Limit of an indeterminate form with a quadratic expression under square root The problem is:
$$ \lim_{x\to0}\frac{\sqrt{1+x+x^2}-1}{x} $$
So far, I've tried substituting $\sqrt{1+x+x^2}-1$ with some variable $t$, but when $x\to0$, $t\to\sqrt{1}$.   
I have also tried to rationalize the numerator, and applied l'hospital.   
I simply can't figure out this limit.   
Any help would be greatly appreciated! Thanks in advance.

$$\lim _{ x\to 0 } \frac { \sqrt { 1+x+x^{ 2 } } -1 }{ x } \cdot \frac { \sqrt { 1+x+x^{ 2 } } +1 }{ \sqrt { 1+x+x^{ 2 } } +1 } =\lim _{ x\to 0 } \frac { x+{ x }^{ 2 } }{ x\sqrt { 1+x+x^{ 2 } } +1 } =\lim _{ x\to 0 } \frac { 1+x }{ \sqrt { 1+x+x^{ 2 } } +1 } =\frac { 1 }{ 2 } $$
