Evaluate $\lim_{x\to0^+} \frac{(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}}{\arctan(x^2)\sin x +x^3 \ln x}$ I'm trying to evaluate
$$\lim_{x\to0^+} \frac{(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}}{\arctan(x^2)\sin x +x^3 \ln x}$$
I've found that it is
$$\lim_{x\to0^+}\frac{-\frac{x^3}{2}+\text{o}(x^3)}{x^3+x^3\ln x +\text{o}(x^3)}$$
My doubt is: which is the dominant term at the denominator? If I write it like this
$$\lim_{x\to0^+}\frac{-\frac{x^3}{2}+\text{o}(x^3)}{x^3\left(1+\ln x +\frac{\text{o}(x^3)}{x^3}\right)}$$
Or like this
$$\lim_{x\to0^+}\frac{-\frac{x^3}{2}+\text{o}(x^3)}{x^3\ln x\left(1+\frac{1}{\ln x} +\frac{\text{o}(x^3)}{x^3\ln x}\right)}$$
I get $0$, which is right; but I'm not sure if they are both correct, because I don't fully understand if the $\ln x$ is irrelevant or not.
Thanks to you all.

What you do is right! Follow your method:
$$\lim_{x\to0^+}\frac{-\frac{x^3}{2}+\text{o}(x^3)}{x^3\left(1+\ln x +\frac{\text{o}(x^3)}{x^3}\right)}=\lim_{x\to0^+}\frac{-\frac{1}{2}+\text{o}(1)}{1+\ln x +\text{o}(1)}=0.$$
Actually, 
$$(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}\sim\frac{-1}{2}x^3.$$
So you also can evaluate this limit as follows:
$$\lim_{x\to0^+} \frac{(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}}{\arctan(x^2)\sin x +x^3 \ln x}=\lim_{x\to0^+} \frac{\frac{-1}{2}x^3}{\arctan(x^2)\sin x +x^3 \ln x}$$
$$=\lim_{x\to0^+} \frac{\frac{-1}{2}}{\frac{\arctan(x^2)\sin x}{x^3}+\ln x}=0.$$
Because $$\lim_{x\to0^+}\frac{\arctan(x^2)\sin x}{x^3}=1$$
and $$\lim_{x\to0^+}\ln x=-\infty.$$
