MAT137Y1 Lecture : MAT 137Y 2007-08Winter Session, Self Generated Solutions to Problem Set 13.pdf
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Mat 137y 2007-08 winter session, solutions to problem set 12 (cid:104) 2. = lim a 0+ (cid:90) 1 (cid:90) dx . 4. dx (cid:90) sin2 2x: since 0 comparison test (10. 7. 2) that x2 x. 1 (cid:90) x2 dx, and sin2 2x x2 (a) we prove by induction on n that lim x 0+ 1 a = 2, so the integral con- (cid:21) 1 x2 dx converges (10. 7. 1), it follows by the integral lim x 0+ xlnx = lim x 0+ lnx. 1/x2 = lim x 0+ ( x) = 0, so the base case is true. Now suppose the statement is true for n = k, i. e. x(lnx)k = 0. H= lim x 0+ so the statement is true for all positive integers n. = (k + 1) lim x 0+ x(lnx)k = 0, (b) once again we apply induction to show that (lnx)n dx = ( 1)nn!. 1/x2 (k + 1)(lnx)k (1/x) (cid:90) 1 (cid:104)
