1. **State the problem:** We want to find the value of $$\ln y = \lim_{x \to 0} \frac{\tan x - \sin x}{3x^2}$$ and then find $$y$$ itself. 2. **Recall series expansions:** For small $$x$$, use the Taylor series expansions: $$\tan x = x + \frac{x^3}{3} + O(x^5)$$ $$\sin x = x - \frac{x^3}{6} + O(x^5)$$ 3. **Substitute expansions into the numerator:** $$\tan x - \sin x = \left(x + \frac{x^3}{3}\right) - \left(x - \frac{x^3}{6}\right) + O(x^5) = \frac{x^3}{3} + \frac{x^3}{6} + O(x^5) = \frac{x^3}{2} + O(x^5)$$ 4. **Form the limit expression:** $$\frac{\tan x - \sin x}{3x^2} = \frac{\frac{x^3}{2} + O(x^5)}{3x^2} = \frac{x^3}{2} \cdot \frac{1}{3x^2} + O(x) = \frac{x}{6} + O(x)$$ 5. **Evaluate the limit as $$x \to 0$$:** $$\lim_{x \to 0} \frac{\tan x - \sin x}{3x^2} = \lim_{x \to 0} \left(\frac{x}{6} + O(x)\right) = 0$$ 6. **Therefore,** $$\ln y = 0$$ 7. **Find $$y$$ by exponentiating both sides:** $$y = e^{\ln y} = e^0 = 1$$ **Final answers:** $$\ln y = 0$$ $$y = 1$$