MATH 152 Midterm: MATH 152 TAMU 2013c Exam 3b Solutions
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= lim n (2013)n+1 (n + 1)! lim n (cid:12)(cid:12)(cid:12)(cid:12) an+1 an (cid:12)(cid:12)(cid:12)(cid:12) 8n3 t |s sn| . R n s s n 3 500 1 . 4000 s t st t r s . F (1) = f (1) = arctan 1 = T s r s r s x = 2 . 4 < x 2 s 2 t . R t t r r s [1, 5] . T y z x = 0 s t r t . S p( 6 0)2 + (0 ( 3))2 + ( 5 ( 5))2 = R (x + 6)2 + y2 + (z + 5)2 = 45 s r t . 21 p( 4)2 + ( 1)2 + 22h 4, 1, 2i = 1 + x(cid:19) dx = c + x) ( 1)nxn+1 n + 1. X = 0, c = ln(1) = 0 r r .
