1. **Problem statement:** Evaluate the limits (i) $$\lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{1}{x} \right)$$ (ii) $$\lim_{x \to \frac{\pi}{2}} (\tan x - \sec x)$$ 2. **Recall important formulas and rules:** - For small $x$, $\sin x \approx x - \frac{x^3}{6}$. - $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. - Use algebraic manipulation and limit properties. --- ### Part (i): 3. Rewrite the expression: $$\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x \sin x}$$ 4. Use the Taylor expansion for $\sin x$ near 0: $$\sin x = x - \frac{x^3}{6} + O(x^5)$$ 5. Substitute into numerator: $$x - \sin x = x - \left(x - \frac{x^3}{6} + O(x^5)\right) = \frac{x^3}{6} + O(x^5)$$ 6. Substitute into denominator: $$x \sin x = x \left(x - \frac{x^3}{6} + O(x^5)\right) = x^2 - \frac{x^4}{6} + O(x^6)$$ 7. So the expression becomes: $$\frac{\frac{x^3}{6} + O(x^5)}{x^2 - \frac{x^4}{6} + O(x^6)} = \frac{x^3/6 + O(x^5)}{x^2 (1 - \frac{x^2}{6} + O(x^4))}$$ 8. Simplify by dividing numerator and denominator by $x^2$: $$\frac{x/6 + O(x^3)}{1 - \frac{x^2}{6} + O(x^4)}$$ 9. As $x \to 0$, numerator $\to 0$ and denominator $\to 1$, so limit is: $$0$$ --- ### Part (ii): 10. Expression: $$\tan x - \sec x = \frac{\sin x}{\cos x} - \frac{1}{\cos x} = \frac{\sin x - 1}{\cos x}$$ 11. Evaluate numerator and denominator as $x \to \frac{\pi}{2}$: - $\sin \frac{\pi}{2} = 1$ - $\cos \frac{\pi}{2} = 0$ 12. This is an indeterminate form $\frac{0}{0}$, so apply L'Hôpital's Rule: 13. Differentiate numerator and denominator w.r.t. $x$: - Numerator derivative: $\cos x$ - Denominator derivative: $-\sin x$ 14. Evaluate at $x = \frac{\pi}{2}$: - Numerator derivative: $\cos \frac{\pi}{2} = 0$ - Denominator derivative: $-\sin \frac{\pi}{2} = -1$ 15. So limit is: $$\lim_{x \to \frac{\pi}{2}} \frac{\cos x}{-\sin x} = \frac{0}{-1} = 0$$ --- **Final answers:** (i) $$0$$ (ii) $$0$$