What is the limit for the radical $\sqrt{x^2+x+1}-x $? I'm trying to find oblique asmyptotes for the function $\sqrt{x^2+x+1}$ and I manage to caclculate that the coefficient for the asymptote when x approaches infinity is 1. 
But when i try to find the m-value for my oblique asymptote by taking the limit of: 
$$
\lim_{x\to\infty}\sqrt{x^2+x+1}-x=m
$$
I'm stuck. 
How do i simplify this expression to find the limit? 
I've tried manipulating the radical by converting it to the denominator: 
$$\lim_{x\to\infty}\frac{x^2+x+1}{\sqrt{x^2+x+1}}-x.$$
Or by multiplying both terms with the conjugate: 
$$\lim_{x\to\infty}\frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}}$$
But in neither case do I know how to take the limit. Any help would be greatly appreciated.

We have
$$\sqrt{x^2 + x + 1} - x = \frac{x^2 + x + 1 - x^2}{\sqrt{x^2 + x + 1} + x} = \frac{x + 1}{\sqrt{x^2 + x + 1} - x}.$$
L'Hopital's rule then gives
$$\lim_{x\to\infty} \sqrt{x^2 + x + 1} - x = \lim_{x\to\infty}\left(1 + \frac{2x+1}{2\sqrt{x^2 + x + 1}}\right)^{-1} = \frac{1}{2}.$$
