Limit of $\lim\limits_{x \rightarrow \infty}{(x-\sqrt \frac{x^3+x}{x+1})}$ - calculation correct? I just want to know if this way of getting the solution is correct.
We calculate $\lim\limits_{x \rightarrow \infty} (x-\sqrt \frac{x^3+x}{x+1}) = \frac {1}{2}$. 
\begin{align}
& \left(x-\sqrt \frac{x^3+x}{x+1}\,\right) \cdot \frac{x+\sqrt \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{x^2 - \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x^2(x+1) - x^3+x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} \\[10pt]
= {} & \frac{\frac{x^2+x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x+1}{1+\frac{1}{x}} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x+1}{1+\frac{1}{x}}}{x+\sqrt x^2 \sqrt \frac{x+\frac{1}{x}}{x+1}}  \\
= {} & \frac{\frac{x+1}{1+\frac{1}{x}} }{x+\sqrt x^2 \sqrt \frac{1+\frac{1}{x^2}}{1+\frac{1}{x}}} = \frac{x+1}{x+ x \sqrt 1} = \frac{x+1}{2x} \to \frac {1}{2}
\end{align}
I used this for symplifying:


*

*${(x-\sqrt \frac{x^3+x}{x+1})} \cdot {{(x+\sqrt \frac{x^3+x}{x+1})}} = x^2 -\frac{x^3+x}{x+1}=> (a-b)(a+b) = a^2-b^2$.

*$\sqrt \frac{x^3+x}{x+1} = \sqrt {x^2\cdot\frac{({x+\frac{1}{x}})}{x+1}} = \sqrt x^2 \sqrt\frac{x+\frac{1}{x}}{x+1}$.


I didn't write everything formally correct but hopefully you'll get the idea. The problem I see is that sometimes I'm applying $\lim\limits_{x \rightarrow \infty}$ to just the lower part of a fraction, when I probably have to apply it to both parts of the fraction? I'm talking about this part especially: $\lim\limits_{x \rightarrow \infty}\frac{x+1}{1-\frac{1}{x}} \sim x+1$.
As always very grateful for any comments/help. Cheers.
 A: I think, there is a little mistake in your computations. 
It should be
\begin{align}
& \left(x-\sqrt \frac{x^3+x}{x+1}\,\right) \cdot \frac{x+\sqrt \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{x^2 - \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x^2(x+1) - x^3-x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} \\[10pt]
= {} & \frac{\frac{x^2-x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x-1}{1+\frac{1}{x}} }{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x-1}{1+\frac{1}{x}}}{x+x \sqrt \frac{x+\frac{1}{x}}{x+1}}  \\
= {} & \frac{\frac{1-\frac{1}{x}}{1+\frac{1}{x}} }{1+\sqrt\frac{1+\frac{1}{x^2}}{1+\frac{1}{x}}}  \to \frac {1}{2}
\end{align}
A: another  solu tion
$$\lim\limits_{x \rightarrow +\infty} (x-\sqrt \frac{x^3+x}{x+1}) $$
$$\lim\limits_{x \rightarrow +\infty} (x-\sqrt{ x^2 - x +2 -\frac{2}{x+1}}) $$
$$\lim\limits_{x \rightarrow +\infty} (x-\sqrt{ (x - 1/2)^2+7/4 -\frac{2}{x+1}}) $$
$$\lim\limits_{x \rightarrow +\infty} (x- x +\frac12 )) = \frac {1}{2}$$
A: As an alternative we have by binomial expansion
$$\sqrt \frac{x^3+x}{x+1}=x\sqrt \frac{x+1/x}{x+1}=x\sqrt \frac{x^2+1}{x^2+x}=x\sqrt \frac{(x+1)^2-2x}{x(x+1)}=x\sqrt{\frac{x+1}{x}-\frac{2}{x+1}}=x\sqrt{1+\frac1x-\frac{2}{x+1}} =x\left( 1+\frac1{2x}-\frac{1}{(x+1)}+o\left(\frac1x\right)\right)=x+\frac12-\frac{x}{(x+1)}+o(1)$$
and therefore
$$x-\sqrt \frac{x^3+x}{x+1}=x-x-\frac12+\frac{x}{(x+1)}+o(1)=-\frac12+\frac{x}{(x+1)}+o(1)\to \frac12$$
