$\lim\limits_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}$ Can anyone help me solve this?
I know the answer is 4, but I don't really know how do I find the biggest power of $x$ when there's a square root.
$$\lim_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}$$

As an alternative, note that
$$\frac{\sqrt{x^4}+3x^2}{x^2-5x}=\frac{4x^2}{x^2-5x}\le\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}\le\frac{\sqrt{(x^2+2)^2}+3x^2}{x^2-5x}=\frac{4x^2+2}{x^2-5x}$$
thus for squeeze theorem
$$\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}\to4$$
