<s>[INST] 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?
$$\lim_{x\to+\infty}(f(x)-g(x))$$
 [/INST] 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*}
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