sol09term1 - MAT 137Y 2008-2009 Winter Session Solutions to...

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Unformatted text preview: MAT 137Y, 2008-2009 Winter Session, Solutions to Term Test 1 1. Evaluate the following limits. (Do not prove them using the formal definition of limit.) (10%) (i) lim x → x sin x 1- cos x . Multiplying top and bottom by 1 + cos x , we have lim x → x sin x ( 1 + cos x ) ( 1- cos x )( 1 + cos x ) = lim x → x sin x ( 1 + cos x ) 1- cos 2 x = lim x → x sin x ( 1 + cos x ) sin 2 x = lim x → x sin x · ( 1 + cos x ) = 1 · 2 = 2 . (10%) (ii) lim x → 3 √ 5- x- √ x 2- 7 √ x + 6- 3 . Here we multiply top and bottom by the conjugates of both expressions to get lim x → 3 √ 5- x- √ x 2- 7 √ x + 6- 3 √ 5- x + √ x 2- 7 √ 5- x + √ x 2- 7 ! √ x + 6 + 3 √ x + 6 + 3 = lim x → 3 [( 5- x )- ( x 2- 7 )]( √ x + 6 + 3 ) [( x + 6 )- 9 ][ √ 5- x + √ x 2- 7 ] = lim x → 3 ( 12- x- x 2 )( √ x + 6 + 3 ) ( x- 3 )( √ 5- x + √ x 2- 7 ) = lim x → 3 ( 3- x )( 4 + x )( √ x + 6 + 3 ) ( x- 3 )( √ 5- x + √ x 2- 7 ) =- lim x → 3 ( 4 + x )( √ x + 6 + 3 ) ( √ 5- x + √ x 2- 7 ) =- 21 √ 2 . 2. (10%) (i) Find all solutions in the interval [ , 2 π ) that satisfy the equation 2sin3 x- 1 = 0. We have 2sin3 x- 1 = ⇐⇒ sin3 x = 1 2 . Solving for 3 x , we have either 3 x = π 6 + 2 k π or 3 x = 5 π 6 + 2 k π . Solving for x gives us the solutions x = π 18 + 2 3 k π , x = 5 π 18 + 2 3 k π ....
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This note was uploaded on 04/09/2010 for the course MAT 137 taught by Professor Uppal during the Spring '08 term at University of Toronto.

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sol09term1 - MAT 137Y 2008-2009 Winter Session Solutions to...

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