The value of $\underset{x\to 0}{Lim}\frac{\sqrt{1+x+x^2}-1}{x}$ is

$\underset{x\to 0}{Lt}\frac{\sqrt{1+x+x^2}-1}{x}=\frac{\sqrt{1+0+0^2}-1}{0}=\frac{0}{0}$indeterminate form. so rationalize numerator.

$\frac{\sqrt{1+x+x^2}-1}{x}=\frac{\sqrt{1+x+x^2}-1}{x}\times \frac{\sqrt{1+x+x^2}+1}{\sqrt{1+x+x^2}+1}$

$=\frac{1+x+x^2-1}{x(\sqrt{1+x+x^2}+1)}=\frac{x(x+1)}{x(\sqrt{1+x+x^2}+1)}$

$=\frac{x+1}{\sqrt{1+x+x^2}+1}$

Now $\underset{x\to 0}{Lt}\frac{\sqrt{1+x+x^2}-1}{x}=\underset{x\to 0}{Lt}\frac{x+1}{\sqrt{1+x+x^2}+1}$

$=\frac{0+1}{\sqrt{1+0+0^2}+1}=\frac{1}{2}$