calculus - How can I find $lim_{x to 0}frac{tan(3x)}{sin(8x)}$ without L'Hospital's Rule

Is there a way to solve $\lim_{x \to 0}\frac{\tan(3x)}{\sin(8x)}$ without using the trig identity $\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-\tan^2(x)}$. I want to know because I had to look up this trig identity in order to solve this limit, but if there is a simpler way, then I would like to know it. I don't want to use L'Hospital's Rule because it hasn't been introduced at this point in the book.

Answer

$$\lim\limits_{x \to 0}\frac{\tan(3x)}{\sin(8x)} = \lim\limits_{x \to 0}\frac{\sin(3x)}{\sin(8x)}\cdot\frac{1}{\cos{(3x)}} = \frac{3}{8}\lim\limits_{x \to 0}\frac{\sin(3x)}{3x}\cdot\frac{8x}{\sin(8x)}\cdot\frac{1}{\cos{(3x)}} = \boxed{\displaystyle\frac{3}{8}}$$

This uses $$\lim_{x\to 0} \frac{\sin(ax)}{ax} = 1$$