sol to term test 1

sol to term test 1 - MAT 137Y, 2008-2009 Winter Session,...

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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 0 x sin x 1 - cos x . Multiplying top and bottom by 1 + cos x , we have lim x 0 x sin x ( 1 + cos x ) ( 1 - cos x )( 1 + cos x ) = lim x 0 x sin x ( 1 + cos x ) 1 - cos 2 x = lim x 0 x sin x ( 1 + cos x ) sin 2 x = lim x 0 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 ² = 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 [ 0 , 2 π ) that satisfy the equation 2sin3 x - 1 = 0. We have 2sin3 x - 1 = 0 ⇐⇒ 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 . Therefore, the solutions in the interval [ 0 , 2 π ) that satisfy the original equation are x = π 18 , 13 π 18 , 25 π 18 , 5 π 18 , 17 π 18 , 29 π 18 . (10%) (ii) Solve the inequality | 2 x | + | x - 3 | < 5 and express your answer as a union of intervals.
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sol to term test 1 - MAT 137Y, 2008-2009 Winter Session,...

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