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evaluate: lim x tends to pi 1+ sec³x/ tan²x

We have, \lim_{x \to π} \frac{1 + \sec^{3} x}{\tan^{2} x} = \lim_{x \to π} \frac{1^{3} + {(\sec x)}^{3}}{\sec^{2} x - 1} = \lim_{x \to π} \frac{(1 + \sec x) (1^{2} - 1 \times \sec x + \sec^{2} x)}{(\sec x + 1) (\sec x - 1)} = \lim_{x \to π} \frac{(\sec x + 1) (1 - \sec x + \sec^{2} x)}{(\sec x + 1) (\sec x - 1)} = \lim_{x \to π} \frac{(1 - \sec x + \sec^{2} x)}{(\sec x - 1)} = \frac{1 - secπ + \sec^{2} π}{secπ - 1} = \frac{1 - (- 1) + {(- 1)}^{2}}{- 1 - 1} = \frac{1 + 1 + 1}{- 2} = - \frac{3}{2}

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