calculus - Evaluate $lim_{x to 0} frac{1-cos(sin(4x))}{sin^2(sin(3x))}$ without L'Hospital

$$\lim_{x \to 0} \frac{1-\cos(\sin(4x))}{\sin^2(\sin(3x))}$$

How can I evaluate this limit without using the L'Hospital Rule? I've expanded $\sin(4x)$ as $\sin(2x+2x)$, $\sin(3x) = \sin(2x + x)$, but none of these things worked.

Answer

The simplest way is to note $\sin x \simeq x$ for small $x$ and $\cos x \simeq 1-\frac{x^2}{2}$ for small $x$. Then, you can obtain
$$\frac{1-\cos\sin 4x}{\sin^2\sin 3x} \simeq \frac{1-\cos 4x}{\sin^2 3x} \simeq \frac{8x^2}{9x^2} = \frac{8}{9}.$$
You need to use Taylor series to formalize this type of argument.