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📅 March 9, 2026 👤 scraface

Mar 09, 2026

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But wait: as $u \to 0^+$, $\frac{1}{u^2} \to \infty$, so $f(u) \to \in...
Let’s test at $x = \frac{\pi}{4}$: $\sin x = \cos x = \frac{\sqrt{2}}{...
So $u = \frac{1}{2}$ at $x = \frac{\pi}{4}$, and $f\left(\frac{1}{2}\r...
Now check the behavior as $u \to 0^+$:
\frac{1}{u^2} \to \infty, \quad \frac{2}{u} \to \infty, \quad 2u \to 0
So $f(u) \to \infty$
But this contradicts intuition — is the expression unbounded?
Wait: let’s recompute at small $x$, say $x = 0.1$ radians:
$\sin x \approx 0.0998$, $\cos x \approx 0.995$, so $\sec x \approx 1....
$\sec x + \csc x \approx 11.025$, square $\approx 121.55$

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