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

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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 ) 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 = 6 + 2 k or 3 x = 5 6 + 2 k . Solving for x gives us the solutions x = 18 + 2 3 k , x = 5 18 + 2 3 k ....
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sol09term1 - MAT 137Y, 2008-2009 Winter Session, Solutions...

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