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 Post subject: limitPosted: Tue, 17 Apr 2012 18:58:17 UTC
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. Let and be the roots of the equation approaches to .

My work i try to find the roots by using quadratic formula and then apply limit . (I have solved this question by taking sum of roots and product ) i want to solve this question by quadratic formula only . irrational part in discriminant is quite cumbersome .how to handle discriminant?

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 Post subject: Re: limitPosted: Tue, 17 Apr 2012 21:47:36 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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mun wrote:
. Let and be the roots of the equation approaches to .

My work i try to find the roots by using quadratic formula and then apply limit . (I have solved this question by taking sum of roots and product ) i want to solve this question by quadratic formula only . irrational part in discriminant is quite cumbersome .how to handle discriminant?

Why not use L'Hopital's rule on the polynomial as a goes to 0 (divide by the leading term first), then use the quadratic formula on the resulting polynomial? iirc that works.

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 Post subject: Re: limitPosted: Wed, 18 Apr 2012 02:13:34 UTC
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sir why direct quadratic formula is not working. i have used binomial theorem in discriminant(on every radical expression) and it is creating problem.

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 Post subject: Re: limitPosted: Wed, 18 Apr 2012 02:16:46 UTC
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mun wrote:
sir why direct quadratic formula is not working. i have used binomial theorem in discriminant(on every radical expression) and it is creating problem.

Well that's probably why, that sounds terrible to deal with. I mean, if you want to use the quadratic formula, that's fine, but then use L'Hopital to calculate the limit.

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