sol1-03 - MAT 137Y 2003-2004 Solutions to Test 1 1 Evaluate...

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MAT 137Y, 2003-2004, Solutions to Test 1 1. Evaluate the following limits. (8%) (i) lim t 1 t 3 - 1 t - 1 . lim t 1 t 3 - 1 t - 1 = lim t 1 ( t - 1 )( t 2 + t + 1 ) t - 1 = lim t 1 t 2 + t + 1 = 3 . (8%) (ii) lim x 3 x 2 - 5 - 1 + x x 2 - 9 . Let L be the limit. Then L = lim x 3 ( x 2 - 5 - 1 + x )( x 2 - 5 + 1 + x ) ( x 2 - 9 )( x 2 - 5 + 1 + x ) = lim x 3 x 2 - 5 - 1 - x ( x 2 - 9 )( x 2 - 5 + 1 + x ) = lim x 3 x 2 - x - 6 ( x 2 - 9 )( x 2 - 5 + 1 + x ) = lim x 3 ( x + 2 )( x - 3 ) ( x + 3 )( x - 3 )( x 2 - 5 + 1 + x ) = lim x 3 ( x + 2 ) ( x + 3 )( x 2 - 5 + 1 + x ) = 5 24 . (12%) 2. Solve the inequality | x - 3 | + x 9 . Express your answer in interval notation. If x 3, then x - 3 + x 9, or x 6. Therefore, a solution is x [ 3 , 6 ] . If x < 3, then 3 - x + x 9, or 3 9, which is true for all x which satisfies the assumption, so x ( - , 3 ) also satisfies the inequality. Combining solutions, we get x ( - , 6 ] . 3. (5%) (a) Give the formal ε , δ definition of the statement lim x a f ( x ) = L . For any
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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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sol1-03 - MAT 137Y 2003-2004 Solutions to Test 1 1 Evaluate...

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