y = \frac{x^2 + 1}{x^2 - 1} = 1 + \frac{2}{x^2 - 1}.
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- As \( y \to 1^- \), \( \frac{y+1}{y-1} \to -\infty \), invalid.
- When \( y < -1 \), \( \frac{y+1}{y-1} > 0 \), e.g., \( y = -2 \Rightarrow x^2 = \frac{-1}{-3} = \frac{1}{3} \), real \( x \).
- But can \( y \) be arbitrary? Let’s solve for \( y \):
- As \( x^2 \to 1^+ \), \( y \to +\infty \); \( x^2 \to 1^- \), \( y \to -\infty \); \( x^2 \to \infty \
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