How to calculate $\lim_{x \to \infty}[(x-1/2)^{2}-x^{4}\ln^2{(1+1/x)}]$ I'm working on real analysis and struggled with the problem for days.
It seems to me that I should factor the formula into $[(x-1/2)-x^2\ln{(1+1/x)}][(x-1/2)+x^2\ln{(1+1/x)}]$ and try a way out though $\lim_{x \to \infty}\frac{\ln{1+1/x}}{1/x}=1$. But I failed.
Would you help me find it out? I'd appreciate your help. Best regards.

$\ln(1+t)=t-\dfrac{t^2}{2}+\dfrac{t^3}{3}-\dfrac{t^4}{4}+\cdots$
$\ln^2(1+t)=\left(t-\dfrac{t^2}{2}+\dfrac{t^3}{3}-\dfrac{t^4}{4}+\cdots\right)^2=t^2-t^3+\dfrac{11}{12}t^4+O(t^5)$
$x^4\ln^2(1+\dfrac1x)=x^2-x+\dfrac{11}{12}+O(\dfrac1x)$
$(x-\dfrac12)^2-x^4\ln^2(1+\dfrac1x)=-\dfrac{2}{3}+O(\dfrac1x)$
