Where is the mistake in finding $c$ and $n$ of $\sqrt{x^3+4x} - \sqrt{x^3+x} \sim cx^n$ for $x \to 0$ and $x \to +\infty$ First we simplify in the following way:
$$\frac{3}{\sqrt{x}\left( (1+\frac{4}{x^2})^{1/2}+(1+\frac{1}{x^2})^{1/2} \right)}$$
For $x \to 0$:
We then postulate that since the above must be equivalent to some constant times $x$ to some power, then the inverted fraction is equivalent to some $\frac{1}{cx^n}$, I actually think that this step is flawed, because for $x \to 0$, $\frac{1}{cx^n}$ is indeterminate for $x \neq 0$. We then have
$$\lim_{x \to 0} \frac{\sqrt{x}}{3}\left(\left(1+\frac{4}{x^2}\right)^{1/2}-1 \right) + \lim_{x \to 0} \left( \left(1+\frac{1}{x^2} \right)^{1/2}-1 \right) = \\ \frac{\sqrt{x}}{6} \left( \frac{4}{x^2} + \frac{1}{x^2} \right)$$
And so $c=6/5$ and $n=1.5$, because we have to invert back.
For $x \to +\infty$:
We invert again, but this time also divide by $cx^n$. So that we end up with:
$$\lim_{x \to +\infty} \frac{c}{3} x^n \sqrt{x} \left(\left( 1+\frac{4}{x^2} \right)^{1/2} + \left( 1+\frac{1}{x^2}\right)^{1/2} \right) = \\
\frac{2c}{3}(x^n \sqrt{x})$$
that means that $c=2/3$ and $n=0.5$.
But one or both of these solutions are, in fact, incorrect.
 A: Your first line is useful when $x\to +\infty$:
$$\frac{3}{\sqrt{x}\left( (1+\frac{4}{x^2})^{1/2}+(1+\frac{1}{x^2})^{1/2} \right)}\sim \frac{3}{\sqrt{x}(1+1)}=\frac{3}{2}\cdot x^{-1/2}.$$
For $x\to 0$, you may simply note that
$$\sqrt{x^3+4x} - \sqrt{x^3+x}=\sqrt{x}(\sqrt{x^2+4} - \sqrt{x^2+1})\sim \sqrt{x} (\sqrt{4}-1)=x^{1/2}.$$
Alternative method.
In both cases, it suffices to use the following fact: as $t\to 0$,
$$\sqrt{1+t}=1+\frac{t}{2}+o(t).$$
1) As $x\to 0$,
$$\begin{align}
\sqrt{x^3+4x} - \sqrt{x^3+x}&=x^{1/2}\left(2\sqrt{1+(x/2)^2} - \sqrt{1+x^2}\right)
\\&=x^{1/2}\left(2(1+\frac{(x/2)^2}{2}+o(x^2)) - (1+\frac{x^2}{2}+o(x^2)\right)\\&=x^{1/2}+o(x^{1/2}).
\end{align}$$
2) As $x\to +\infty$,
$$\begin{align}\sqrt{x^3+4x} - \sqrt{x^3+x}&=x^{3/2}\left(\sqrt{1+4/x^2} - \sqrt{1+1/x^2}\right)
\\&=x^{3/2}\left((1+\frac{4/x^2}{2}+o(1/x^2)) - (1+\frac{1/x^2}{2}+o(1/x^2)\right)
\\&=x^{3/2}\left(\frac{(4-1)/x^2}{2}+o(1/x^2)\right)\\&=\frac{3}{2}\cdot x^{-1/2}+o(x^{-1/2}).\end{align}$$
A: For $x\to0$, $$\sqrt{x^3+4x} - \sqrt{x^3+x}=\frac{3x}{ \sqrt{x^3+4x} +\sqrt{x^3+x}}=x^{1/2}\frac3{ \sqrt{x^2+4} +\sqrt{x^2+1}}\\\to x^{1/2}$$
because the denominator of the fraction tends to $3$ (the $x^2$ become negligible).
And for $x\to\infty$, $$\sqrt{x^3+4x} - \sqrt{x^3+x}=\frac{3x}{ \sqrt{x^3+4x} +\sqrt{x^3+x}}=\frac32x^{-1/2}\frac{2x^{3/2}}{ \sqrt{x^3+4} +\sqrt{x^3+1}}\\\to \frac32x^{-1/2}$$
because the denominator of the fraction tends to $2x^{3/2}$ (the additive constant become negligible; you can as well consider $y:=\dfrac1x\to0^+$).
A: For $x\to 0$, recall that firstly we can estimate
$$\sqrt{x^3+4x}\sim \sqrt{4x}\left(1+\frac14x^2\right)^\frac12=2\sqrt{x}+\frac14\sqrt{x}x^2$$
$$\sqrt{x^3+x} \sim \sqrt{x}\left(1+x^2\right)^\frac12=\sqrt{x}+\frac12\sqrt{x}x^2$$
therefore it seems that
$$\sqrt{x^3+4x} - \sqrt{x^3+x} \sim  \sqrt{x}$$
to prove that we need to show that
$$\frac{\sqrt{x^3+4x} - \sqrt{x^3+x}}{\sqrt{x}}\to 1$$
For $x\to \infty$, recall that firstly we can estimate
$$\sqrt{x^3+4x}\sim \sqrt{x^3}\left(1+2\frac1{x^2}\right)^\frac12=\sqrt{x^3}+\frac{2}{\sqrt x}$$
$$\sqrt{x^3+x} \sim \sqrt{x^3}\left(1+\frac1{2x^2}\right)^\frac12=\sqrt{x^3}+\frac{1}{2\sqrt x}$$
therefore it seems that
$$\sqrt{x^3+4x} - \sqrt{x^3+x} \sim \frac32 \frac{1}{\sqrt x}$$
to prove that we need to show that
$$\frac{\sqrt{x^3+4x} - \sqrt{x^3+x}}{\frac32 \frac{1}{\sqrt x}}\to 1$$
