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 $0$
 A: $x^{\frac{1}{x}}-1 = \exp\left(\frac{\log x}{x}\right)-1 = \frac{\log x}{x}+O\left(\frac{\log^2 x}{x^2}\right)$ 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 $\color{red}{0}$.
A: Noting that
$$ \frac1{x^x}=1-\ln x+O(x^2\ln^2x), \sqrt{\frac1{1+x}+x^2}=1-\frac x2+O(x^2) $$
and letting $x\to\frac1x$, then we have
\begin{eqnarray}
&&\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}\\
&=&\lim _{x\to 0}\:\frac{\left(\sqrt{1+\frac{1}{x^2(1+x)}}-\frac{1}{x}\right)\ln \frac1x}{\frac1x\left(\frac1{x^x}-1\right)+\frac1{\sqrt{x}}\ln^2\frac1x}\\
&=&-\lim _{x\to 0}\:\frac{\left(\sqrt{x^2+\frac{1}{1+x}}-1\right)\ln x}{\left(\frac1{x^x}-1\right)+\sqrt{x}\ln^2x}\\
&=&-\lim _{x\to 0}\:\frac{\left(-\frac{x}2+O(x^2)\right)\ln x}{\left(-\ln x+O(x^2\ln^2x)\right)+\sqrt{x}\ln^2x}\\
&=&-\lim _{x\to 0}\:\frac{-\frac{x}2+O(x^2)}{\left(-1+O(x^2\ln x)\right)+\sqrt{x}\ln x}\\
&=&0.
\end{eqnarray}
A: As usual I prefer the elementary way. We have
\begin{align}
L &= \lim_{x \to \infty}\dfrac{\left(\sqrt{1 + \dfrac{x^{3}}{x + 1}} - x\right)\log x}{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x}\notag\\
&= \lim_{x \to \infty}\dfrac{\left(1 + \dfrac{x^{3}}{x + 1} - x^{2}\right)\log x}{\{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x\}\left(\sqrt{1 + \dfrac{x^{3}}{x + 1}} + x\right)}\notag\\
&= \lim_{x \to \infty}\dfrac{(1 + x - x^{2})\log x}{x(x + 1)\{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x\}\left(\sqrt{\dfrac{1}{x^{2}} + \dfrac{x}{x + 1}} + 1\right)}\notag\\
&= \frac{1}{2}\lim_{x \to \infty}\dfrac{(1 + x - x^{2})\log x}{x(x + 1)\{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x\}}\notag\\
&= \frac{1}{2}\lim_{x \to \infty}\dfrac{\left(\dfrac{1}{x^{2}} + \dfrac{1}{x} - 1\right)\log x}{\left(1 + \dfrac{1}{x}\right)\{x(x^{1/x} - 1) + \sqrt{x}\log^{2}x\}}\notag\\
&= -\frac{1}{2}\lim_{x \to \infty}\dfrac{\log x}{x(\exp((\log x)/x) - 1) + \sqrt{x}\log^{2}x}\notag\\
&= -\frac{1}{2}\lim_{x \to \infty}\dfrac{1}{\dfrac{\exp((\log x)/x) - 1}{(\log x)/x} + \sqrt{x}\log x}\notag\\
&= -\frac{1}{2}\lim_{x \to \infty}\dfrac{\dfrac{1}{\sqrt{x}\log x}}{\dfrac{1}{\sqrt{x}\log x}\cdot\dfrac{\exp((\log x)/x) - 1}{(\log x)/x} + 1}\notag\\
&= -\frac{1}{2}\cdot\dfrac{0}{0\cdot 1 + 1}\notag\\
&= 0\notag
\end{align}
In the last step we have used the fact that $\sqrt{x}\log x \to \infty, (\log x)/x \to 0$ as $x \to \infty$ and $(\exp(t) - 1)/t \to 1$ as $t \to 0$.
