MAT137Y1 Midterm: MAT 137Y, 2007–2008Winter Session, Self Generated Solutions to Term Test 1.pdf

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Mat 137y, 2007 2008 winter session, solutions to term test 1 (10%) (i) lim x 0: evaluate the following limits algebraically. |x2 6x + 8| x2 x 12 lim x 4 . Factoring the numerator, x2 6x + 8 = (x 4)(x 2), so. Since x < 4, it follows that |x 4| = 4 x and |x 2| = x 2 (for values x > 2), so we have. Lim x 4 x 2 x + 3 lim x 4 . 7: solve the following inequalities; express your answer as a union of intervals. (10%) (i) x. Moving everything to one side, we have x x + 1. > 0 = x(3x + 2) x + 1. We can now draw a chart of signs to determine where x(3x + 2)/(x + 1) is positive or negative: x < 1 1 < x < 2. Therefore the inequality holds when x ( , 1) ( 2 x+1.

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