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.

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.
