Question: 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.

$$\lim\_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}$$

Answer: 
$$\lim\_{x\to\infty}\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x} = \lim\_{x\to\infty}\frac{\sqrt{4/x^2+1}+3}{1-5/x} $$

Now as $x\to\infty$ $4/x^2,5/x$ will go to $0$. Thus we have $4$.
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

$$\frac{\sqrt{4x^2+x^4}+3x^2}{x^2-5x}\to4$$