Tiêu đề MATH 21 LE 1 ANSWER KEY.pdf ✅

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