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.
 A: I don't know a way Mathematica is going to solve that directly for you, but it can help you understand what happens. First, let's ask it what the general expression for the limit is:
Limit[(Sqrt[c + d*x] - Sqrt[3])/x, x -> 0]

The answer is:
DirectedInfinity[-Sqrt[3] + Sqrt[c]]

So, this limit is an infinity, and has the same sign as $\sqrt c-\sqrt 3$. If you want the limit to be finite, you have only one special case you can try: what happens if $\sqrt c-\sqrt 3=0$, i.e. if $c=3$?
Limit[(Sqrt[3 + d*x] - Sqrt[3])/x, x -> 0]

The results is:
d/(2 Sqrt[3])

which is finite, so all is well. Now, you need this to be equal to $\sqrt 3$, which you can sure solve yourself, but just to be thorough, let's ask Mathematica to do it:
Solve[Limit[(Sqrt[3 + d*x] - Sqrt[3])/x, x -> 0] == Sqrt[3], d]


Conclusion: no silver bullet, but definitely a way to help you understand this equation if your math skills fail you!
A: Method I
Since we have a fraction going to a non-vanishing value given its denominator is going to 0 we have to assume that its numerator also tends to 0, therefore we should solve :
Reduce[Limit[Numerator[(Sqrt[c + d*x] - Sqrt[3])/x], x -> 0] == 0, {c, d}]


c == 3


We could use Solve as well with the option : InverseFunctions -> True, now taking the value into the system we have :
Reduce[ c == 3 && Limit[(Sqrt[3 + d*x] - Sqrt[3])/x, x -> 0] == Sqrt[3], {c, d}]


c == 3 && d == 6


Method II
Another way to tackle the problem is to expand the expression in a power series around x == 0 : 
Normal @ Series[ (Sqrt[c + d*x] - Sqrt[3])/x, {x, 0, 3}]


Now you can find limits of every term separately :
Limit[#, x -> 0] & /@ List @@ Normal@Series[(Sqrt[c + d*x] - Sqrt[3])/x, {x, 0, 5}]


{d/(2 Sqrt[c]), (-Sqrt[3] + Sqrt[c]) Infinity, 0, 0, 0, 0, 0}


The only problem comes from the second term, however we can get rid of it by assuming a priori its value by setting c == 3 or simply solving this system :
Solve[ {( -Sqrt[3] + Sqrt[c])/x == 0,  d/(2 Sqrt[c]) == Sqrt[3]}, {c, d}, 
         InverseFunctions -> True]


 {{c -> 3, d -> 6}}


A: $$\lim_{x\to 0}\frac{\sqrt{c+dx}-\sqrt{3}}{x}=\sqrt{3}$$
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 $d=6$.
