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Find the values of the real constants $c$ and $d$ such that

$\lim_{x\to 0}\frac{\sqrt{c+dx} - \sqrt{3}}{x} = \sqrt{3} $

I really have no clue how to even get started.

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    fair enough ^.^ I'm new here sorry. Writing my eqn out in latext now.2012-09-13

2 Answers 2

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$\lim_{x\to 0}\frac{\sqrt{c+dx}-\sqrt{3}}{x}=\lim_{x\to 0}\frac{(\sqrt{c+dx}-\sqrt{3})(\sqrt{c+dx}+\sqrt{3})}{x(\sqrt{c+dx}+\sqrt{3})}=\lim_{x\to 0}\frac{c+dx-3}{x(\sqrt{c+dx}+\sqrt{3})}$. If this limit wants to be $\sqrt{3}$ so, we have to eliminate $x$ from the denominator. This makes $c=3$ and $d=6$. Check it.

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    Thanks!!! Thanks so much. I hope i'm not asking for too much, but I was also hoping for another solution by the l hopital theorem.2012-09-13
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In the comment to the other answer you ask for a method using L'Hôpital's rule. Note first that the only way the limit is going to exists is if $c = 3$. By L'Hôpital then you have $ \lim_{x \to 0}\frac{\sqrt{3 + dx} - \sqrt{3}}{x} = \lim_{x\to 0} \frac{d}{2\sqrt{3+dx}}. $

The only say that is going to equal $\sqrt{3}$ is if $d = 6$.