Question

$\underset{x\to 0}{\lim }\frac{e^x-x-1}{2}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{e^x-x-1}{2}\right] \\ {e}^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.....\infty \\ =\underset{x\to 0}{\lim }\left[\frac{\left(1+{\displaystyle \frac{x}{1!}}+{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^3}{3!}}....\infty \right)-x-1}{2}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^3}{3!}}+....\infty }{2}\right] \\ =0 \end{gathered}$$