<s>[INST] Evaluating $\lim_{x \rightarrow +\infty} \frac{\sqrt x + x^3 -\sqrt{x^6-2x^4} -x}{\log(1+{e}^{3\sqrt x})}$ 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? 
 [/INST] If you multiply/divide by the conjugate $\sqrt x+x^3+\sqrt{x^6-2x^4}-x$, the numerator becomes
$$(\sqrt x+x^3-x)^2-(x^6-2x^4)=2x^{7/2}+\cdots$$
It is enough to keep the leading term. 
The conjugate can be written as $$2x^3+\cdots$$ by pulling $x^3$ out of the radical. 
Then with the denominator $3\sqrt x$, the limit is
$$\frac{2x^{7/2}}{2x^33\sqrt x}\to\frac13.$$
</s>