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.

I think there's a mistake in your last expression. We have 
$$ \sqrt{x^2+x+1}-x = \frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x} = \frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x} + \frac{1}{x^2}} + 1} $$
Therefore we have $$\lim(\sqrt{x^2+x+1}-x)= \lim\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x} + \frac{1}{x^2}} + 1} = \frac{1}{2} $$
