Finding the values of the real constants such that the limit exists

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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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3 Answers

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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]

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Conclusion: no silver bullet, but definitely a way to help you understand this equation if your math skills fail you!

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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}}

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

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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