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, $$(x^2+1)^{1/2}-(x^3+1)^{1/3}=\frac{((x^2+1)^{1/2}-(x^3+1)^{1/3})((x^2+1)^{1/2}+(x^3+1)^{1/3})}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}$$
$$=\frac{x^2+1-(x^3+1)^{2/3}}{(x^2+1)^{1/2}+(x^3+1)^{1/3}}=\frac{(x^2+1)^2-(x^3+1)^{4/3}}{(x^2+1)^{3/2}+(x^2+1)(x^3+1)^{1/3}+x^3+1+(x^3+1)^{2/3}(x^2+1)}=??$$
 A: You can do it with Taylor expansions:
$x\left(1+x^{2}\right)^{1/2}-x\left(1+x^{3}\right)^{1/3}=
x^{2}\left(1+\frac{1}{x^{2}}\right)^{1/2}-x^{2}\left(1+\frac{1}{x^{3}}\right)^{1/3}
 =x^{2}\left(1+\frac{1}{2}\cdot\frac{1}{x^{2}}+o\left(\frac{1}{x^{2}}\right)\right)-x^{2}\left(1+\frac{1}{3}\cdot\frac{1}{x^{3}}+o\left(\frac{1}{x^{3}}\right)\right)
 =\frac{1}{2}-\frac{1}{3x}+o\left(1\right)
 \to\frac{1}{2}$
A: 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 $x\to \infty$.
A: 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 $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}$$
A: Hint: Your expression equals
$$x^2[(1+1/x^2)^{1/2} - (1+1/x^3)^{1/3}].$$
Now use the fact that $(1+h)^p = 1 + ph +o(h)$ as $h\to 0.$ (This fact is equivalent to the statement that the derivative of $(1+x)^p$ at $x=1$ is $p.$) 
A: 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 :
$$
\lim_{x \to \infty} { x (\sqrt{x^2+1}-(x^3+1)^{1/3})} = 
\lim_{x \to \infty} { x ({1 \over {2x}}-{1 \over {3x^2}}-...)}= {1 \over 2}
$$
