1. **Stating the problem:** We want to find the limit as $x \to -\infty$ of the function $$f(x) = x - \sqrt{x^2 - 2}$$ 2. **Rewrite the function:** Notice that $$f(x) = x - \sqrt{x^2 - 2} = \frac{(x - \sqrt{x^2 - 2})(x + \sqrt{x^2 - 2})}{x + \sqrt{x^2 - 2}} = \frac{x^2 - (x^2 - 2)}{x + \sqrt{x^2 - 2}} = \frac{2}{x + \sqrt{x^2 - 2}}$$ 3. **Analyze the denominator as $x \to -\infty$:** Since $x$ is very large negative, $\sqrt{x^2 - 2} \approx |x| = -x$ (because $x$ is negative, $-x$ is positive). So, $$x + \sqrt{x^2 - 2} \approx x + (-x) = 0$$ But more precisely, since $\sqrt{x^2 - 2} > |x| - \text{small positive}$, the denominator tends to $0$ from the positive side, actually it tends to $+\infty$ because the dominant term is $-x + \sqrt{x^2 - 2}$ when rewritten. 4. **Calculate the limit:** $$\lim_{x \to -\infty} \frac{2}{-x + \sqrt{x^2 - 2}} = 0$$ 5. **Interpretation:** The function behaves like $2x$ for large negative $x$, so $$\lim_{x \to -\infty} (f(x) - 2x) = 0$$ which means the line $y = 2x$ is an oblique asymptote as $x \to -\infty$. **Final answer:** $$\boxed{\lim_{x \to -\infty} f(x) = 2x \text{ (oblique asymptote) }}$$ This completes the study of the function $f(x) = x - \sqrt{x^2 - 2}$ including its domain, limits, derivative, variations, and asymptotes.