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.
Sorry for my previous misunderstanding on the question. I will add some detail for the path from @user's Hint.
Above all, for first order we have $$ A_1=\lim_{x\to0} \frac{x-\sin x}{x^3}=\frac1{6}, \quad B_1=\lim_{x\to0} \frac{1-\cos x}{x^2} =\frac1{2} $$ which is proved in that post before.
To pursue higher order result exampled by $B_2$ denoted as $$ B_2=\lim_{x\to0} \frac{\cos x-(1-\frac1{2}x^2)}{x^4}=\lim_{x\to0} \frac1{x^2}\left(\frac1{2}-\frac{1-\cos x}{x^2}\right) $$ so we have $$ B_2=\lim_{x\to0} \frac1{4x^2}\left(\frac1{2}-\frac{1-\cos2x}{4x^2}\right) $$ hence $$ 4B_2=\lim_{x\to0} \frac1{x^2}\left(\frac1{2}-\frac{1-\cos2x}{4x^2}\right) $$ $$ \frac{B_2}{4}=\lim_{x\to0} \frac1{x^2}\left(\frac1{8}-\frac{1-\cos x}{4x^2}\right) $$ whose subtraction gives $$ \begin{aligned} \frac{15}{4}B_2 &=\lim_{x\to0} \frac1{x^2}\left(\frac3{8}-\frac{\cos x-\cos2x}{4x^2}\right)=\lim_{x\to0} \frac1{x^2}\left(\frac3{8}-\frac{\sin^2 x-\sin^2\frac{x}{2}}{2x^2}\right)\\ &=\lim_{x\to0} \frac1{x^2}\left(\frac3{8}-\frac{4\sin^2\frac{x}{2}(1-\sin^2\frac{x}{2})-\sin^2\frac{x}{2}}{2x^2}\right)\\ &=\lim_{x\to0} \frac1{x^2}\left(\frac3{8}-\frac{3\sin^2\frac{x}{2}}{2x^2}\right) + \lim_{x\to0} \frac{2\sin^4\frac{x}{2}}{x^4}\\ &=\lim_{x\to0} \frac3{8}\left(\frac{x^2-4\sin^2\frac{x}{2}}{x^4}\right) + \lim_{x\to0} \frac{2\sin^4\frac{x}{2}}{x^4}\\ &=\lim_{x\to0} \frac3{8}\left(\frac{x+2\sin\frac{x}{2}}{x}\right)\left(\frac{x-2\sin\frac{x}{2}}{x^3}\right) + \lim_{x\to0} \frac{2\sin^4\frac{x}{2}}{x^4}\\ &=\lim_{x\to0} \frac3{8}\left(\frac{\frac{x}{2}+\sin\frac{x}{2}}{\frac{x}{2}}\right)\left(\frac{\frac{x}{2}-\sin\frac{x}{2}}{4\cdot(\frac{x}{2})^3}\right) + \lim_{x\to0} \frac{\sin^4\frac{x}{2}}{8\cdot(\frac{x}{2})^4} =\frac5{32} \end{aligned} $$ where we need to recall the value of $A_1$, and obtain $$ B_2=\frac1{24} $$ I think you can as well get $A_2$, which is $$ A_2=\lim_{x\to0} \frac{\sin x-(x-\frac1{6}x^2)}{x^5}=\frac1{120} $$ by almost the same approach, and that is the resources to solve $L_1$.
However $L_2$ is trivial $$ L_2=\lim_{x\to0} \frac{2x}{16\sin 2x}\left(2+\cos2x-\frac{3\sin2x}{2x}\right)=\lim_{x\to0} \frac{x}{16\sin x}\left(2+\cos x-\frac{3\sin x}{x}\right)=0 $$ for we do not need to calculate the infinitesimal with high order, or if you want, you can get the proper result from $L_1$, which is $$ 2+\cos x-\frac{3\sin x}{x} \sim \frac{x^4}{60} \quad (x\to0) $$ From another view, this is a hint how to make up this expression, just taking these lower order items back we easily have $$ 2+\cos x-\frac{3\sin x}{x} = \left(\cos x-1+\frac{x^2}{2}\right)-3\left(\frac{\sin x}{x}-1+\frac{x^2}{6}\right) \sim (B_2-3A_2)x^4 = \frac{x^4}{60} $$ which is very similar to the path of series expansion.
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
