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What is sqrt(x^2 + 5x + 8) - x as x tends to infinity?

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xetaalpha2 | Student | (Level 2) Honors

Posted April 15, 2011 at 1:18 PM via web

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What is sqrt(x^2 + 5x + 8) - x as x tends to infinity?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 15, 2011 at 1:28 PM (Answer #1)

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We need to find lim x--> inf. [sqrt(x^2 + 5x + 8) - x]

sqrt(x^2 + 5x + 8) - x

=> [sqrt(x^2 + 5x + 8) - x]*[sqrt(x^2 + 5x + 8) + x]/[sqrt(x^2 + 5x + 8) + x]

=> [sqrt(x^2 + 5x + 8)]^2 - x^2 / [sqrt(x^2 + 5x + 8) + x]

=> x^2 + 5x + 8 - x^2 / [sqrt(x^2 + 5x + 8) + x]

=> 5x + 8 / [sqrt(x^2 + 5x + 8) + x]

divide all the terms by x

=> (5 + 8/x) / [sqrt (1 + 5/x + 8/x) + 1]

lim x--> inf. [sqrt(x^2 + 5x + 8) - x]

=> lim (1/x)-->0 [(5 + 8/x) / [sqrt (1 + 5/x + 8/x) + 1]]

substitute 1/x = 0

=> (5 + 0)/[sqrt 1 + 0 +0 + 1)

=> 5/2

The required limit is 5/2.

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