Evaluate $\lim_{x \to 0} \frac{\sin(3x)}{\sin(4x)}$ without using L'Hopital's rule. I'm trying to evaluate $\lim_{x \to 0} \frac{\sin(3x)}{\sin(4x)}$  without using L'Hopital's rule. I've managed to get $\lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{3x}{\sin(4x)} = 1 \cdot \lim_{x \to 0} \frac{3x}{\sin(4x)}$ but can't figure out how to evaluate that limit.
 A: You had the right idea, but you needed to apply this idea twice. Consider: $$\frac{\sin3x}{\sin4x}=\frac{\sin3x}{3x}\cdot\frac{4x}{\sin4x}\cdot\frac{3}{4}$$
A: The application of the  limit $\lim_{x\to 0}{\sin x\over x}=1$ can be avoided. The solution below makes use of  the continuity of the cosine function at $0$ and trigonometry formulas.  Moreover  the units of measure for angles are not essential.
By the way, the continuity of the cosine function at $0$ follows easily from the symmetry $\cos(-x)=\cos x$ and an elementary analysis of the right triangle.
We have
$$\sin 3x=\sin(x+2x)=\sin x\cos 2x+\cos x \sin 2x\\
=\sin x\cos 2x +2\cos^2x\sin x =\sin x(\cos 2x+2\cos^2x) $$
Also
$$\sin 4x=2\sin(2x)\cos(2x)=4\sin x\cos x\cos 2x$$
Therefore
$${\sin 3x\over \sin 4x}={\cos 2x+2\cos^2x\over 4\cos x\cos 2x}\underset{x\to 0}{\longrightarrow} {3\over 4}$$
The method is painful for coefficients larger than $3$ and $4.$ Then one can rely on  formulas with complex numbers
$$\sin nx={e^{inx}-e^{-inx}\over 2i},\quad \cos nx={e^{inx}+e^{-inx}\over 2}$$
