sol1-0607 - MAT 137Y, 20062007 Winter Session, Solutions to...

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Unformatted text preview: MAT 137Y, 20062007 Winter Session, Solutions to Term Test 1 1. Evaluate the following limits. Do not use LHopitals Rule to evaluate the limit. (7%) (i) lim x ( x- 1 ) 2- 1 x 2 + 6 x L = lim x x 2- 2 x + 1- 1 x ( x + 6 ) = lim x x 2- 2 x x ( x + 6 ) = lim x x ( x- 2 ) x ( x + 6 ) = lim x x- 2 x + 6 =- 1 3 . (7%) (ii) lim t sin 2 ( 5 t ) 3 t 2 Using the property that lim x sin x x = 1 and (as a result of making the substitution x = 5 t ) we have lim t sin5 t 5 t = 1, lim t sin 2 ( 5 t ) 3 t 2 = 1 3 lim t sin5 t t sin5 t t = 25 3 lim t sin5 t 5 t sin5 t 5 t = 25 3 1 1 = 25 3 . (7%) (iii) lim x 4 + ( 4- x ) | 3 x- 14 | | 4- x | . If x 4 + , then x > 4, so | 4- x | = | x- 4 | = x- 4. Therefore, lim x 4 + ( 4- x ) | 3 x- 14 | | 4- x | = lim x 4 + ( 4- x ) | 3 x- 14 | x- 4 = lim x 4 +-| 3 x- 14 | =- 2 . (7%) (iv) lim x 3- 9- x 2 x 2 . Multiplying top and bottom by the conjugate, we have lim x 3- 9- x 2 x 2 = lim x ( 3- 9- x 2 )( 3 + 9- x 2 ) x 2 ( 3 + 9- x 2 ) = lim x 9- ( 9- x 2 ) x 2 ( 3 + 9- x 2 ) = lim x x 2 x 2 ( 3 + 9- x 2 ) = lim x 1 3 + 9- x 2 = 1 6 . 2. (7%) (i) Solve the inequality x 2- 3 x x 4- 1 0. Express your answer as a union of intervals....
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sol1-0607 - MAT 137Y, 20062007 Winter Session, Solutions to...

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