$\lim_{x\rightarrow 0} \frac{2\cdot \ln(1+x)+x^2-2x}{x^3}$ 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.
 A: Note that
$$\int_0^x{-t^3\over1+t}dt=\int_0^x\left({1\over1+t}-(1-t+t^2) \right)dt=\ln(1+x)-x+{1\over2}x^2-{1\over3}x^3$$
and thus
$$\lim_{x\to0}{2\ln(1+x)+x^2-2x\over x^3}={2\over3}+\lim_{x\to0}\left({-2\over x^3}\int_0^x{t^3\over1+t}dt\right)$$
For $|x|\lt1$, we have
$$0\le\left|{-2\over x^3}\int_0^x{t^3\over1+t}dt \right|\le{2\over |x|^3}\int_0^{|x|}{|x|^3\over1-|x|}dt={2|x|\over1-|x|}\to0\quad\text{as }x\to0$$
so by the Squeeze Theorem, we wind up with
$$\lim_{x\to0}{2\ln(1+x)+x^2-2x\over x^3}={2\over3}$$
Remark: The two keys here are 1) the integral definition of the natural logarithm, and 2) the expansion ${1\over1+t}=1-t+t^2-t^3+\cdots$, which can be truncated, with the appropriate adjustment, so as to give the opening expression.
A: Use that
$$
2\ln(1+x)=2x-x^2+2\frac{x^3}{3}+o\left(x^3\right)
$$
Hence

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

A: 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.
$$\frac{2\cdot \ln(1+x)+x^2-2x}{x^3}=\frac23\frac{3\ln(1+x)+\frac32x^2-3x}{x^3}=\frac23\frac{\ln(1+3x+3x^2+x^3)-(3x+3x^2+x^3)}{(3x+3x^2+x^3)^2}\frac{(3x+3x^2+x^3)^2}{x^3}+\frac23\frac{x^3+\frac92x^2}{x^3}$$
