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Unformatted text preview: Math 115 001 Test 1  Solutions to Sample Problems Fall 2011 (1) lim x → x 3 1 x does not exist. lim x →∞ x 3 2 x 2 + x 3 x 3 + 5 x = lim x →∞ 1 (2 /x ) + (1 /x 2 ) 3 + (5 /x 2 ) = 1 0 + 0 3 + 0 = 1 3 . lim x →∞ x 2 + 2 x + 1 5 x 4 + 10 x 2 = lim x →∞ (1 /x 2 ) + (2 /x 3 ) + (1 /x 4 ) 5 + 10 /x 2 = 5 = 0. [Note that the preceding was done by cancelling the highest power of x , namely x 4 , that appears in either the numerator or denominator. In class, we cancelled out the highest common power, which would be x 2 . Both methods work.] lim x → sin x 2 x = 1 2 lim x → sin x x = 1 2 (1) = 1 2 . [What if the problem is: Find lim x → sin2 x x ] (2) f ( x ) = lim h → √ x + h + 1 √ x + 1 h = lim h → √ x + h + 1 √ x + 1 h √ x + h + 1 + √ x + 1 √ x + h + 1 + √ x + 1 = lim h → ( x + h + 1) ( x + 1) h ( √ x + h + 1 + √ x + 1) = lim h → h h ( √ x + h + 1 + √ x + 1) = lim h → 1 √ x + h + 1 + √ x + 1 = 1 2 √ x + 1 (3) f ( x ) = x 3 1 x 2 + 1 (2 x ) +3 x 2 ln( x 2 +1) = 2 x 4 ( x 2 + 1) x 2 + 1 +3 x 2 ln( x 2 +1), using the product rule and the chain rule....
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 Fall '09
 Math, Calculus, Derivative, lim, Mathematical analysis

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