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, $$(x^2+1)^{1/2}-(x^3+1)^{1/3}=\frac{((x^2+1)^{1/2}-(x^3+1)^{1/3})((x^2+1)^{1/2}+(x^3+1)^{1/3})}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}$$
$$=\frac{x^2+1-(x^3+1)^{2/3}}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}=\frac{(x^2+1)^2-(x^3+1)^{4/3}}{(x^2+1)^{3/2}+(x^2+1)(x^3+1)^{1/3}+x^3+1+(x^3+1)^{2/3}(x^2+1)}=??$$

I like the Taylor expansion approach, but wondered if there was a more "Cal I" approach.
Using
$$
A-B = (A^{1/6} - B^{1/6})(A^{5/6} + A^{4/6}B^{1/6} + A^{3/6}B^{2/6} + \cdots + B^{5/6})
$$
with $A = (x^2+1)^3$ and $B = (1+x^3)^2$ gives
$$
\begin{multline*}
(x^2+1)^{1/2} - (x^3+1)^{1/3}\\
=\frac{(x^2+1)^3 - (x^3+1)^2}{(x^2+1)^{5/2}+ (x^2+1)^{2}(1+x^3)^{1/3} + \cdots + (1+x^3)^{5/3}}\\
=\frac{3x^4+2x^3+3x^2}{(x^2+1)^{5/2}+ (x^2+1)^{2}(1+x^3)^{1/3} + \cdots + (1+x^3)^{5/3}}
\end{multline*}
$$
Now
$$
\begin{multline*}
x((x^2+1)^{1/2} - (x^3+1)^{1/3})\\
=\frac{x^5(3+2/x+3/x^2)}{x^5((1+1/x^2)^{5/2}+ (1+1/x^2)^{2}(1+1/x^3)^{1/3} + \cdots + (1+1/x^3)^{5/3})}\\
\end{multline*}
$$
which goes to $3/6$ as $x\to \infty$.
