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Unformatted text preview: Math 138 Assignment 6 Solutions Fall 2008 1. a) Use the δ definition of lim x → a f ( x ) = L to prove the limit: lim x → 5 4 3 x 5 = 7 . Solution Given any > 0, we must show that there is a δ such that 4 3 x 5 7 < for all x such that <  x + 5  < δ . 4 3 x 5 7 = 3 3 x 5 =   3  1 + x 5 = 3 5  5 + x  . Thus 4 3 x 5 7 = 3 5  5 + x  < if  x + 5  < 5 3 . Choosing δ = 5 3 for the given , we thus have 4 3 x 5 7 = 3 5  x + 5  < 3 5 5 3 = if  x + 5  < δ = 5 3 , as required. b) The definition for the limit of f ( x ) as x tends to infinity is: lim x →∞ f ( x ) = b means that for all > 0 there exists R > such that for all x , if x > R then  f ( x ) b  < . Use this definition to prove that lim x →∞ x 2 + x + 1 x 2 + 1 = 1. Solution For all x > 0 we have  f ( x ) 1  = x 2 + x + 1 x 2 + 1 1 = x x 2 + 1 = x x 2 + 1 < x x 2 = 1 x and so given > 0 we can choose R = 1 , and then we have x > R = ⇒  f ( x ) 1  < 1 x < 1 R = , as required....
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 Fall '07
 Anoymous
 Math, Calculus, common ratio

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