<s>[INST] Finding the values of the real constants such that the limit exists 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.
 [/INST] $$\lim_{x\to 0}\frac{\sqrt{c+dx}-\sqrt{3}}{x}=\sqrt{3}$$
Since the denominator goes to $0$, the limit cannot exist unless the numerator also goes to $0$.  The numerator is $\sqrt{c+dx}-\sqrt{3}$, so that would have to go to $0$ as $x$ goes to $0$.  But it goes to $\sqrt{c+d\cdot0} - \sqrt{3}$.  Hence $c+d\cdot0$ must be $3$.  That tells you $c$ is $3$, and you've got
$$\lim_{x\to 0}\frac{\sqrt{3+dx}-\sqrt{3}}{x}=\sqrt{3}.$$
Now rationalize the numerator:
$$
\frac{\sqrt{3+dx}-\sqrt{3}}{x} = \frac{\left(\sqrt{3+dx}-\sqrt{3}\right)\left(\sqrt{3+dx}+\sqrt{3}\right)}{x\left(\sqrt{3+dx}+\sqrt{3}\right)}=  \frac{dx}{x\left(\sqrt{3+dx}+\sqrt{3}\right)}.
$$
The $x$s cancel and we get
$$
\frac{d}{\sqrt{3+dx}+\sqrt{3}}.
$$
The limit of that as $x\to0$ is $d/(2\sqrt{3})$.  So you want $d/(2\sqrt{3}) = \sqrt{3}$.  Multiply both sides by $2\sqrt{3}$ and you get $d=6$.
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