1. We want to find the limit\n\nb) \lim_{x \to 0} \frac{\sin 3x \cdot \tan 4x}{2x^2}\n\n2. Recall the standard limits: \lim_{x \to 0} \frac{\sin ax}{ax} = 1 and \lim_{x \to 0} \frac{\tan bx}{bx} = 1 for constants $a$ and $b$.\n\n3. Rewrite the expression to use these limits:\n\n$$\frac{\sin 3x \cdot \tan 4x}{2x^2} = \frac{\sin 3x}{x} \cdot \frac{\tan 4x}{x} \cdot \frac{x^2}{2x^2} = \frac{\sin 3x}{x} \cdot \frac{\tan 4x}{x} \cdot \frac{1}{2}$$\n\n4. Now express each term:\n\n$$\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x} \to 3 \cdot 1 = 3$$\n$$\frac{\tan 4x}{x} = 4 \cdot \frac{\tan 4x}{4x} \to 4 \cdot 1 = 4$$\n\n5. Putting everything together:\n\n$$ \lim_{x \to 0} \frac{\sin 3x \cdot \tan 4x}{2x^2} = \frac{1}{2} \times 3 \times 4 = \frac{12}{2} = 6 $$\n\nFinal answer: 6