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:** - Near zero, $$\sin x \approx x - \frac{x^3}{6}$$ (Taylor expansion) - $$\tan x = \frac{\sin x}{\cos x}$$ and $$\sec x = \frac{1}{\cos x}$$ - Limits involving trigonometric functions often use expansions or algebraic manipulation. 3. **Evaluate (i):** Start with $$\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x \sin x}$$ Using the expansion $$\sin x = x - \frac{x^3}{6} + O(x^5)$$, $$x - \sin x = x - \left(x - \frac{x^3}{6} + O(x^5)\right) = \frac{x^3}{6} + O(x^5)$$ Denominator: $$x \sin x = x \left(x - \frac{x^3}{6} + O(x^5)\right) = x^2 - \frac{x^4}{6} + O(x^6)$$ So, $$\frac{x - \sin x}{x \sin x} = \frac{\frac{x^3}{6} + O(x^5)}{x^2 - \frac{x^4}{6} + O(x^6)} = \frac{\frac{x^3}{6} + O(x^5)}{x^2 (1 - \frac{x^2}{6} + O(x^4))}$$ Divide numerator and denominator by $$x^2$$: $$= \frac{\frac{x^3}{6} + O(x^5)}{x^2} \cdot \frac{1}{1 - \frac{x^2}{6} + O(x^4)} = \frac{x}{6} + O(x^3)$$ As $$x \to 0$$, this tends to 0. **Answer for (i):** $$0$$ 4. **Evaluate (ii):** $$\lim_{x \to \frac{\pi}{2}} (\tan x - \sec x)$$ Rewrite in terms of $$\cos x$$ and $$\sin x$$: $$\tan x - \sec x = \frac{\sin x}{\cos x} - \frac{1}{\cos x} = \frac{\sin x - 1}{\cos x}$$ As $$x \to \frac{\pi}{2}$$, $$\sin x \to 1$$ and $$\cos x \to 0$$. Use substitution $$x = \frac{\pi}{2} - h$$ with $$h \to 0$$: $$\sin\left(\frac{\pi}{2} - h\right) = \cos h \approx 1 - \frac{h^2}{2}$$ $$\cos\left(\frac{\pi}{2} - h\right) = \sin h \approx h$$ So, $$\frac{\sin x - 1}{\cos x} = \frac{\cos h - 1}{\sin h} \approx \frac{\left(1 - \frac{h^2}{2}\right) - 1}{h} = \frac{-\frac{h^2}{2}}{h} = -\frac{h}{2}$$ As $$h \to 0$$, this tends to 0. **Answer for (ii):** $$0$$