How to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$? I try to calculate $\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}}$. I've got $\frac{\sqrt{1 + x + x^2} - 1}{x} = \sqrt{\frac{1}{x^2} + \frac{1}{x} + x} - \frac{1}{x}$ but I don't know what to do next.
 A: Hint:
Multiply the numerator and the denominator by $$\sqrt{1+x+x^2}+1.$$
A: Hint: $\dfrac{\sqrt{1 + x+ x^2} -1}{x} = \dfrac{x + x^2}{x (\sqrt{1 + x+ x^2} + 1)} = \dfrac{1+x}{\sqrt{1 + x+ x^2} + 1}$
A: $$\sqrt{1 + x + x^2} = 1 + \dfrac{1}{2}(x+x^2) + \cdots = 1 + \dfrac{1}{2}x + \cdots$$ so the 
$$\lim \limits_{x \to 0}{\frac{\sqrt{1 + x + x^2} - 1}{x}} = 
\lim \limits_{x \to 0}\dfrac{1 + 1/2 x + \cdots - 1}{x} = \dfrac{1}{2}$$
A: Here are the steps
$$
\lim \limits_{x \to 0}\left[{\frac{\sqrt{1 + x + x^2} - 1}{x}}\right]
$$
$$
=\lim \limits_{x \to 0}\left[{\frac{\sqrt{1 + x + x^2} - 1}{x}}\right] \left[{\frac{\sqrt{1 + x + x^2} + 1}{\sqrt{1 + x + x^2} + 1}}\right]
$$
$$
=\lim \limits_{x \to 0}\left[{\frac{1 + x + x^2 - 1}{ x\left(\sqrt{1 + x + x^2} + 1\right)}}\right]
$$
$$
=\lim \limits_{x \to 0}\left[{\frac{x\left(1+ x\right)}{x\left(\sqrt{1 + x + x^2} + 1\right)}}\right]
$$
$$
=\lim \limits_{x \to 0}\left[{\frac{1+ x}{\sqrt{1 + x + x^2} + 1}}\right]
$$
$$
={\frac{1+ 0}{\sqrt{1 + 0+ 0} + 1}}
$$
$$
={\frac{1}{\sqrt{1} + 1}}
$$
$$
=\frac{1}{2}
$$
