<s>[INST] Solving $\lim_{x\to 0}\left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)$ without series expansion or L'Hopital's rule I want to find the following limit without using series expansion or L'Hopital's rule. I tried replacing $x$ with $2x$.

$$\begin{aligned}L_{1}&=\lim_{x\to 0}\left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)\end{aligned}$$

$$\begin{aligned}L_{2}&=\lim_{x\to 0}\left(\frac{4x+2x(\cos^2x-\sin^2x)-6\sin x\cos x}{32\sin x\cos x}\right)\end{aligned}$$
How to proceed. Any hints are appreciated. Thanks.
 [/INST] HINT
To simplify the derivation we can proceed starting to prove the followings
$$L_1: \frac{\cos x-1}{x^2}\to \frac12 \implies L_2: \frac{x\cos x-x+\frac12x^3}{x^5}\to \frac1{24}$$
$$L_3:\frac{x-\sin x}{x^3}\to \frac16 \implies L_4:\frac{3(x-\sin x)-\frac12 x^3}{x^5}\to -\frac1{40}$$
and then use that
$$\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{x}{\sin x}\left(\frac{x\cos x-x+\frac12x^3}{x^5}+\frac{3(x-\sin x)-\frac12 x^3}{x^5}\right)$$
Refer to the related

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*Are all limits solvable without L'Hôpital Rule or Series Expansion
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