Calculating $\lim_{x\to+\infty}(\sqrt{x^2-3x}-x)$

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?

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

Your second one is incorrect, since $\sqrt{x^2}=-x$ for all $x<0$. Hence: $$ \lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{x}=\lim_{x\to-\infty}\frac{\sqrt{x^2-3x}}{-\sqrt{x^2}}=\lim_{x\to-\infty}-\sqrt{1-\frac{3}{x}}=-1 $$

For the last one, we multiply by the conjugate: \begin{align*} \lim_{x \to \infty} [f(x)-g(x)] &= \lim_{x \to \infty} \left[ \sqrt{x^2-3x}-x \right] \\ &= \lim_{x \to \infty} \left[\left(\sqrt{x^2-3x}-x\right) \cdot \dfrac{\sqrt{x^2-3x}+x}{\sqrt{x^2-3x}+x} \right] \\ &= \lim_{x \to \infty} \dfrac{(x^2-3x)-x^2}{\sqrt{x^2-3x}+x} \\ &= \lim_{x \to \infty} \dfrac{-3x}{\sqrt{x^2-3x}+x} \\ &= \lim_{x \to \infty} \dfrac{-3x}{\sqrt{x^2(1-\frac3x)}+x} \\ &= \lim_{x \to \infty} \dfrac{-3x}{x\sqrt{1-\frac3x}+x} \qquad \text{since }\sqrt{x^2}=x \text{ for all }x>0\\ &= \lim_{x \to \infty} \dfrac{-3}{\sqrt{1-\frac3x}+1} \\ &= \dfrac{-3}{\sqrt{1-0}+1} \\ &= \dfrac{-3}{2} \\ \end{align*}

$$\lim_{x\to\infty}\{f(x)-g(x)\}=\lim_{x\to\infty}\{\sqrt{x^2-3x}-x\}$$

$$=\lim_{h\to0}\frac{\sqrt{1-3h}-1}h(\text{ putting } x=\frac1h)$$

Now this can be handled in at least three ways :

Method $1:$

Taylor expansion: $$\lim_{h\to0}\frac{(1-3h)^{\frac12}-1}h=\lim_{h\to0}\frac{1-3h\cdot\frac12+O(h^2)-1}h=-\frac32$$

Method $2:$

L'Hosiptals' Rule :

$$\lim_{h\to0}\frac{(1-3h)^{\frac12}-1}h=\lim_{h\to0}\frac{\frac12\cdot\frac1{\sqrt{1-3h}}\cdot(-3)}1=-\frac32$$

Method $3:$

Rationalizing the numerator like Adriano, $$\lim_{h\to0}\frac{\sqrt{1-3h}-1}h=\lim_{h\to0}\frac{(1-3h)-1}{h(\sqrt{1+3h}+1)}$$

$$=\lim_{h\to0}\frac{-3}{\sqrt{1+3h}+1}(\text{ Cancelling } h\text{ as }h\ne0\text{ as }h\to0 )$$

$$=-\frac32$$