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? 
 A: First note that $\sqrt{x^6-2x^4} = x^3 \sqrt{1 - 2/x^2} = x^3 \left(1 - 1/x^2 + O(1/x^4)\right) = x^3 - x + O(1/x)$ so
$$\sqrt x + x^3 -\sqrt{x^6-2x^4} - x = \sqrt x + x^3 - \left( x^3 - x + O(1/x) \right) - x \\ = \sqrt x + O(1/x) \sim \sqrt x$$
for large $x$.
Also, for large $x$,
$\log\left(1+e^{3\sqrt x}\right) \sim 3\sqrt x$.
Therefore,
$$\frac{\sqrt x + x^3 -\sqrt{x^6-2x^4} - x}{\log(1+{e}^{3\sqrt x})}
\sim \frac{\sqrt x}{3\sqrt x} = \frac{1}{3}.$$
Thus the limit is $\frac13$.
A: You are proceeding in the right direction. Your last expression can be rewritten as $$\frac{1}{3}+\frac{\sqrt{x}(x^{2}-\sqrt{x^{4}-2x^{2}}-1)}{3}$$ Thus we need to deal with the numerator of the last term. We will show that it tends to $0$ as $x\to\infty$. To do so in an easy manner we can put $x=1/t$ so that $t\to 0^{+}$ and then the numerator is transformed into $$\frac{1-t^{2}-\sqrt{1-2t^{2}}} {t^{2}\sqrt{t}}$$ and then we need the rationalization trick to express it as $$\frac{(1-t^{2})^{2}-(1-2t^{2})}{t^{5/2}(1-t^{2}+\sqrt{1-2t^{2}})}$$ which is further simplified as $$\frac{t^{3/2}}{1-t^{2}+\sqrt {1-2t^{2}}}$$ and this tends to $0$ because numerator tends to $0$ and denominator tends to $2$.
The answer to the problem is thus $1/3$.
A: HINT: Break it up as $$\frac{\sqrt{x}}{3\sqrt{x}} + \frac{x^3 - \sqrt{x^6 - 2x^4} - x}{3\sqrt{x}}.$$
To deal with the numerator of the second term, write it as $$x^3\left(1 - \sqrt{1 - 2/x^2} - 1/x^2\right)$$
and use the Taylor expansion of $\sqrt{1 - y}$.
A: 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
$$\frac{2x^{7/2}}{2x^33\sqrt x}\to\frac13.$$
