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?
$$\lim_{x\to+\infty}(f(x)-g(x))$$
 A: 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*}
A: $$\lim_{x\to\infty}\{f(x)-g(x)\}=\lim_{x\to\infty}\{\sqrt{x^2-3x}-x\}$$
$$=\lim_{h\to0}\frac{\sqrt{1-3h}-1}h(\text{ putting } x=\frac1h)$$
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,
 $$\lim_{h\to0}\frac{\sqrt{1-3h}-1}h=\lim_{h\to0}\frac{(1-3h)-1}{h(\sqrt{1+3h}+1)}$$
$$=\lim_{h\to0}\frac{-3}{\sqrt{1+3h}+1}(\text{ Cancelling } h\text{ as }h\ne0\text{ as }h\to0  )$$
$$=-\frac32$$
A: 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:
$\lim \limits_{x \to \infty} \frac{\left(\frac{3/x^2}{2 \sqrt{1-3/x}}\right)}{(-1/x^2)}=\lim \limits_{x \to \infty} \frac{-3}{2\sqrt{1-\frac{3}{x}}}=\frac{-3}{2}$
A: As
$$f(x)-g(x)=\sqrt{x^2-3x}-x$$
$$=\frac{x^2-3x-x^2}{\sqrt{x^2-3x}+x}$$
$$=\frac{-3x}{\sqrt{x^2-3x}+x}$$
$$=\frac{-3}{\sqrt{1-3/x}+1}$$
Now when $\lim_{x\to\infty}$ we get
$$=\frac{-3}{\sqrt{1-0}+1}$$
$$=\frac{-3}{2}$$
