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

Let's put $x=1/t$ so that $t\to 0^{+}$ and the expression under limit is transformed into $$\frac{\sqrt{1+1/t^{2}}-\sqrt[3]{1+1/t^{3}}}{t}=\frac{\sqrt{1+t^{2}}-\sqrt[3]{1+t^{3}}}{t^{2}}$$ and this can be expressed as a difference of two standard limits $$\lim_{t\to 0^{+}}\frac{(1+t^{2})^{1/2}-1}{t^{2}}-t\cdot \frac{(1+t^{3})^{1/3}-1}{t^{3}}$$ which equals $1/2-0(1/3)=1/2$. The standard limit formula $$\lim_{x\to 0}\frac{(1+x)^{n}-1}{x}=n$$ used above is a direct consequence of the more general limit formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$
