Limits at infinity

Hint:$$\frac{1}{\sqrt{x^2-2x}-x}\cdot\frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x}$$

The points that we can add to Adi's post are:

$$\frac{1}{\sqrt{x^2-2x}-x}\times\frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x}=\frac{\sqrt{x^2-2x}+x}{(x^2-2x)-x^2}=\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}$$ Now if $x\to+\infty$ so $|x|=x$ and so $$\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}=\frac{x\sqrt{1-2/x}+x}{-2x}=\frac{x\times \sqrt{1-0}+x}{-2x}\longrightarrow -1$$ And if $x\to-\infty$ so $|x|=-x$ and so $$\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}\longrightarrow 0$$

What you did is correct but you did not go to the end.

You know that, when $y$ is small, $\sqrt{1-y}$ is close to $1 - y /2$ (this is the begining of the Taylor series); so, the denominator of your last fraction is approximated by : $(1 - 2 / (2 x) - 1) = - 1 / x$.

Dividing the numerator by this last result gives the limit of $-1$ you were expecting.