[Math] Limits at infinity

calculusinfinitylimits

I'm working with limits at infinity and stumbled upon this exercise where I want to evaluate the indicated limit:

$$\lim_{x \to \infty} \frac{1}{\sqrt{x^2-2x}-x}$$

I tried to solve it by doing the following:

$$\lim_{x \to \infty} \frac{1}{\sqrt{x^2-2x}-x} = \lim_{x \to \infty} \frac{1}{\sqrt{x^2} \sqrt{1-\frac{2}{x}}-x} = \lim_{x \to \infty} \frac{1}{x \sqrt{1-\frac{2}{x}}-x} = \lim_{x \to \infty} \frac{\frac{1}{x}}{\sqrt{1-\frac{2}{x}}-1}$$

But the answer should be $-1$, so what I did must be wrong. How do you evaluate this limit the best way possible?

Best Answer

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