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 expansions:** - For small $x$, $\sin x \approx x - \frac{x^3}{6} + O(x^5)$ - $\tan x \approx x + \frac{x^3}{3} + O(x^5)$ - $\sec x = \frac{1}{\cos x} \approx 1 + \frac{x^2}{2} + O(x^4)$ 3. **Evaluate (i):** Start with the expression: $$\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x \sin x}$$ Using the expansion for $\sin x$: $$\sin x = x - \frac{x^3}{6} + O(x^5)$$ So, $$x - \sin x = x - \left(x - \frac{x^3}{6} + O(x^5)\right) = \frac{x^3}{6} + O(x^5)$$ 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, $$\frac{x - \sin x}{x \sin x} = \frac{\frac{x^3}{6} + O(x^5)}{x^2 - \frac{x^4}{6} + O(x^6)} = \frac{\frac{x^3}{6} + O(x^5)}{x^2 \left(1 - \frac{x^2}{6} + O(x^4)\right)}$$ 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):** $$\lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{1}{x} \right) = 0$$ 4. **Evaluate (ii):** Use expansions: $$\tan x \approx x + \frac{x^3}{3} + O(x^5)$$ $$\sec x \approx 1 + \frac{x^2}{2} + O(x^4)$$ So, $$\tan x - \sec x \approx \left(x + \frac{x^3}{3}\right) - \left(1 + \frac{x^2}{2}\right) + O(x^4) = -1 + x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$$ Taking the limit as $x \to 0$: $$\lim_{x \to 0} (\tan x - \sec x) = -1$$ **Final answers:** (i) 0 (ii) -1