Evaluate limit of $(2\sin x\log \cos x + x^{3})/x^{7}$ as $x \to 0$

While trying to solve this question, I came across the following limit $$\lim_{x \to 0}\frac{2\sin x\log \cos x + x^{3}}{x^{6}}\tag{1}$$ Using some algebraic manipulation (and L'Hospital's Rule) I was able to show that $$\lim_{x \to 0}\frac{2\sin x\log \cos x + x^{3}}{x^{5}} = 0\tag{2}$$ From the fact that the numerator in the above limit expression is an odd function, I guessed that the limit in $(1)$ would also be $0$ (it would be great if this guess can be supported by a proof). However I was not able to do this via simple algebraic manipulation. Also note that evaluating $(1)$ is equivalent to solving the linked question (without the assumption of existence of limit).

I think it is better to go one step ahead and establish that $$\lim_{x \to 0}\frac{2\sin x \log \cos x + x^{3}}{x^{7}} = -\frac{1}{40}\tag{3}$$ It is possible to evaluate the above limit via Taylor's series very easily, but I would prefer to have a solution of either $(1)$ or $(3)$ without using Taylor's series.

Update: I provide an evaluation of limit $(2)$ as an illustration of the kind of answer I would prefer. We have \begin{align} L &= \lim_{x \to 0}\frac{2\sin x\log \cos x + x^{3}}{x^{5}}\notag\\ &= \lim_{x \to 0}\frac{\sin x\log (1 - \sin^{2}x) + x^{3}}{x^{5}}\notag\\ &= \lim_{x \to 0}\frac{\sin x\log (1 - \sin^{2}x) + \sin^{3}x + x^{3} - \sin^{3}x}{x^{5}}\notag\\ &= \lim_{x \to 0}\frac{\sin x\log (1 - \sin^{2}x) + \sin^{3}x}{x^{5}} + \frac{x^{3} - \sin^{3}x}{x^{5}}\notag\\ &= \lim_{x \to 0}\frac{\sin x\log (1 - \sin^{2}x) + \sin^{3}x}{\sin^{5}x}\cdot\frac{\sin^{5}x}{x^{5}} + \lim_{x \to 0}\frac{x - \sin x}{x^{3}}\cdot\frac{x^{2} + x\sin x + \sin^{2}x}{x^{2}}\notag\\ &= \lim_{x \to 0}\frac{\log (1 - \sin^{2}x) + \sin^{2}x}{\sin^{4}x}\cdot 1 + \lim_{x \to 0}\frac{1 - \cos x}{3x^{2}}\cdot (1 + 1 + 1)\text{ (via LHR)}\notag\\ &= \lim_{t \to 0}\frac{\log (1 - t) + t}{t^{2}} + \frac{1}{2}\notag\\ &= \lim_{t \to 0}\dfrac{-\dfrac{1}{1 - t} + 1}{2t} + \frac{1}{2}\text{ (via LHR)}\notag\\ &= -\frac{1}{2} + \frac{1}{2} = 0\notag \end{align} Further Update: I have finally found a solution which uses algebraic manipulation and L'Hospital's Rule. The rule has been applied 4 times in total and resulting expressions are simple. See the details in my answer.

I finally found a simple way to evaluate the limit $(3)$ using some algebraic manipulation and L'Hospital's Rule. Let the desired limit $(3)$ in question be $L$ so that $$L = \lim_{x \to 0}\frac{2\sin x\log \cos x + x^{3}}{x^{7}}\tag{1}$$ The existence of $L$ is dependent on the existence of certain limits which will be evaluated in what follows.

