Find $\lim \limits_{x\to 0}{\sin{42x} \over \sin{6x}-\sin{7x}}$ I want to find
$$\lim \limits_{x\to 0}{\sin{42x} \over \sin{6x}-\sin{7x}}$$
without resorting to L'Hôpital's rule. Numerically, this computes as $-42$. My idea is to examine two cases: $x>0$ and $x<0$ and use ${\sin{42x} \over \sin{6x}}\to 7$ and ${\sin{42x} \over \sin{7x}}\to 6$. I can't find the appropriate inequalities to use the squeeze theorem, though. Do you have suggestions?

Hint:
$$\frac{\sin42x}{\sin6x-\sin7x}=\cfrac{\frac{\sin42x}{42x}}{\frac17\frac{\sin6x}{6x}-\frac16\frac{\sin7x}{7x}}$$
