Let $f(x) = \sqrt{x^2-3x}$ and $g(x) = x$. Calculate the following limits, showing all working.

I've done the first two - 

$$\lim\_{x\to0}f(x)\\
=\lim\_{x\to0}\sqrt{x^2-3x}=0$$

$$\lim\_{x\to-\infty}\frac{f(x)}{g(x)}\\
=\lim\_{x\to-\infty}\frac{\sqrt{x^2-3x}}{x}=\sqrt{1-\frac{3}{x}}=1$$

How do I calculate this one?

$$\lim\_{x\to+\infty}(f(x)-g(x))$$