Answer
$\dfrac{-1}{2}$
Work Step by Step
Write the Taylor series for the function as: $\sqrt {1+2x}=1+x-\dfrac{(x)^2}{2}+......$
Now, we have:
$\lim\limits_{x \to 0} \dfrac{\sqrt {1+2x}-1-x}{x^2}=\lim\limits_{x \to 0} \dfrac{1+x-\dfrac{(x)^2}{2}-1-x}{x^2}
\\=\lim\limits_{x \to 0} \dfrac{\dfrac{-x^2}{2}}{x^2}\\=\dfrac{-1}{2}$
