Find limit $\lim{x \to 0}\frac{\arctan 3x}{\ln((1-x)^2)}$ I have to find the following limit:
$$\lim_{x \to 0}\frac{\arctan3x}{\ln((1-x)^2)}$$
L’Hôpital’s rule only makes things worse. Transforming the numerator and the denominator in taylor series doesn’t help too. I’ve seen in another question that $\arctan \left(\frac{x}{x}\right) \to 1$, but then I don’t see what to do with 
$x/(\ldots)$.
Thanks for your help!
Edit: I think I made an error while differentiating, l’Hopital’s rule works. Thanks for your help.
 A: By L'Hospital 's rule, one has
$$
\lim_{x \to 0}\frac{\arctan(3x)}{\ln((1-x)^2)}=\lim_{x \to 0}\frac{\frac 3{1+9x^2}}{\frac {-2}{1-x}}
$$Can you fnish it?
A: L'Hospital doesn't make things worse.
$$\lim_{x\to0}\frac{\dfrac3{1+9x^2}}{-\dfrac{2(1-x)}{(1-x)^2}}=-\frac32.$$

Note that it is wiser to draw the exponent $2$ out of the logarithm first.
A: "Transforming the numerator and the denominator in taylor series doesn’t help too"
I do not agree
$$\frac{\tan ^{-1}(3 x)}{\log \left((1-x)^2\right)}=\frac{\tan ^{-1}(3 x)}{2\log \left(1-x\right)}=\frac 12\frac{3 x-9 x^3+O\left(x^4\right)}{-x-\frac{x^2}{2}-\frac{x^3}{3}+O\left(x^4\right) }$$ Simplify and use long division to find the limit and how it is approached.
A: $\lim{x \to 0}\frac{arctan(3x)}{ln^2((1-x))}$
Take the Limit from the left:
$\lim{x \to 0^-}\frac{arctan(3x)}{ln^2((1-x))}$
With l`Hoital's rule
\begin{equation}
\lim{x \to 0^-}\frac{arctan(3x)}{ln^2((1-x))}=\lim{x \to 0^-}\frac{3(x-1)}{2(9x^2+1)log(1-x)}
\end{equation}
Now
\begin{equation}
\lim{x \to 0^-}\frac{3(x-1)}{2(9x^2+1)log(1-x)}=\frac{3\lim{x \to 0^-}\frac{x-1}{9x^2+1}\lim{x \to 0^-\frac{1}{log(1-x)}}}{2}=\frac{3(-1)\lim{x \to 0^-\frac{1}{log(1-x)}}}{2}=\frac{-3(\infty)}{2}=-\infty
\end{equation}
Following the same steps for the limit from the right you will get to
\begin{equation}
\lim{x \to 0^+}\frac{3(x-1)}{2(9x^2+1)log(1-x)}=\frac{3\lim{x \to 0^+}\frac{x-1}{9x^2+1}\lim{x \to 0^+\frac{1}{log(1-x)}}}{2}=\frac{3(-1)\lim{x \to 0^+\frac{1}{log(1-x)}}}{2}=\frac{-3(-\infty)}{2}=\infty
\end{equation}
Since $\lim{x\to 0^+}\frac{1}{log(1-x)}=-\infty$ and $\lim{x\to 0^-}\frac{1}{log(1-x)}=\infty$
Hence two sided limit does not exist.
