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what is the time of canada now - Feedz API

📅 May 18, 2026 👤 scraface

May 18, 2026

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But we must check if all such \( y \) are attainable. From \( x^2 = \f...
But when \( y = -1 \), \( x^2 = 0 \Rightarrow x = 0 \), valid.
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 \Rightar...
But can \( y \) be arbitrary? Let’s solve for \( y \):
y = \frac{x^2 + 1}{x^2 - 1} = 1 + \frac{2}{x^2 - 1}.
As \( x^2 \to 1^+ \), \( y \to +\infty \); \( x^2 \to 1^- \), \( y \to...
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