$\lim_{x\rightarrow 0} \frac{2\cdot \ln(1+x)+x^2-2x}{x^3}$ 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.

As shown, without l'Hopital, the limit can be solved easily by Taylor's expansion that is given the two following limits


*

*$\frac{\ln(1+x)-x}{x^2}\to -\frac12$ (as already mentioned)


and


*

*$\frac{\ln(1+x)-x+\frac{x^2}2}{x^3}\to \frac13$ (which correspond to to your second limit)


indeed by those limits, as you noticed, it follow that
$$\frac{(x+2)\cdot \ln(1+x)-2x}{x^3}=\frac{\ln(1+x)-x}{x^2}+\frac{2\cdot \ln(1+x)+x^2-2x}{x^3}\to-\frac12+\frac23=\frac16$$
Note that I've also tried by various manipulation as the following but I can't reach the result.
$$\frac{2\cdot \ln(1+x)+x^2-2x}{x^3}=\frac23\frac{3\ln(1+x)+\frac32x^2-3x}{x^3}=\frac23\frac{\ln(1+3x+3x^2+x^3)-(3x+3x^2+x^3)}{(3x+3x^2+x^3)^2}\frac{(3x+3x^2+x^3)^2}{x^3}+\frac23\frac{x^3+\frac92x^2}{x^3}$$
