Calculating the limit $\lim_{x \to 0} \frac{1 + \sin x - \cos x + \ln(1-x)}{x^3}$ without using l'Hospital rule or Taylor series

Can anyone please solve the question below, without using l'Hospital rule or Taylor series expansion?

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

I tried very hard using all possible methods, but was unable to come up with any suitable solution. The solution at the end of my book also used taylor expansion to solve the problem. But my maths teacher gave us as homework to solve the question without l'Hospital rule or Taylor series. Can anyone please help me out?

Solution 1:

Assumption: if the limit ($L$) exists, \begin{align} L&=\lim_{x \to 0} \frac{1 + \sin x - \cos x + \ln(1-x)}{x^3}\tag1\\ (x\rightarrow -x)\,&=\lim_{x \to 0} \frac{1 - \sin x - \cos x + \ln(1+x)}{-x^3}\\ &=\lim_{x \to 0} \frac{-1 + \sin x + \cos x - \ln(1+x)}{x^3}\tag2\\\\ \frac{(1)+(2)}2&=L\\ &=\lim_{x \to 0} \frac{\sin x + \frac12\ln(1-x)-\frac12\ln(1+x)}{x^3}\tag3\\ (x\rightarrow 3x)\,&=\lim_{x \to 0} \frac{\sin 3x + \frac12\ln(1-3x)-\frac12\ln(1+3x)}{27x^3}\\ &=\lim_{x \to 0} \frac{3\sin x - 4\sin^3x + \frac12\ln(1-3x)-\frac12\ln(1+3x)}{27x^3}\tag4\\\\ 27\times(4)-&3\times(3)\rightarrow\\ 24L&=\lim_{x \to 0} \frac{3\sin x - 4\sin^3x + \frac12\ln(1-3x)-\frac12\ln(1+3x)}{x^3}\\ &+\lim_{x \to 0} \frac{-3\sin x - \frac32\ln(1-x)+\frac32\ln(1+x)}{x^3}\\ &=\lim_{x \to 0} \frac{-8\sin^3x + \ln(1-3x)-\ln(1+3x)-3\ln(1-x)+3\ln(1+x)}{2x^3}\\ &=\lim_{x \to 0} \frac{-8\sin^3x}{2x^3}+\lim_{x \to 0} \frac{\ln\left[\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}\right]}{2x^3}\\ &=-4+\lim_{x \to 0} \frac{\ln\left[1+\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}-1\right]}{\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}-1}\frac{\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}-1}{2x^3}\\ &=-4+\lim_{y \to 0}\frac{\ln(1+y)}{y}\cdot\lim_{x \to 0}\frac{\dfrac{(1-3x)(1+x)^3}{(1+3x)(1-x)^3}-1}{2x^3}\\ &=-4+1\cdot\lim_{x \to 0}\frac{(1-3x)(1+x)^3-(1+3x)(1-x)^3}{2x^3(1+3x)(1-x)^3}\\ &=-4+\lim_{x \to 0}\frac{-16x^3}{2x^3}\\ &=-12\\ \therefore\,L&=-\frac12 \end{align}

Solution 2:

$\textbf{Hint :}$

We can prove using some integration by parts that, $ \left(\forall x\in\mathbb{R}^{*}\right) $ : \begin{aligned}\frac{1+\sin{x}-\cos{x}+\ln{\left(1-x\right)}}{x^{3}}=-\int_{0}^{1}{\left(1-y\right)^{2}\left(\frac{1}{\left(1-xy\right)^{3}}+\frac{\cos{\left(xy\right)} + \sin{\left(xy\right)}}{2}\right)\mathrm{d}y}\end{aligned}

What happens when $ x\to 0 $ ?