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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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 Spring '08
 UPPAL
 Limits

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