How to evaluate $\lim _{x\to \infty }\:\frac{\left(\sqrt{1+\frac{x^3}{x+1}}-x\right)\ln x}{x\left(x^{\frac{1}{x}}-1\right)+\sqrt{x}\ln^2x}$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\:\frac{\left(\sqrt{1+\frac{x^3}{x+1}}-x\right)\ln x}{x\left(x^{\frac{1}{x}}-1\right)+\sqrt{x}\ln^2x}$$
The result should be $-\frac{1}{2}$, but wolfram says that is $0$

$x^{\frac{1}{x}}-1 = \exp\left(\frac{\log x}{x}\right)-1 = \frac{\log x}{x}+O\left(\frac{\log^2 x}{x^2}\right)$ for $x\to +\infty$, so the denominator behaves like $x^{1/2}\log(x)+O(\log x)$. On the other hand:
$$ \sqrt{1+\frac{x^3}{x+1}}-x = \frac{1+\frac{x^3}{x+1}-x^2}{x+\sqrt{1+\frac{x^3}{x+1}}}=\frac{x+1+x^3-x^2(x+1)}{x(x+1)+\sqrt{(x+1)^2+(x+1)x^3}}$$
behaves like $-\frac{1}{2}+O\left(\frac{1}{x}\right)$ for $x\to +\infty$, hence the wanted limit is just $\color{red}{0}$.
