## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{1}{2}$
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, $=\dfrac{1}{2}$