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# sol1-04 - MAT 137Y 2004-2005 Solutions to Term Test 1 1...

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MAT 137Y 2004-2005 Solutions to Term Test 1 1. Evaluate the following limits, or show that the limit does not exist. (10%) (i) lim x 0 1 + x - 1 - x x . lim x 0 1 + x - 1 - x x = lim x 0 1 + x - 1 - x x 1 + x + 1 - x 1 + x + 1 - x = lim x 0 ( 1 + x ) - ( 1 - x ) x ( 1 + x + 1 - x ) = lim x 0 2 x x ( 1 + x + 1 - x ) = lim x 0 2 1 + x + 1 - x = 1 . (10%) (ii) lim x 0 ( tan x )( tan2 x ) x tan3 x . lim x 0 ( tan x )( tan2 x ) x tan3 x = lim x 0 sin x x · sin2 x 2 x 3 x sin3 x 2 x 3 x cos3 x ( cos x )( cos2 x ) = 1 · 1 · 1 · 2 3 · 1 ( 1 )( 1 ) = 2 3 . (10%) (iii) lim x 4 | 4 - x | x 2 - x - 12 . By definition, | 4 - x | = x - 4 , x 4 , 4 - x , x < 4 . Taking cases on x , we have lim x 4 + | 4 - x | x 2 - x - 12 = lim x 4 + x - 4 ( x - 4 )( x + 3 ) = lim x 4 + 1 x + 3 = 1 7 . Similarly, lim x 4 - | 4 - x | x 2 - x - 12 = lim x 4 - 4 - x ( x - 4 )( x + 3 ) = lim x 4 - - 1 x + 3 = - 1 7 . Since lim x 4 - | 4 - x | x 2 - x - 12 = lim x 4 + | 4 - x | x 2 - x - 12 , it follows the limit does not exist. 2. (8%) (i) Solve the inequality x 2 - 9 3 x + 1 < 1 . Moving all terms to one side, we solve x 2 - 9 3 x + 1 - 1 < 0 ⇐⇒ x 2 - 9 - ( 3 x + 1 ) 3 x + 1 < 0 ⇐⇒ x 2 - 3 x - 10 3 x + 1 < 0 ⇐⇒ ( x - 5 )( x + 2 ) 3 x + 1 < 0 .

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