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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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It is highly recommended that instead of putting links you type with LaTeX the actual expressions in the body of the question, otherwise it becomes cumbersome trying to answer without seeing the question. This can be fine when there's a huge question or perhaps a diagram. – DonAntonio Sep 13 '12 at 10:22
fair enough ^.^ I'm new here sorry. Writing my eqn out in latext now. – kingboonz Sep 13 '12 at 10:25

$\lim_{x\to 0}\frac{\sqrt{c+dx}-\sqrt{3}}{x}=\lim_{x\to 0}\frac{(\sqrt{c+dx}-\sqrt{3})(\sqrt{c+dx}+\sqrt{3})}{x(\sqrt{c+dx}+\sqrt{3})}=\lim_{x\to 0}\frac{c+dx-3}{x(\sqrt{c+dx}+\sqrt{3})}$. 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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Another way to eliminate x is to let c=3 and d = 1 right? Why not use that instead? Thanks by the way!@! – kingboonz Sep 13 '12 at 10:22
Yeah the reason I'm confused is how does c=3 and d=6 makes the limit specifically \sqrt(3)? – kingboonz Sep 13 '12 at 10:23
First you must get rid of the $x$ in the denominator. Thus you want your numerator being a multiple of $x$. That forces $c=3$ to eliminate the constant terms. Once you've done so, you are left with $\lim_{x\rightarrow 0} \frac{d}{\sqrt{3+dx}+\sqrt{3}}$. As $x\rightarrow 0$, the limit is $\frac{d}{2\sqrt{3}}$, which must be equal to $\sqrt{3}$. That gives you $d=6$. – Andrea Orta Sep 13 '12 at 10:28
Why $d=1$? :) If you put $c=3, d=1$ then you have $\frac{x}{x(\sqrt{3+x}+\sqrt{3})}=\frac{1}{(\sqrt{3+x}+\sqrt{3})}=\frac{1}{2\sqr‌​t{3}}$ near $x=0$ and this is not $\sqrt{3}$. – Babak S. Sep 13 '12 at 10:31
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. – kingboonz Sep 13 '12 at 10:36

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 $d = 6$.

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