$$\lim_{x\to 0^{+}} (\ln x)^3\left(\arctan\left(\ln\left(x+x^2\right)\right) + \frac{\pi}{2}\right) + (\ln x)^2$$
I have this limit in my sheet and the answer is $\frac 13$. But I don't know how to approach any step using taylor expansion near zero (this is our lesson by the way). Please help me with the simplest way.
On
When $x\to 0^+$, $$ \ln(x+x^2) = \ln x + \ln(1+x^2) = \ln x + o(x) \tag{1} $$ and this goes to $-\infty$ as $x\to 0^+$. To continue, we would like to have $\arctan u$ with $u\to 0$, though. When $x<0$, we can use the fact that $$ \arctan x + \arctan \frac{1}{x} = -\frac{\pi}{2} \tag{2} $$ to rewrite (for $x$ small enough) $$ \arctan \ln(x+x^2) = \arctan(\ln x + o(x)) = -\frac{\pi}{2} - \arctan\frac{1}{\ln x + o(x)} $$ which also us to write, as $\frac{1}{\ln x + o(x)}\xrightarrow[x\to 0^+]{} 0^-$, $$\begin{align} \arctan \ln(x+x^2) + \frac{\pi}{2} &= - \arctan\frac{1}{\ln x + o(x)}\\ &= - \arctan\frac{1}{\ln x}\left(\frac{1}{1 + o\left(\frac{x}{\ln x}\right)}\right)\\ &= - \arctan\left( \frac{1}{\ln x}+o\left(\frac{x}{\ln^2 x}\right)\right)\\ &= -\frac{1}{\ln x}+\frac{1}{3\ln^3 x}+o\left(\frac{1}{\ln^3 x}\right) .\tag{3} \end{align}$$
From there, we get $$\ln^3 x\left(\arctan \ln(x+x^2) + \frac{\pi}{2}\right) = -\ln^2 x+\frac{1}{3}+o\left(1\right) $$ and, finally, $$\ln^3 x\left(\arctan \ln(x+x^2) + \frac{\pi}{2}\right) + \ln^2 x= \frac{1}{3}+o\left(1\right) \xrightarrow[x\to 0^+]{} \boxed{\frac{1}{3}}. $$
As $x \to 0^+$, one has $\ln x<0$, giving $$ \arctan\left(\ln\left(x+x^2\right)\right)=-\frac \pi2-\arctan\left(\frac1{\ln\left(x+x^2\right)}\right), $$ and using $\dfrac1{\ln\left(x+x^2\right)} \to 0$ one gets, by applying a standard taylor series expansion, $$ \arctan\left(\frac1{\ln\left(x+x^2\right)}\right)=\frac1{\ln\left(x+x^2\right)}-\frac1{3\ln^3\left(x+x^2\right)}+O\left( \frac1{\ln^5\left(x+x^2\right)}\right) $$ and $$ \begin{align} &(\ln x)^3\left(\arctan\left(\ln\left(x+x^2\right)\right) + \frac{\pi}{2}\right) + (\ln x)^2 \\\\&= (\ln x)^2-\frac{(\ln x)^3}{\ln\left(x+x^2\right)}+\frac{(\ln x)^3}{3\ln^3\left(x+x^2\right)}+O\left( \frac{(\ln x)^3}{\ln^5\left(x+x^2\right)}\right) \\\\&= (\ln x)^2-\frac{(\ln x)^3}{\ln x+O(x)}+\frac{(\ln x)^3}{3\ln^3x+O(x)}+O\left( \frac{(\ln x)^3}{\ln^5x+O(x)}\right) \\\\&= (\ln x)^2-(\ln x)^2+\frac13+O\left( \frac1{\ln^2x}\right)+O(x) \\\\& \to \frac13 \end{align} $$ as announced.