I'm trying to calculate the following limit:
$$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log(1+{e}^{3\sqrt x})}$$
For WolframAlpha the result is: $\frac 13$, I've seen its step by step process but I didn't understand the background logic.
Before I got stuck, I did this step:
$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log[{e}^{3\sqrt x}({e}^{-3\sqrt x}+1)]}$ and then $\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle \log({e}^{3\sqrt x})+\log({e}^{-3\sqrt x}+1)}$
so $$\lim_{x \rightarrow +\infty} \frac{\displaystyle\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\displaystyle {3\sqrt x}}$$
Someone could give me a hint for solve it?