Tiêu đề MATH 21 LE 1 ANSWER KEY.pdf ✅

Math 21 Sample 1st Long Exam First Semester, A.Y. 2025-2026 MBAN 109 This sample long exam and accompanying answer key are intended as reviewers for the actual long exam. General Instructions: This exam is good for 80 minutes. Provide neat and complete solutions. Calculators are not allowed. 1. Calculate the following limits. (a) lim x→−5 x 2 − 25 −5 − x (b) lim x→−1− −x 3x 2 + 10x + 7 + 1 (x + 1)2 (c) lim x→−∞ p 4x 2 + 3x + 2x (d) lim x→0+ cos−1 1 ln(x) (e) limx→0 x cos2 (5x) sin(−3x) + 1 − cos(x) x (f) lim x→+∞ sin(6x − 9) xex Solution: We calculate the limits as follows. (a) lim x→−5 x 2 − 25 −5 − x = lim x→−5 (x − 5)(x + 5) −(x + 5) 0 0 = lim x→−5 − (x − 5) = 10 (b) lim x→−1− −x 3x 2 + 10x + 7 + 1 (x + 1)2 = lim x→−1− −x (3x + 7)(x + 1) + 1 (x + 1)2 (−∞ + ∞) = lim x→−1− − x 2 − 2x − 7 (3x + 7)(x + 1)2 − (−4) (4)(0−) 2 =+∞ (c) lim x→−∞ p 4x 2 + 3x + 2x = lim x→−∞ √ 4x 2 + 3x + 2x 1 · √ 4x 2 + 3x − 2x √ 4x 2 + 3x − 2x √ ∞ − ∞ − ∞ = lim x→−∞ 4x 2 + 3x − 4x 2 √ 4x 2 + 3x − 2x · √ 1 x2 √ 1 x2 = lim x→−∞ −3 q 4 + 3 x + 2 √ x 2 = |x| = −x when x < 0 = − 3 √ 4 + 2 =− 3 4 (d) Note that lim x→0+ ln(x) = −∞ and lim y→−∞ 1 y = 0−, so that lim x→0+ cos−1 1 ln(x) = lim y→−∞ cos−1 1 y = lim z→0− cos−1 (z) = cos−1 (0) = π 2 .
Math 21 Sample 1st Long Exam First Semester, A.Y. 2025-2026 MBAN 109 (e) limx→0 x cos2 (5x) sin(−3x) + 1 − cos(x) x =limx→0 x cos2 (5x) sin(−3x) + limx→0 1 − cos(x) x =limx→0 x cos2 (5x) sin(−3x) + 0 Recall: limx→0 1 − cos(x) x = 0 =limx→0 x cos2 (5x) sin(−3x) · −3 −3 =limx→0 −3x sin(−3x) · cos2 (5x) −3 =(1) · cos2 (0) −3 Recall: limx→0 sin(x) x = 1 =− 1 3 (f) Note first that −1 ≤ sin(6x − 9) ≤ 1 =⇒ − 1 xex ≤ sin(2x − 3) xex ≤ 1 xex . Now, since lim x→+∞ − 1 xex = lim x→+∞ 1 xex = 0, the Squeeze Theorem implies that lim x→+∞ sin(6x − 9) xex = 0 as well. 2. Define the function f(x) =    J2x − 1K if x < 1 x 2 − 7x + 12 |x − 3| if 1 ≤ x < 3 cosh(sinh(x − 3)) if x ≥ 3. Is f(x) continuous at x = 1 and x = 3? If it is discontinuous on either point, identify the type of discontinuity. Solution: We examine the one-sided limits of f(x) about x = 1 and x = 3. At x = 1: lim x→1− f(x) = lim x→1− J2x − 1K = J1 −K = 0, lim x→1+ f(x) = lim x→1+ x 2 − 7x + 12 |x − 3| = 1 − 7 + 12 |1 − 3| = 3. Since lim x→1− f(x) ̸= lim x→1+ f(x), f has a jump discontinuity at x = 1. At x = 3: lim x→3− f(x) = lim x→3− (x − 3)(x − 4) −(x − 3) = lim x→3− − (x − 4) = 1, lim x→3+ f(x) = lim x→3+ cosh(sinh(x − 3)) = cosh(sinh 0) = 1. Since f(3) = 1 and both one-sided limits equal 1, f is continuous at x = 3. 3. Use the Intermediate Value Theorem to show that the equation e x = −10x + 7 has a solution in the interval [0, 1]. (Hint: Define h(x) = e x + 10x − 7.) Page 2