MATH 152 Midterm: MATH 152 TAMU 2013c Exam 2b Solutions
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4 lim (ln|4a 2| ln|2|) + lim a 1/2+ a 1/2 . R s s lim a 1/2 ln|4a 2| = . 4x 2 (ln|10| ln|4a 2|) a. T s r s st t t s x 4 = 4 tan . T t x s r = y s t t . Ds = s1 +(cid:18) 1 r1 + r2 x2 dx . R s ts t r y = 0 x2 x2 = 36 r x = 6 . 2 x s t t t r s s = An = ln(cid:18) n2 + 1. S n , |an| 0 s an 0 . T rst t r s a1 t r t r t r s n + 1 t t r t s . Q t n = 1 s 2n + 1 r r an = n + 1. S r s s tr t a = 4 r =
