1. **Problem 13:** Find the value of $a$ and the limit if $$\lim_{x \to 0} \frac{a \sin x - \sin 2x}{\tan^3 x}$$ is finite. 2. **Step 1:** Recall the Taylor expansions near $x=0$: $$\sin x = x - \frac{x^3}{6} + O(x^5), \quad \sin 2x = 2x - \frac{(2x)^3}{6} + O(x^5) = 2x - \frac{8x^3}{6} + O(x^5),$$ $$\tan x = x + \frac{x^3}{3} + O(x^5) \implies \tan^3 x = \left(x + \frac{x^3}{3} + O(x^5)\right)^3 = x^3 + O(x^5).$$ 3. **Step 2:** Substitute expansions into numerator: $$a \sin x - \sin 2x = a\left(x - \frac{x^3}{6}\right) - \left(2x - \frac{8x^3}{6}\right) + O(x^5) = (a x - 2x) - \left(\frac{a x^3}{6} - \frac{8 x^3}{6}\right) + O(x^5) = (a - 2)x - \frac{a - 8}{6} x^3 + O(x^5).$$ 4. **Step 3:** The denominator is $\tan^3 x = x^3 + O(x^5)$. 5. **Step 4:** Form the limit expression: $$\frac{a \sin x - \sin 2x}{\tan^3 x} = \frac{(a - 2)x - \frac{a - 8}{6} x^3 + O(x^5)}{x^3 + O(x^5)} = \frac{(a - 2)x}{x^3} - \frac{a - 8}{6} \frac{x^3}{x^3} + O(x^2) = (a - 2) \frac{1}{x^2} - \frac{a - 8}{6} + O(x^2).$$ 6. **Step 5:** For the limit to be finite as $x \to 0$, the term with $\frac{1}{x^2}$ must vanish, so: $$a - 2 = 0 \implies a = 2.$$ 7. **Step 6:** Substitute $a=2$ back to find the limit: $$\lim_{x \to 0} \frac{2 \sin x - \sin 2x}{\tan^3 x} = - \frac{2 - 8}{6} = - \frac{-6}{6} = 1.$$ --- 8. **Problem 14:** Evaluate values of $a$, $b$, and $c$ such that $$\lim_{x \to 0} \frac{(a + b \cos x) x - c \sin x}{x^3}$$ is finite. 9. **Step 1:** Use expansions near $x=0$: $$\cos x = 1 - \frac{x^2}{2} + O(x^4), \quad \sin x = x - \frac{x^3}{6} + O(x^5).$$ 10. **Step 2:** Substitute into numerator: $$(a + b \cos x) x - c \sin x = (a + b(1 - \frac{x^2}{2} + O(x^4))) x - c \left(x - \frac{x^3}{6} + O(x^5)\right) = (a + b) x - \frac{b}{2} x^3 - c x + \frac{c}{6} x^3 + O(x^5).$$ 11. **Step 3:** Simplify numerator: $$((a + b) - c) x + \left(- \frac{b}{2} + \frac{c}{6}\right) x^3 + O(x^5).$$ 12. **Step 4:** The denominator is $x^3$. 13. **Step 5:** Form the limit expression: $$\frac{((a + b) - c) x + \left(- \frac{b}{2} + \frac{c}{6}\right) x^3 + O(x^5)}{x^3} = ((a + b) - c) \frac{1}{x^2} + \left(- \frac{b}{2} + \frac{c}{6}\right) + O(x^2).$$ 14. **Step 6:** For the limit to be finite, the $\frac{1}{x^2}$ term must vanish: $$(a + b) - c = 0 \implies c = a + b.$$ 15. **Step 7:** The limit then is: $$\lim_{x \to 0} \frac{(a + b \cos x) x - c \sin x}{x^3} = - \frac{b}{2} + \frac{c}{6} = - \frac{b}{2} + \frac{a + b}{6} = \frac{a}{6} - \frac{b}{3}.$$ 16. **Summary:** - For problem 13: $a=2$, limit $=1$. - For problem 14: $c = a + b$, limit $= \frac{a}{6} - \frac{b}{3}$ (finite for any $a,b$).