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?

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}