First we can see that \begin{align} A &= \lim_{x \to 0}\frac{2\log\cos x + x^{2}}{x^{4}}\notag\\ &= \lim_{x \to 0}\frac{\log(1 - \sin^{2} x) + \sin^{2}x + x^{2} - \sin^{2}x}{x^{4}}\notag\\ &= \lim_{x \to 0}\frac{\log(1 - \sin^{2} x) + \sin^{2}x}{x^{4}} + \lim_{x \to 0}\frac{x - \sin x}{x^{3}}\cdot\frac{x + \sin x}{x}\notag\\ &= \lim_{x \to 0}\frac{\log(1 - \sin^{2} x) + \sin^{2}x}{\sin^{4}x}\cdot\frac{\sin^{4}x}{x^{4}} + \frac{1}{6}\cdot(1 + 1)\notag\\ &= \frac{1}{3} + \lim_{x \to 0}\frac{\log(1 - \sin^{2} x) + \sin^{2}x}{\sin^{4}x}\notag\\ &= \frac{1}{3} + \lim_{t \to 0}\frac{\log(1 - t) + t}{t^{2}}\notag\\ &= \frac{1}{3} - \frac{1}{2}\text{ (see the derivation in question)}\notag\\ &= -\frac{1}{6}\tag{2} \end{align} Multiplying this limit with the the equation (which is easily obtained via one application of LHR, see the derivation in question) $$\lim_{x \to 0}\frac{\sin x - x}{x^{3}} = -\frac{1}{6}$$ we get $$\lim_{x \to 0}\frac{2\sin x\log \cos x - 2x\log \cos x + x^{2}\sin x - x^{3}}{x^{7}} = \frac{1}{36}\tag{3}$$ and thus we have $$L + \lim_{x \to 0}\frac{x^{2}\sin x - 2x\log \cos x - 2x^{3}}{x^{7}} = \frac{1}{36}$$ or $$\lim_{x \to 0}\frac{x\sin x - 2\log\cos x - 2x^{2}}{x^{6}} = \frac{1}{36} - L = B\text{ (say)}\tag{4}$$ The above limit will now be calculated via algebraic manipulation and application of L'Hospital's Rule. \begin{align} B &= \lim_{x \to 0}\frac{x\sin x - 2\log\cos x - 2x^{2}}{x^{6}}\notag\\ &= \lim_{x \to 0}\frac{\sin x + x\cos x+ 2\tan x - 4x}{6x^{5}}\text{ (via LHR)}\notag\\ &= \lim_{x \to 0}\frac{2\cos x - x\sin x + 2\sec^{2}x - 4}{30x^{4}}\text{ (via LHR)}\notag\\ \Rightarrow 30B &= \lim_{x \to 0}\frac{2\cos^{3} x - x\sin x\cos^{2}x + 2 - 4\cos^{2}x}{x^{4}\cos^{2}x}\notag\\ &= \lim_{x \to 0}\frac{2\cos^{3} x - x\sin x\cos^{2}x + 2 - 4\cos^{2}x}{x^{4}}\notag\\ &= \lim_{x \to 0}\frac{2\cos x(1 - \sin^{2}x) - x\sin x + x\sin^{3}x - 2 + 4\sin^{2}x}{x^{4}}\notag\\ &= \lim_{x \to 0}\frac{2\cos x(1 - \sin^{2}x) - x\sin x - 2 + 4\sin^{2}x}{x^{4}} + \frac{\sin^{3}x}{x^{3}}\notag\\ &= 1 + \lim_{x \to 0}\frac{2\cos x(1 - \sin^{2}x) - x\sin x - 2 + 4\sin^{2}x}{x^{4}}\notag\\ &= 1 + \lim_{x \to 0}\frac{2(1 - 2\sin^{2}(x/2))(1 - \sin^{2}x) - x^{2} + x^{2} - x\sin x - 2 + 4\sin^{2}x}{x^{4}}\notag\\ &= 1 + \lim_{x \to 0}\frac{- 4\sin^{2}(x/2) + 4\sin^{2}(x/2)\sin^{2}x - x^{2} + 2\sin^{2}x}{x^{4}} + \lim_{x \to 0}\frac{x - \sin x}{x^{3}}\notag\\ &= \frac{7}{6} + \lim_{x \to 0}\frac{- 4\sin^{2}(x/2) - x^{2} + 2\sin^{2}x}{x^{4}} + \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{\sin^{2}(x/2)}{(x/2)^{2}}\notag\\ &= \frac{13}{6} + \lim_{x \to 0}\frac{- 4\sin^{2}(x/2) - x^{2} + 8\sin^{2}(x/2)\cos^{2}(x/2)}{x^{4}}\notag\\ &= \frac{13}{6} + \lim_{x \to 0}\frac{- 4\sin^{2}(x/2) - x^{2} + 8\sin^{2}(x/2)(1 - \sin^{2}(x/2))}{x^{4}}\notag\\ &= \frac{13}{6} + \lim_{x \to 0}\frac{4\sin^{2}(x/2) - x^{2} - 8\sin^{4}(x/2)}{x^{4}}\notag\\ &= \frac{13}{6} + \lim_{x \to 0}\frac{4\sin^{2}(x/2) - x^{2}}{x^{4}} - \frac{1}{2}\lim_{x \to 0}\frac{\sin^{4}(x/2)}{(x/2)^{4}}\notag\\ &= \frac{5}{3} + \lim_{x \to 0}\frac{4\sin^{2}(x/2) - x^{2}}{x^{4}}\notag\\ &= \frac{5}{3} + \frac{1}{4}\lim_{t \to 0}\frac{\sin^{2}t - t^{2}}{t^{4}}\text{ (by putting }x = 2t)\notag\\ &= \frac{5}{3} + \frac{1}{4}\lim_{t \to 0}\frac{\sin t - t}{t^{3}}\cdot\frac{\sin t + t}{t}\notag\\ &= \frac{5}{3} - \frac{1}{4}\cdot\frac{1}{6}\cdot(1 + 1)\notag\\ &= \frac{19}{12}\notag \end{align} and therefore $B = 19/360$ and finally from $(4)$ we have $$L = \frac{1}{36} - B = -\frac{9}{360} = -\frac{1}{40}$$ The above derivation looks long because of detailed steps, but does not use anything more than $(\sin x)/x \to 1$ or $(x - \sin x)/x^{3} \to 1/6$.

If you use L'Hôpital rule, if there is a limit, because of the $x^7$, you would need to differentiate seven times the numerator. After these seven differentiations (have fun !), the denominator will be $7!=5040$ and, after a long series of successive simplifications, the seventh derivative of numerator would write $$-2 \left(720 \sec ^7(x)-864 \sec ^5(x)+204 \sec ^3(x)-4 \sec (x)+\cos (x) (\log (\cos (x))+7)\right)$$ the value of which being $-126$ for $x=0$. Then, your result.