Answer
$$\frac{-3}{2}$$
Work Step by Step
Given $$\lim_{x\to \infty} (\sqrt{x^2-3x}-x)$$
Then
\begin{align*}
\lim_{x\to \infty} (\sqrt{x^2-3x}-x)&=\lim_{x\to \infty} (\sqrt{x^2-3x}+x)\frac{\sqrt{x^2+3}+x}{\sqrt{x^2+3}+x}\\
&=\lim_{x\to \infty} \frac{x^2-3x -x^2}{\sqrt{x^2-3x}+x}\\
&=\lim_{x\to \infty} \frac{ -3x }{\sqrt{x^2-3x}+x} \\
&=\lim_{x\to \infty} \frac{ -3x/x }{\sqrt{x^2/x^2-3x/x^2}+x/x} \\
&=\lim_{x\to \infty} \frac{ -3 }{\sqrt{1-3 /x}+1} \\
&=\frac{-3}{1+1}\\
&=\frac{-3}{2}
\end{align*}
