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Evaluate the limit using L'Hopital's Rule limx→49, (7)/(√x−7)−(98)(x−49)?

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Evaluate the limit using L'Hopital's Rule

limx→49, (7)/(√x−7)−(98)(x−49)

limits
-lhopitals-rule

asked Apr 20, 2013 in CALCULUS by payton Apprentice

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1 Answer

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$italic lim with x rightwards arrow 49 below fraction numerator 7 over denominator square root of x minus 7 end fraction minus 98 open parentheses x minus 49 close parentheses equal fraction numerator 7 over denominator square root of 49 minus 7 end fraction minus 98 open parentheses 49 minus 49 close parentheses$ (Substitute x = 49 )

$space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equal fraction numerator 7 over denominator 7 minus 7 end fraction minus 98 open parentheses 0 close parentheses$

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equal 7 over 0 minus open parentheses 0 close parentheses

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equal 7 over 0

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equal infinity

answered Apr 26, 2013 by diane Scholar

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