Limit with sin indeterminate

Hint: divide numerator and denominator by $x$.

Just for the sake of variety, without dividing by $x$,

$$\frac{3x - \sin x}{x + \sin x} = \frac{3(x+ \sin x) - 4\sin x}{x + \sin x} = 3 - \frac{4\sin x}{x + \sin x}$$

For the second expression, the numerator is bounded while the denominator tends to positive infinity, so $\displaystyle \lim_{x \to \infty}\frac{4\sin x}{x + \sin x} = 0$, hence the original limit is $3$.

One has

$$0\leq\left\vert{\sin{x}\over x}\right\vert\leq{1\over |x|}$$

And so ${\sin{x}\over x}\to 0$ as $x\to\infty$. Now in the fraction divide the numerator and denominator by $x$ to get

$${3-{\sin{x}\over x}\over 1-{\sin{x}\over x}}\to 3$$