1. **State the problem:** We need to analyze the function $h(x) = \frac{x}{\cos x - 1}$. 2. **Recall the formula and important rules:** The function is a rational function where the numerator is $x$ and the denominator is $\cos x - 1$. 3. **Identify domain restrictions:** The denominator cannot be zero, so we find where $\cos x - 1 = 0$. 4. **Solve for domain restrictions:** $$\cos x - 1 = 0 \implies \cos x = 1$$ The cosine function equals 1 at $x = 2k\pi$ for all integers $k$. 5. **Domain:** The function is defined for all real $x$ except $x = 2k\pi$ where $k$ is any integer. 6. **Simplify near $x=0$ to understand behavior:** Using the Taylor expansion for cosine near 0: $$\cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$$ So, $$\cos x - 1 \approx -\frac{x^2}{2} + \frac{x^4}{24} - \cdots$$ 7. **Rewrite $h(x)$ near 0:** $$h(x) = \frac{x}{\cos x - 1} \approx \frac{x}{-\frac{x^2}{2} + \frac{x^4}{24}} = \frac{x}{x^2(-\frac{1}{2} + \frac{x^2}{24})}$$ 8. **Cancel common factor $x$ in numerator and denominator:** $$h(x) \approx \frac{\cancel{x}}{\cancel{x} x(-\frac{1}{2} + \frac{x^2}{24})} = \frac{1}{x(-\frac{1}{2} + \frac{x^2}{24})}$$ 9. **Simplify denominator:** $$-\frac{1}{2} + \frac{x^2}{24} = -\frac{1}{2} \left(1 - \frac{x^2}{12}\right)$$ 10. **Final approximate form near 0:** $$h(x) \approx \frac{1}{x \cdot -\frac{1}{2} \left(1 - \frac{x^2}{12}\right)} = \frac{1}{-\frac{x}{2} \left(1 - \frac{x^2}{12}\right)} = -\frac{2}{x \left(1 - \frac{x^2}{12}\right)}$$ 11. **Interpretation:** As $x \to 0$, $h(x)$ behaves like $-\frac{2}{x}$, which tends to $\pm \infty$ depending on the direction. **Final answer:** The function $h(x) = \frac{x}{\cos x - 1}$ is defined for all real $x$ except $x = 2k\pi$, where it has vertical asymptotes. Near zero, it behaves like $-\frac{2}{x}$.