MAT137Y1 Lecture : MAT 137Y 2007-08Winter Session, Self Generated Solutions to Problem Set 4.pdf
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Mat 137y 2007-08 winter session, solutions to problem set 4. Let b = max(|l |,|l + |), then it follows that. | f (x)| < b, as required. (cid:40) 2 (a) lim x 2 x2 4 x3 8. = lim x 2 (x + 2)(x 2) (x 2)(x2 + 2x + 4) (b) by the de nition of absolute value, = lim x 2 x + 2 x2 + 2x + 4. Therefore, for values close to zero, |5x + 1| = 5x + 1 and |5x 1| = 5x + 1. Hence lim x 0 (5x + 1) ( 5x + 1) 10x x x x (c) multiplying top and bottom by the conjugate, we have lim x 3 x + 6 x x3 3x2 = lim x 3. 54 x + 6 x2 (x3 3x2)( x + 6 + x) (x 3)(x + 2) x2(x 3)( x + 6 + x)
