Find $c$ and $p$ such that $\sqrt{x^3+4x}-\sqrt{x^3+x} \sim cx^p$ for $x \to +\infty$

Question

I have managed to solve this for $x \to 0$, or at least I think so. My answers are $c=-0.375$ and $p=1$. However, I am stuck for the case when we go to positive infinity. Here is my fruitless attempt:

$$\lim_{x \to + \infty} \frac{x^{3/2}\left(\left(1+\frac{4}{x^2}\right)^{1/2}-\left(1+\frac{1}{x^2}\right)^{1/2} \right)}{cx^n} = 1$$

We can get rid of the numerator no problem. But then the numerator converges to zero. So this is not the form that we should be looking at. If I take out the $4$ from the first brackets, then I am still getting $1-1$ in the numerator.

EDIT:

for $x \to 0$:

$$\sqrt{x} \left( (1+x^2/4)^{1/2} - (1+x^2)^{1/2} \right) = \\ \sqrt{x} \left( ((1+x^2/4)^{1/2}-1)-((1+x^2)^{1/2}-1) \right) \sim \sqrt{x} \left(\frac{x^2}{8}-\frac{x^2}{2}\right)$$

I think I missed a 4 in this case.

Answer 1

Observe that

$$\sqrt{x^3+4x}-\sqrt{x^3+x}=\frac{3}{\sqrt x\left(\sqrt{1+\frac4{x^2}}+\sqrt{1+\frac1{x^2}}\right)}\xrightarrow[x\to\infty]{}0$$

Answer 2

Let $1/x=h^2,h>0$ to find

$$\lim_{h\to0}\dfrac{\sqrt{1+4h^4}-\sqrt{1+h^4}}{h^3}$$

$$=\lim...\dfrac{1+4h^4-(1+h^4)}{h^3}\cdot\lim...\dfrac1{\sqrt{1+4h^4}+\sqrt{1+h^4}}=?$$