1. **Problem statement:** Evaluate the limits (i) $$\lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{1}{x} \right)$$ (ii) $$\lim_{x \to 0} (\tan x - \sec x)$$ 2. **Recall important formulas and rules:** - As $x \to 0$, $\sin x \approx x - \frac{x^3}{6}$ - As $x \to 0$, $\tan x \approx x + \frac{x^3}{3}$ - As $x \to 0$, $\sec x = \frac{1}{\cos x} \approx 1 + \frac{x^2}{2}$ - Use series expansions to simplify expressions near zero. 3. **Evaluate (i):** $$\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x \sin x}$$ Using the expansion $\sin x = x - \frac{x^3}{6} + O(x^5)$, $$x - \sin x = x - \left(x - \frac{x^3}{6} + O(x^5)\right) = \frac{x^3}{6} + O(x^5)$$ Also, $x \sin x = x \left(x - \frac{x^3}{6} + O(x^5)\right) = x^2 - \frac{x^4}{6} + O(x^6)$ Therefore, $$\frac{x - \sin x}{x \sin x} = \frac{\frac{x^3}{6} + O(x^5)}{x^2 - \frac{x^4}{6} + O(x^6)} = \frac{\frac{x^3}{6} + O(x^5)}{x^2 \left(1 - \frac{x^2}{6} + O(x^4)\right)}$$ Simplify denominator: $$= \frac{\frac{x^3}{6} + O(x^5)}{x^2} \cdot \frac{1}{1 - \frac{x^2}{6} + O(x^4)} = \frac{x}{6} + O(x^3)$$ As $x \to 0$, this tends to 0. 4. **Evaluate (ii):** Using expansions: $$\tan x - \sec x \approx \left(x + \frac{x^3}{3}\right) - \left(1 + \frac{x^2}{2}\right) = x + \frac{x^3}{3} - 1 - \frac{x^2}{2}$$ As $x \to 0$, the dominant term is $-1$, so the limit is: $$\lim_{x \to 0} (\tan x - \sec x) = -1$$ **Final answers:** (i) $$0$$ (ii) $$-1$$