Evaluating $\lim_{x \to 0}{\frac{\sin(4x)}{\sin(3x)}}$ without L'Hospital I need to evaluate
$$\lim_{x \to 0}{\frac{\sin(4x)}{\sin(3x)}}$$
I solved this using L'Hospital's Theorem and I got 4/3 
However, is there a way to do this without applying this theoerm?
 A: Rewrite as
$$
\frac{4}{3}\lim_{x \to 0}\frac{\sin(4x)}{4x}\frac{3x}{\sin(3x)}
$$
A: Since $$\sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}+\ldots=\sum_{k=0}^{\infty}\frac{(-1)^kt^{2k+1}}{(2k+1)!}$$
we have
\begin{align}
\frac{\sin(4x)}{\sin(3x)}&=\frac{4x-\frac{(4x)^3}{3!}+\frac{(4x)^5}{5!}+\ldots}{3x-\frac{(3x)^3}{3!}+\frac{(3x)^5}{5!}+\ldots}\\
&=\frac{4-\frac{4^3}{3!}x^2+\frac{4^5}{5!}x^4+\ldots}{3-\frac{3^3}{3!}x^2+\frac{3^5}{5!}x^4+\ldots}\\
\lim_{x\to 0}\frac{\sin(4x)}{\sin(3x)}&=\frac{4}{3}
\end{align}
A: By elementary trigonometry, with obvious shorthands:
$$\frac{\sin(4x)}{\sin(3x)}=\frac{\sin(2(2x))}{\sin(x+2x)}=\frac{2(2sc)(2c^2-1)}{s(2c^2-1)+c(2sc)}=\frac ss\frac{4c(2c^2-1)}{4c^2-1}\to\frac43.$$

More generally, one can use
$$L_{m+1}=\lim_{x\to0}\frac{\sin(x+mx)}{\sin(x)}=\lim_{x\to0}\frac{\sin(x)\cos(mx)}{\sin(x)}+\lim_{x\to0}\frac{\cos(x)\sin(mx)}{\sin(x)}=1+L_m.$$
Together with $L_0=0$, this gives $L_m=m$.
[It is easy to generalize to $m$ rational, by rescaling $x$, and even to $m$ real by squeezing $m$ between rationals.]
Of course,
$$\lim_{x\to0}\frac{\sin(px)}{\sin(qx)}=\lim_{x\to0}\frac{\sin(px)}{\sin(x)}\frac{\sin(x)}{\sin(qx)}=\lim_{x\to0}\frac{\sin(px)}{\sin(x)}\lim_{x\to0}\frac{\sin(x)}{\sin(qx)}=\frac pq.$$
A: As $$\lim_{x \to 0}\frac{\sin(x)}{x}=1$$
$$\lim_{x \to 0}{\frac{\sin(4x)}{\sin(3x)}}$$ can be written as 
$$
\frac{4}{3}\lim_{x \to 0}\frac{\sin(4x)}{4x}\frac{3x}{\sin(3x)}
$$
$$=\frac{4}{3}$$
A: By Taylor's expansion around $0$, $x\mapsto\sin (tx)=tx+O(t^2)$, where $t \in \mathbb{R}$. Can you conclude?
