<s>[INST] 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.
 [/INST] $$\lim_{x->\infty}\sqrt{x^2+x+1}-x$$
$$=\lim_{x->\infty}\left(\sqrt{x^2+x+1}-x\right)\cdot\frac{\sqrt{x^2+x+1}+x}{\sqrt{x^2+x+1}+x}$$
$$=\lim_{x->\infty}\frac{1+x}{\sqrt{x^2+x+1}+x}$$
The limit of a sum is the sum of the limits
$$=\lim_{x->\infty}\frac{1}{\sqrt{x^2+x+1}+x} + \lim_{x->\infty}\frac{x}{\sqrt{x^2+x+1}+x}$$
The limit of a quotient is the quotient of the limits
The limit of a constant is the constant
$$=\frac{1}{\displaystyle\lim_{x->\infty}\sqrt{x^2+x+1}+x} + \lim_{x->\infty}\frac{x}{\sqrt{x^2+x+1}+x}$$
$$=\frac{1}{\displaystyle\lim_{x->\infty}\sqrt{x^2+x+1}+x} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
The limit of a sum is the sum of the limits
$$=\frac{1}{\infty+\displaystyle\lim_{x->\infty}\sqrt{x^2+x+1}} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
Use the power law
$$=\frac{1}{\infty+\sqrt{\displaystyle\lim_{x->\infty}x^2+x+1}} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
$$=\frac{1}{\infty+\infty} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
$$=\frac{1}{\infty} + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
$$=0 + \lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
$$=\lim_{x->\infty}\frac{1}{1+\frac{\sqrt{x^2+x+1}}x}$$
The limit of a quotient is the quotient of the limits
The limit of a constant is the constant
The limit of a sum is the sum of the limits
$$=\frac{1}{1+\displaystyle\lim_{x->\infty}\frac{\sqrt{x^2+x+1}}x}$$
$$=\frac{1}{1+\displaystyle\lim_{x->\infty}\sqrt{\frac{x^2+x+1}{x^2}}}$$
Use the power law
$$=\frac{1}{1+\sqrt{\displaystyle\lim_{x->\infty}\frac{x^2+x+1}{x^2}}}$$
Use L'Hopital
$$=\frac{1}{1+\sqrt{\displaystyle\lim_{x->\infty}1+\frac{1}{2x}}}$$
$$=\frac{1}{1+\sqrt{\displaystyle1+\frac12\cdot\lim_{x->\infty}\frac{1}{x}}}$$
$$=\frac{1}{1+\sqrt{\displaystyle1+\frac12\cdot0}}$$
$$=\color{lightgray}{\boxed{\color{black}{\dfrac{1}{2}}}}$$
</s>