calculus - How to solve $limlimits_{x to -infty} left(xleft(sqrt{x^2-x}-sqrt{x^2-1}right)right)$?

I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?

$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$

Answer

$$x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)=\frac{x\left(\left(\sqrt{x^2-x}\right)^2-\left(\sqrt{x^2-1}\right)^2\right)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$

$$=\frac{x(1-x)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$

Since we're searching for the limit as $x\to -\infty$, let $x<0$. Then:

$$=\frac{\frac{1}{-x}(x(1-x))}{\sqrt{\frac{x^2}{(-x)^2}-\frac{x}{(-x)^2}}+\sqrt{\frac{x^2}{(-x)^2}-\frac{1}{(-x)^2}}}=\frac{x-1}{\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}}}\stackrel{x\to -\infty}\to -\infty$$

Because $\sqrt{1-\frac{1}{x}}\stackrel{x\to -\infty}\to 1$ and $\sqrt{1-\frac{1}{x^2}}\stackrel{x\to -\infty}\to 1$ and $x-1\stackrel{x\to -\infty}\to -\infty$.