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.

This problem is simpler and does not require the use of $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$. We have $$\begin{aligned}L &= \lim_{x \to 0}\frac{\tan 3x}{\sin 8x}\\
&= \lim_{x \to 0}\frac{\tan x(3 - \tan^{2}x)}{1 - 3\tan^{2}x}\cdot\frac{1}{2\sin 4x\cos 4x}\\
&= \lim_{x \to 0}\frac{3 - \tan^{2}x}{1 - 3\tan^{2}x}\cdot\frac{\sin x}{\cos x}\cdot\frac{1}{4\sin 2x\cos 2x\cos 4x}\\
&= \lim_{x \to 0}\frac{3 - \tan^{2}x}{1 - 3\tan^{2}x}\cdot\frac{\sin x}{\cos x}\cdot\frac{1}{8\sin x\cos x\cos 2x\cos 4x}\\
&= \lim_{x \to 0}\frac{3 - \tan^{2}x}{1 - 3\tan^{2}x}\cdot\frac{1}{\cos x}\cdot\frac{1}{8\cos x\cos 2x\cos 4x}\\
&= \frac{3}{1}\cdot\frac{1}{1}\cdot\frac{1}{8\cdot 1\cdot 1\cdot 1} = \frac{3}{8}\end{aligned}$$ We have used the limits $$\lim_{x \to 0}\tan x = 0, \lim_{x \to 0}\cos x = 1$$ which are on slightly simpler level compared to $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$.
