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.
 A: Don't do the difference, divide!
$\lim_{x->\infty}\dfrac{\sqrt{x^2+x+1}}{x}=1$ very easily
For $\dfrac{\sqrt{x^2+x+1}}{x}=\dfrac{\sqrt{x^2+x+1}}{\sqrt{x^2}}=\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x²}}$
EDIT: It is not over, of course, since the asymptote does not necessarily goes through the origin... You have to make the actual difference between the function and $x+a$. But by using the same strategy, you will find easily the result. 
EDIT2: in order to be more complete.
$\sqrt{x^2+x+1}-x=x\times(\sqrt{1+\frac{1}{x}+\frac{1}{x²}}-1)=\dfrac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x²}}+1}$
Whose limit is clearly $\dfrac{1}{2}$.
A: Your last expression simplifies to
$${x+1\over\sqrt{x^2+x+1}}$$
There are now various ways to proceed.  One is to divide numerator and denominator by $x$, bringing the $x$ in the denominator inside as an $x^2$:
$${x+1\over\sqrt{x^2+x+1}}={1+{1\over x}\over\sqrt{1+{1\over x}+{1\over x^2}}}$$
Do you see where this is going?
Added later:  I answered too quickly, without checking how you got to the last expression.  It turns out it's missing a "$+x$" in the denominator: it should have been
$${x+1\over\sqrt{x^2+x+1}+x}$$
so when you do the trick I advised you get
$${x+1\over\sqrt{x^2+x+1}}={1+{1\over x}\over\sqrt{1+{1\over x}+{1\over x^2}}+1}$$
A: $$\lim_{x->\infty}\sqrt{x^2+x+1}-x$$
$$=\lim_{x->\infty}\left(\sqrt{x^2+x+1}-x\right)\cdot\frac{\sqrt{x^2+x+1}+x}{\sqrt{x^2+x+1}+x}$$
$$=\lim_{x->\infty}\frac{1+x}{\sqrt{x^2+x+1}+x}$$
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
$$=\frac{1}{1+\sqrt{\displaystyle\lim_{x->\infty}1+\frac{1}{2x}}}$$
$$=\frac{1}{1+\sqrt{\displaystyle1+\frac12\cdot\lim_{x->\infty}\frac{1}{x}}}$$
$$=\frac{1}{1+\sqrt{\displaystyle1+\frac12\cdot0}}$$
$$=\color{lightgray}{\boxed{\color{black}{\dfrac{1}{2}}}}$$
A: 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 $$\lim(\sqrt{x^2+x+1}-x)= \lim\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x} + \frac{1}{x^2}} + 1} = \frac{1}{2} $$
A: Easiest way is to complete to square $$\lim_{x\to\infty}\sqrt{x^2+x+1}-x=\sqrt{(x+1/2)^2+3/4}-x=\sqrt{(x+1/2)^2}-x=\sqrt{(x+1/2)^2(1+\frac{3}{4(x+1/2)^2})}-x=(x+1/2)-x=1/2$$
A: 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
$$\lim_{x\to\infty} \sqrt{x^2 + x + 1} - x = \lim_{x\to\infty}\left(1 + \frac{2x+1}{2\sqrt{x^2 + x + 1}}\right)^{-1} = \frac{1}{2}.$$
