We are asked to find the limit of the expression $\lim_{x\to 0} \frac{x+ \sin 3x}{x+ \sin x}$.

1. Problem Description

We are asked to find the limit of the expression $\lim_{x\to 0} \frac{x+ \sin 3x}{x+ \sin x}$.

2. Solution Steps

To evaluate the limit, we can divide both the numerator and the denominator by $x$.

$$\lim_{x\to 0} \frac{x+ \sin 3x}{x+ \sin x} = \lim_{x\to 0} \frac{1+ \frac{\sin 3x}{x}}{1+ \frac{\sin x}{x}}$$

We know that $\lim_{x\to 0} \frac{\sin x}{x} = 1$. Therefore, $\lim_{x\to 0} \frac{\sin 3x}{x} = \lim_{x\to 0} \frac{\sin 3x}{3x} \cdot 3 = 1 \cdot 3 = 3$.

$$\lim_{x\to 0} \frac{1+ \frac{\sin 3x}{x}}{1+ \frac{\sin x}{x}} = \frac{1+3}{1+1} = \frac{4}{2} = 2$$

3. Final Answer

The final answer is 2.