MAT137Y1 Midterm: 2006 Test 1 solution
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MAT137Y1 Full Course Notes
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Mat 137y, 2006 2007 winter session, solutions to term test 1: evaluate the following limits. Do not use l"h opital"s rule to evaluate the limit. (7%) (i) lim x 0 (x 1)2 1 x2 + 6x x2 2x + 1 1 x(x + 6) L = lim x 0 sin2(5t) (7%) (ii) lim t 0. = 1, x2 2x x(x + 6) x(x 2) x(x + 6) x 2 x + 6. = 1 and (as a result of making the substitution x = 5t) we have sin2(5t) 3 (4 x)|3x 14| (7%) (iii) lim x 4+ If x 4+, then x > 4, so |4 x| = |x 4| = x 4. Multiplying top and bottom by the conjugate, we have. 2. (7%) (i) solve the inequality x2 3x x4 1. Express your answer as a union of intervals. x(x 3)