1. **State the problem:** Determine if the function $f(x) = \sqrt{x^2 - 1}$ has any asymptotes. 2. **Recall the definition of asymptotes:** - A vertical asymptote occurs where the function approaches infinity as $x$ approaches a certain value. - A horizontal asymptote is a horizontal line that the function approaches as $x \to \pm \infty$. - An oblique asymptote is a slant line the function approaches as $x \to \pm \infty$. 3. **Analyze the domain:** The expression inside the square root must be non-negative: $$x^2 - 1 \geq 0 \implies x \leq -1 \text{ or } x \geq 1$$ 4. **Check for vertical asymptotes:** Vertical asymptotes occur where the function is undefined but the limit tends to infinity. At $x = \pm 1$, the function is defined as $f(\pm 1) = \sqrt{0} = 0$, so no vertical asymptotes. 5. **Check for horizontal asymptotes:** Evaluate the limit as $x \to \infty$: $$\lim_{x \to \infty} \sqrt{x^2 - 1} = \lim_{x \to \infty} |x| \sqrt{1 - \frac{1}{x^2}} = \infty$$ Similarly, as $x \to -\infty$: $$\lim_{x \to -\infty} \sqrt{x^2 - 1} = \infty$$ No finite horizontal asymptotes. 6. **Check for oblique asymptotes:** Rewrite: $$f(x) = \sqrt{x^2 - 1} = |x| \sqrt{1 - \frac{1}{x^2}}$$ For $x \to \infty$, $|x| = x$, so: $$f(x) = x \left(1 - \frac{1}{2x^2} + \cdots \right) = x - \frac{1}{2x} + \cdots$$ As $x \to \infty$, $f(x) \sim x$, so the oblique asymptote is $y = x$. For $x \to -\infty$, $|x| = -x$, so: $$f(x) = -x \left(1 - \frac{1}{2x^2} + \cdots \right) = -x + \frac{1}{2x} + \cdots$$ As $x \to -\infty$, $f(x) \sim -x$, so the oblique asymptote is $y = -x$. **Final answer:** The function $f(x) = \sqrt{x^2 - 1}$ has no vertical or horizontal asymptotes but has two oblique asymptotes: $$y = x \text{ as } x \to \infty$$ $$y = -x \text{ as } x \to -\infty$$