<s>[INST] Find: $\lim\limits_{x\to +\infty}x\left(\sqrt{x^{2}+1}-\sqrt[3]{x^{3}+1}\right)$ Calculate the following limit: 

$$\displaystyle\lim_{x\to +\infty}x\left(\sqrt{x^{2}+1}-\sqrt[3]{x^{3}+1}\right)$$

I need find this limit without l'Hospital or Taylor series. 
Wolfram alpha gives $\frac{1}{2}$ 
My try is: 
Let: $a=\sqrt{1+x^{2}}$ and $b=\sqrt[3]{1+x^{3}}$ 
And we know that: 
$a-b=\frac{a^{3}-b^{3}}{a^{2}+b^{2}+ab}$ 
But after applied this I find again the problems $0.+\infty$ indeterminate 
 [/INST] Note 
$$\sqrt{x^{2}+1}-x=x \left(\sqrt{1+\frac1{x^2}}-1\right)=\frac{\frac1x}{\sqrt{1+\frac1{x^2}}+1}$$
$$\sqrt[3]{x^{3}+1}-x=x \left(\sqrt[3]{1+\frac1{x^3}}-1\right)=\frac{\frac1{x^2}}{\sqrt[3]{\left(1+\frac1{x^2}\right)^2}+\sqrt[3]{1+\frac1{x^2}}+1}$$
Thus, 
$$\displaystyle\lim_{x\to +\infty}x\left(\sqrt{x^{2}+1}-\sqrt[3]{x^{3}+1}\right)
=\displaystyle\lim_{x\to +\infty}x\left[\left(\sqrt{x^{2}+1}-x\right) -\left(\sqrt[3]{x^{3}+1}-x \right)\right]$$
$$ =\displaystyle\lim_{x\to +\infty} \frac{1}{\sqrt{1+\frac1{x^2}}+1}
-\frac{\frac1x}{\sqrt[3]{\left(1+\frac1{x^2}\right)^2}+\sqrt[3]{1+\frac1{x^2}}+1}$$
$$=\frac12-0=\frac12$$
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