1. **Stating the problem:** We want to find the limit $$\lim_{x \to 0} \frac{4x + \sin 3x}{6x - \tan 4x}$$ 2. **Formula and important rules:** - For small $x$, use the approximations $\sin x \approx x$ and $\tan x \approx x$. - Limits involving $\frac{0}{0}$ form can be solved by applying these approximations or L'Hôpital's Rule. 3. **Apply approximations:** - $\sin 3x \approx 3x$ - $\tan 4x \approx 4x$ 4. **Rewrite the expression using approximations:** $$\frac{4x + 3x}{6x - 4x} = \frac{7x}{2x}$$ 5. **Simplify:** $$\frac{7x}{2x} = \frac{7}{2}$$ 6. **Conclusion:** $$\lim_{x \to 0} \frac{4x + \sin 3x}{6x - \tan 4x} = \frac{7}{2}$$ This means as $x$ approaches 0, the value of the expression approaches $\frac{7}{2}$.