Answer
$\dfrac{1}{2}$
Work Step by Step
Taylor series for $ e^x $ can be defined as: $ e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+....$
Now, $\lim\limits_{x \to 0}\dfrac{e^x-(1+x)}{x^2}=\lim\limits_{x \to 0} \dfrac{[(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+....)-(1+x)]}{x^2}$
or, $=\lim\limits_{x \to 0} \dfrac{\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+...}{x^2}$
or, $=\lim\limits_{x \to 0} [\dfrac{1}{2}+\dfrac{x}{3!}+\dfrac{x^2}{4!}+.....]$
or, $=\dfrac{1}{2}$
