1− x 1 limx→∞ 4 = 2. √ √ 2 x− x2 −4x (f) limx→−∞ √1 x+ x2 −4x = limx→−∞ x+√x2 −4x · x−√x2 −4x = limx→−∞ x2 −(x2 −4x) = 1 x− x −4x q √ √ q √ 1 2 −4x 1+ 1 x2 −4x 1+ 1 (x2 −4x) x− x2 −4x 1− x x x2 x2 limx→−∞ = limx→−∞ = limx→−∞ = limx→−∞ = √ 4 4x 4 √ 4 4 1+ 1− x 1 2 ). limx→−∞ 4 = 2 (Note: when x is negative, x = − x √ √ f (5+h)−f (5) 4. [4] f (5) = limh→0 = limh→0 1 √ 1 − √1 = limh→0 1 √4− 4+h √ h h 5+h−1 5−1 h 4+h · 4 2 √ √ √ √ (4)−(4+h) = limh→0 1√4− 4+h √ √4+√4+h = limh→0 1 √ √ √ √ h 4+h · 4 4+ 4+h h 4+h · 4 · ( 4+ 4+h) = limh→0 √4+h · √4 −1 4+√4+h) = √4 · √4−1 4+√4) = − 16 ·( √ ( √ 1 (which agrees with use of the power rule). 5. Using the deﬁnition of |x|, rewrite the function f (x) = x · |x| as x · (−x) = −x2 if x < 0 f (x) = x · x = x2 if x ≥ 0. We carefully investigate the derivative at x = 0. As instructed, we use the deﬁnition of the derivative. (CAUTION: You don’t always get the correct answer by simply taking the limits of f (x) as x → 0± . In this question it would work, but in the next question it does not! You can try it out.) Does limh→0 f (0+h)−f (0) exist? We need to look at the right- and left-hand h limits separately (why?). f (h)−f (0) 2 The right-hand limit is limh→0+ h = limh→0+ h = limh→0+ h = 0. h f (h)−f (0) 2 The left-hand limit is limh→0− h = limh→0− (−h ) = limh→0− (−h) = h 0. Since the right- and left-hand limits both exist, and agree, the limit exists and its value is the derivative of f (x) at x = 0: thus f (0) = 0. 6. [7] (a) By Theorem 4, p. 171, the function must at least be continuous to be diﬀerentiable, so from Question 1 we must have C = 0. To check diﬀerentiability at x = 0 (using C = 0), we need to check: Does limh→0 f (0+h)−f (0) = limh→0 f (h)−f (0) exist? Since the formula for h h f (h) depends on whether h < 0 or h > 0, we need to consider left- and right-hand limits. For h < 0, f (0 + h) − f (0) Ah2 + Bh lim = lim = B, h→0− h h→0− h f (h)−f (0) noting that f (0) = C = 0. On the other hand, for h > 0 we have limh→0+ h = 3/2 √ limh→0+ h cos(1/h) = limh→0+ h cos(1/h), and we have h √ √ √ −1 ≤ cos(1/h) ≤ 1, − h ≤ h cos(1/h) ≤ h for h > 0, 3 √ √ and the limits of both − h and h are 0, as h → 0+ . Therefore by the Squeeze Theorem for limits, f (0 + h) − f (0) √ lim+ = lim+ h cos(1/h) = 0. h→0 h h→0 For the limit of the diﬀerence quotient to exist, the left- and right-hand limits must agree, so B=0 is required. The derivative f (0) = 0 is the common value of the left- and right-hand limits of the diﬀerence quotient when B = 0. In summary, for f (0) to exist, we need B = 0, C = 0, A can be any real number. 4

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