<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] To find $\lim \limits_{x \to \infty} \sqrt{x^2-3x}-x:$
Rewrite $\lim \limits_{x \to \infty} \sqrt{x^2-3x}-x=\lim \limits_{x \to \infty} x \left(\sqrt{1-\frac{3}{x}}-1\right)=\lim \limits_{x \to \infty} \frac{\sqrt{1-\frac{3}{x}}-1}{1/x}.$ This gives an indeterminate $\frac{0}{0}$ form. Applying L'Hopital gives us:
$\lim \limits_{x \to \infty} \frac{\left(\frac{3/x^2}{2 \sqrt{1-3/x}}\right)}{(-1/x^2)}=\lim \limits_{x \to \infty} \frac{-3}{2\sqrt{1-\frac{3}{x}}}=\frac{-3}{2}$
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