By using L'Hôpital's rule I can clearly see the answer is $1$. But when I tried without using L'Hôpital's rule, I somehow ended up getting $a/b$ as answer. Here's what I did:
$$\begin{align} \lim_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx)+ax} &= \frac{\lim_{x \to 0} (\sin(ax) + bx)}{\lim_{x \to 0} (\sin(bx) + ax)} \\ &= \frac{\lim_{x \to 0} \sin(ax) + \lim_{x \to 0} bx}{\lim_{x \to 0} \sin(bx) + \lim_{x \to 0} ax} \\ &= \frac{\lim_{x \to 0} \sin(ax) + 0}{\lim_{x \to 0} \sin(bx) + 0} \\ &= \frac{\lim_{x \to 0} \frac{\sin(ax)}{ax} \times ax}{\lim_{x \to 0} \frac{\sin(bx)}{bx} \times bx} \\ &= \lim_{x \to 0} \frac{\frac{\sin(ax)}{ax} \times ax}{\frac{\sin(bx)}{bx} \times bx} \\ &= \frac ab \frac{\lim_{x \to 0} \frac{\sin(ax)}{ax}}{\lim_{x \to 0} \frac{\sin(bx)}{bx}} \\ &= \frac ab \end{align} $$
Best Answer
Assuming $a\ne -b$ $$\lim\limits_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx)+ax}=\lim\limits_{x \to 0} \frac{\frac{\sin(ax)}{ax}\cdot ax + bx}{\frac{\sin(bx)}{bx}\cdot bx+ax}=\\ =\lim\limits_{x \to 0} \frac{\frac{\sin(ax)}{ax}\cdot a + b}{\frac{\sin(bx)}{bx}\cdot b+a}=\frac{a+b}{a+b}=1$$ when $a,b \ne 0$, though in this case initial fraction doesn't make sense.
Addition.
Full picture, (thanking comment), can be viewed from one expression using Taylor's series/expansion. Using, for example, $\sin x =x-\frac{x^3}{3!}+o(x^4) $ with $o$-little. Then, assuming $a^2+b^2\ne 0$, we have $$\lim\limits_{x \to 0} \frac{\sin(ax) + bx}{\sin(bx)+ax}=\lim\limits_{x \to 0} \frac{(a+ b)x-\frac{a^3x^3}{3!}+o(x^4)}{(b+a)x -\frac{b^3x^3}{3!}+o(x^4)}=\\ =\lim\limits_{x \to 0} \frac{(a+ b)-\frac{a^3x^2}{3!}+o(x^3)}{(b+a) -\frac{b^3x^2}{3!}+o(x^3)}=\begin{cases}1, & a\ne-b \\ -1, & a=-b\end{cases}$$