<s>[INST] 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.
 [/INST] 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 $y:=\dfrac1x\to0^+$).
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