How to calculate the limit $\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}$ which involves rational functions? 
Find $$\lim_{x\to +\infty}x\sqrt{x^2+1}-x(1+x^3)^{1/3}.$$ 

I have tried rationalizing but there is no pattern that I can observe. 
Edit:
So we forget about the $x$ that is multiplied to both the functions and try to work with the expression $(x^2+1)^{1/2}-(x^3+1)^{1/3}.$ Thus we have, $$(x^2+1)^{1/2}-(x^3+1)^{1/3}=\frac{((x^2+1)^{1/2}-(x^3+1)^{1/3})((x^2+1)^{1/2}+(x^3+1)^{1/3})}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}$$
$$=\frac{x^2+1-(x^3+1)^{2/3}}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}=\frac{(x^2+1)^2-(x^3+1)^{4/3}}{(x^2+1)^{3/2}+(x^2+1)(x^3+1)^{1/3}+x^3+1+(x^3+1)^{2/3}(x^2+1)}=??$$

This limit would be equal to $1/2$. See my demo : 
$$
\lim_{x \to \infty} {x\sqrt{x^2+1}-x (x^3+1)^{1/3}} = 
\lim_{x \to \infty} { x (\sqrt{x^2+1}-(x^3+1)^{1/3})}
$$
Now, let us see how much is each member, by Laurent-Taylor expansion :
$$
\sqrt{x^2+1} = x+ {1 \over {2x}} - {1 \over {8x^3}}+{1 \over {16x^5}}-...
$$
$$
(x^3+1)^{1/3} = x+ {1 \over {3x^2}} - {1 \over {9x^5}}+{5 \over {81x^8}}-...
$$
Resting this 2 members :
$$
\sqrt{x^2+1}-(x^3+1)^{1/3} = {1 \over {2x}}-{1 \over {3x^2}} - {1 \over {8x^3}}+({1 \over 16}+{1 \over 9}){1 \over x^5}-...
$$
Now we substitute this in our limit :
$$
\lim_{x \to \infty} { x (\sqrt{x^2+1}-(x^3+1)^{1/3})} = 
\lim_{x \to \infty} { x ({1 \over {2x}}-{1 \over {3x^2}}-...)}= {1 \over 2}
$$
