$\lim_{x\rightarrow 0} \frac{2\cdot \ln(1+x)+x^2-2x}{x^3}$ without 'Hopital Rule Compute $$\lim_{x\rightarrow 0} \frac{(x+2)\cdot \ln(1+x)-2x}{x^3}$$ without L'Hopital Rule.
I proved before that  that $$\lim_{x\rightarrow 0} \frac{ \ln(1+x)-x}{x^2}=-\frac{1}{2}$$
I tried to use it and I have to compute $$\lim_{x\rightarrow 0} \frac{2\cdot \ln(1+x)+x^2-2x}{x^3}$$
This limit looks easier to compute.

Use that
$$
2\ln(1+x)=2x-x^2+2\frac{x^3}{3}+o\left(x^3\right)
$$
Hence

$$\frac{2\ln(1+x)+x^2-2x}{x^3}=\frac{2x^3+o\left(x^3\right)}{3x^3}\underset{x \rightarrow 0}{\rightarrow}\frac{2}{3}$$

